| blob | 7b843127f2627300f62da098a5be385385415436 |
1 ;; -*- Mode: Lisp; Package: Maxima; Syntax: Common-Lisp; Base: 10 -*- ;;;;
2 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3 ;;; The data in this file contains enhancements. ;;;;;
4 ;;; ;;;;;
5 ;;; Copyright (c) 1984,1987 by William Schelter,University of Texas ;;;;;
6 ;;; All rights reserved ;;;;;
7 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
8 ;;; (c) copyright 1982 massachusetts institute of technology ;;;
9 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
15 ;;; this is the definite integration package.
16 ;; defint does definite integration by trying to find an
17 ;;appropriate method for the integral in question. the first thing that
18 ;;is looked at is the endpoints of the problem.
19 ;;
20 ;; i(grand,var,a,b) will be used for integrate(grand,var,a,b)
22 ;; References are to "Evaluation of Definite Integrals by Symbolic
23 ;; Manipulation", by Paul S. Wang,
24 ;; (http://www.lcs.mit.edu/publications/pubs/pdf/MIT-LCS-TR-092.pdf;
25 ;; a better copy might be: https://maxima.sourceforge.io/misc/Paul_Wang_dissertation.pdf)
26 ;;
27 ;; nointegrate is a macsyma level flag which inhibits indefinite
28 ;;integration.
29 ;; abconv is a macsyma level flag which inhibits the absolute
30 ;;convergence test.
31 ;;
32 ;; $defint is the top level function that takes the user input
33 ;;and does minor changes to make the integrand ready for the package.
34 ;;
35 ;; next comes defint, which is the function that does the
36 ;;integration. it is often called recursively from the bowels of the
37 ;;package. defint does some of the easy cases and dispatches to:
38 ;;
39 ;; dintegrate. this program first sees if the limits of
40 ;;integration are 0,inf or minf,inf. if so it sends the problem to
41 ;;ztoinf or mtoinf, respectively.
42 ;; else, dintegrate tries:
43 ;;
44 ;; intsc1 - does integrals of sin's or cos's or exp(%i var)'s
45 ;; when the interval is 0,2 %pi or 0,%pi.
46 ;; method is conversion to rational function and find
47 ;; residues in the unit circle. [wang, pp 107-109]
48 ;;
49 ;; ratfnt - does rational functions over finite interval by
50 ;; doing polynomial part directly, and converting
51 ;; the rational part to an integral on 0,inf and finding
52 ;; the answer by residues.
53 ;;
54 ;; zto1 - i(x^(k-1)*(1-x)^(l-1),x,0,1) = beta(k,l) or
55 ;; i(log(x)*x^(x-1)*(1-x)^(l-1),x,0,1) = psi...
56 ;; [wang, pp 116,117]
57 ;;
58 ;; dintrad- i(x^m/(a*x^2+b*x+c)^(n+3/2),x,0,inf) [wang, p 74]
59 ;;
60 ;; dintlog- i(log(g(x))*f(x),x,0,inf) = 0 (by symmetry) or
61 ;; tries an integration by parts. (only routine to
62 ;; try integration by parts) [wang, pp 93-95]
63 ;;
64 ;; dintexp- i(f(exp(k*x)),x,a,inf) = i(f(x+1)/(x+1),x,0,inf)
65 ;; or i(f(x)/x,x,0,inf)/k. First case hold for a=0;
66 ;; the second for a=minf. [wang 96-97]
67 ;;
68 ;;dintegrate also tries indefinite integration based on certain
69 ;;predicates (such as abconv) and tries breaking up the integrand
70 ;;over a sum or tries a change of variable.
71 ;;
72 ;; ztoinf is the routine for doing integrals over the range 0,inf.
73 ;; it goes over a series of routines and sees if any will work:
74 ;;
75 ;; scaxn - sc(b*x^n) (sc stands for sin or cos) [wang, pp 81-83]
76 ;;
77 ;; ssp - a*sc^n(r*x)/x^m [wang, pp 86,87]
78 ;;
79 ;; zmtorat- rational function. done by multiplication by plog(-x)
80 ;; and finding the residues over the keyhole contour
81 ;; [wang, pp 59-61]
82 ;;
83 ;; log*rat- r(x)*log^n(x) [wang, pp 89-92]
84 ;;
85 ;; logquad0 log(x)/(a*x^2+b*x+c) uses formula
86 ;; i(log(x)/(x^2+2*x*a*cos(t)+a^2),x,0,inf) =
87 ;; t*log(a)/sin(t). a better formula might be
88 ;; i(log(x)/(x+b)/(x+c),x,0,inf) =
89 ;; (log^2(b)-log^2(c))/(2*(b-c))
90 ;;
91 ;; batapp - x^(p-1)/(b*x^n+a)^m uses formula related to the beta
92 ;; function [wang, p 71]
93 ;; there is also a special case when m=1 and a*b<0
94 ;; see [wang, p 65]
95 ;;
96 ;; sinnu - x^-a*n(x)/d(x) [wang, pp 69-70]
97 ;;
98 ;; ggr - x^r*exp(a*x^n+b)
99 ;;
100 ;; dintexp- see dintegrate
101 ;;
102 ;; ztoinf also tries 1/2*mtoinf if the integrand is an even function
103 ;;
104 ;; mtoinf is the routine for doing integrals on minf,inf.
105 ;; it too tries a series of routines and sees if any succeed.
106 ;;
107 ;; scaxn - when the integrand is an even function, see ztoinf
108 ;;
109 ;; mtosc - exp(%i*m*x)*r(x) by residues on either the upper half
110 ;; plane or the lower half plane, depending on whether
111 ;; m is positive or negative.
112 ;;
113 ;; zmtorat- does rational function by finding residues in upper
114 ;; half plane
115 ;;
116 ;; dintexp- see dintegrate
117 ;;
118 ;; rectzto%pi2 - poly(x)*rat(exp(x)) by finding residues in
119 ;; rectangle [wang, pp98-100]
120 ;;
121 ;; ggrm - x^r*exp((x+a)^n+b)
122 ;;
123 ;; mtoinf also tries 2*ztoinf if the integrand is an even function.
132 ;;;rsn* is in comdenom. does a ratsimp of numerator.
133 ;expvar
135 ;impvar
137 context
138 ;;LIMITP T Causes $ASKSIGN to do special things
139 ;;For DEFINT like eliminate epsilon look for prin-inf
140 ;;take realpart and imagpart.
141 integer-info
142 ;;If LIMITP is non-null ask-integer conses
143 ;;its assumptions onto this list.
144 ))
149 "When @code{true}, definite integration tries to find poles in the integrand
152 ;; Currently, if true, $solvetrigwarn is set to true. No additional
153 ;; debugging information is displayed.
158 nil
159 "When NIL, print a message that the principal value of the integral has
163 nil
164 "When non-NIL, a divergent integral will throw to `divergent.
172 ;; Set to true when OSCIP-VAR returns true in DINTEGRATE.
190 ;; Distribute $defint over equations, lists, and matrices.
222 (merror (intl:gettext "defint: variable of integration must be a simple or subscripted variable.~%defint: found ~M") ivar))
251 ;; If antideriv can't do it, returns nil
252 ;; use limit to evaluate every answer returned by antideriv.
256 ;;;Hack the expression up for exponentials.
259 ;; Is this expr a candidate for SININT ?
268 t)))
269 ;;ERF type things go here.
274 t))))))
294 ;; We have some subscripted variable that's not our variable
295 ;; (I think), so it's deg-lessp.
296 ;;
297 ;; FIXME: Is this the best way to handle this? Are there
298 ;; other cases we're mising here?
299 t)))
309 ;; This routine tries to take a limit a couple of ways.
315 ans
318 ivar
319 0
328 ;; Test whether fun2 is inverse of fun1 at val.
334 ;; integration change of variable
342 ())
344 ;; This is a hack! If nv is of the form b*x^n+a, we can
345 ;; solve the equation manually instead of using solve.
346 ;; Why? Because solve asks us for the sign of yx and
347 ;; that's bogus.
349 ;; Solve yx = b*x^n+a, for x. Any root will do. So we
350 ;; have x = ((yx-a)/b)^(1/n).
352 d
358 ))))
375 ;; d: original variable (ivar) as a function of 'yx
376 ;; ind: boolean flag
377 ;; nv: new variable ('yx) as a function of original variable (ivar)
386 ;; converts limits of integration to values for new variable 'yx
389 ;; rewrites exp, the integrand in terms of ivar, the
390 ;; integrand in terms of 'yx, and returns the new
391 ;; integrand.
396 ll1 ul1)
405 ;; wrapper around limit, returns nil if
406 ;; limit not found (nounform returned), or undefined ($und or $ind)
413 ans))))
430 ;; D (not used? -- cwh)
443 ;; Look for integrals which involve log and exp functions.
444 ;; Maxima has a special algorithm to get general results.
455 %tan %sinh %cosh %tanh
456 %log %asin %acos %atan
457 %cot %acot %sec
458 %asec %csc %acsc
460 ;; Call ANTIDERIV with logabs disabled,
461 ;; because the Risch algorithm assumes
462 ;; the integral of 1/x is log(x), not log(abs(x)).
463 ;; Why not just assume logabs = false within RISCHINT itself?
464 ;; Well, there's at least one existing result which requires
465 ;; logabs = true in RISCHINT, so try to make a minimal change here instead.
506 ans)
509 ans)
522 ;;Should be a PROG inside of unwind-protect, but Multics has a compiler
523 ;;bug wrt. and I want to test this code now.
531 ;;;This seems((and (and (eq *ul* '$inf)
532 ;;;fairly losing (setq exp (subin (m+ *ll* ivar) exp))
533 ;;; (setq *ll* 0.))
534 ;;; (ztoinf exp ivar)))
539 ;; ((and (and (equal *ul* 1.)
540 ;; (equal *ll* 0.)) (zto1 exp)))
548 ;;;NOT COMPLETE for sin's and cos's.
570 ;; Call ANTIDERIV with logabs disabled,
571 ;; because the Risch algorithm assumes
572 ;; the integral of 1/x is log(x), not log(abs(x)).
573 ;; Why not just assume logabs = false within RISCHINT itself?
574 ;; Well, there's at least one existing result which requires
575 ;; logabs = true in RISCHINT, so try to make a minimal change here instead.
581 ;;;Recursion stopper
604 ;; adds up integrals of ranges between each pair of poles.
605 ;; checks if whole thing is divergent as limits of integration approach poles.
607 ;;; calling $logcontract causes antiderivative of 1/(1-x^5) to blow up
608 ;; (setq anti-deriv (cond ((involve anti-deriv '(%log))
609 ;; ($logcontract anti-deriv))
610 ;; (t anti-deriv)))
619 ivar))))
622 ;;Return section.
647 pole-list)
649 ;; Assumes EXP is a rational expression with no polynomial part and
650 ;; converts the finite integration to integration over a half-infinite
651 ;; interval. The substitution y = (x-a)/(b-x) is used. Equivalently,
652 ;; x = (b*y+a)/(y+1).
653 ;;
654 ;; (I'm guessing CV means Change Variable here.)
657 ;; FIXME! This is a hack. We apply the transformation with
658 ;; symbolic limits and then substitute the actual limits later.
659 ;; That way method-by-limits (usually?) sees a simpler
660 ;; integrand.
661 ;;
662 ;; See Bugs 938235 and 941457. These fail because $FACTOR is
663 ;; unable to factor the transformed result. This needs more
664 ;; work (in other places).
668 ;; If the limit is a number, use $substitute so we simplify
669 ;; the result. Do we really want to do this?
677 ()))
679 ;; Integrate rational functions over a finite interval by doing the
680 ;; polynomial part directly, and converting the rational part to an
681 ;; integral from 0 to inf. This is evaluated via residues.
684 ;; PQR divides the rational expression and returns the quotient
685 ;; and remainder
692 ;; No polynomial part
695 ans
698 ;; Only polynomial part
701 ;; A non-zero quotient and remainder. Combine the results
702 ;; together.
706 ans))
711 ;; I think this takes a rational expression E, and finds the
712 ;; polynomial part. A cons is returned. The car is the quotient and
713 ;; the cdr is the remainder.
737 ;;;If leadop isn't plus don't do anything.
746 t)
752 t)
807 ;; *Z* is a "zero parameter" for this package.
808 ;; Its also used to transform.
809 ;; limit(exp,var,val,dir) -- limit(exp,tvar,0,dir)
816 ;; EPSILON is used in principal value code to denote the familiar
817 ;; mathematical entity.
822 ;;; PRIN-INF Is a special symbol in the principal value code used to
823 ;;; denote an end-point which is proceeding to infinity.
846 ; We have minf <= *ll* = *ul* <= inf
847 )
849 ; We have minf <= *ll* < *ul* = inf
850 )
852 ; We have minf = *ll* < *ul* < inf
853 ;
854 ; Now substitute
855 ;
856 ; ivar -> -ivar
857 ; *ll* -> -*ul*
858 ; *ul* -> inf
859 ;
860 ; so that minf < *ll* < *ul* = inf
866 ; We have minf <= *ul* < *ll*
867 ;
868 ; Now substitute
869 ;
870 ; exp -> -exp
871 ; *ll* <-> *ul*
872 ;
873 ; so that minf <= *ll* < *ul*
876 t)))
890 ;; Substitute a and b into integral e
891 ;;
892 ;; Looks for discontinuties in integral, and works around them.
893 ;; For example, in
894 ;;
895 ;; integrate(x^(2*n)*exp(-(x)^2),x) ==>
896 ;; -gamma_incomplete((2*n+1)/2,x^2)*x^(2*n+1)*abs(x)^(-2*n-1)/2
897 ;;
898 ;; the integral has a discontinuity at x=0.
899 ;;
906 ivar a b)))))
920 total))))))
922 ;; look for terms with a negative exponent
923 ;;
924 ;; recursively traverses exp in order to find discontinuities such as
925 ;; erfc(1/x-x) at x=0
937 ;; returns list of places where exp might be discontinuous in ivar.
938 ;; list begins with *ll* and ends with *ul*, and include any values between
939 ;; *ll* and *ul*.
940 ;; return '$no or '$unknown if no discontinuities found.
957 ;; (multiplicity (cdar dummy)) ;; not used
962 ;; Carefully substitute the integration limits A and B into the
963 ;; expression E.
973 ;; Try easy substitutions. Return NIL if we can't.
976 ;; Infinite limits aren't easy
977 nil)
981 %gamma_incomplete %expintegral_ei)))
983 ;; It's easy if we have a polynomial. I (rtoy) think
984 ;; it's also easy if the denominator is free of the
985 ;; integration variable and also if the expression
986 ;; doesn't involve inverse functions.
987 ;;
988 ;; gamma_incomplete and expintegral_ie
989 ;; included because of discontinuity in
990 ;; gamma_incomplete(0, exp(%i*x)) and
991 ;; expintegral_ei(exp(%i*x))
992 ;;
993 ;; XXX: Are there other cases we've forgotten about?
994 ;;
995 ;; So just try to substitute the limits into the
996 ;; expression. If no errors are produced, we're done.
1001 ;; no-err-sub has returned T. An error was catched.
1002 nil)
1016 ;; check for divergent integral
1028 ;;;This function works only on things with ATAN's in them now.
1030 ;; POLES-IN-INTERVAL doesn't know about the poles of tan(x). Call
1031 ;; trigsimp to convert tan into sin/cos, which POLES-IN-INTERVAL
1032 ;; knows how to handle.
1033 ;;
1034 ;; XXX Should we fix POLES-IN-INTERVAL instead?
1035 ;;
1036 ;; XXX Is calling trigsimp too much? Should we just only try to
1037 ;; substitute sin/cos for tan?
1038 ;;
1039 ;; XXX Should the result try to convert sin/cos back into tan? (A
1040 ;; call to trigreduce would do it, among other things.)
1043 ;;POLES -> ((mlist) ((mequal) ((%atan) foo) replacement) ......)
1044 ;;We can then use $SUBSTITUTE
1049 ul-ans)
1051 nil)))
1082 llist)))))))))))))))
1097 ;; There are no calls where fn1 is ever equal to
1098 ;; 'mtorat. Hence this case is never true. What is
1099 ;; this testing for?
1106 nil)
1125 ;; e is of form poly(x)*exp(m*%i*x)
1126 ;; s is degree of denominator
1127 ;; adds e to *bptu* or *bptd* according to sign of m
1144 ;; Check term is of form poly(x)*exp(m*%i*x)
1145 ;; n is degree of denominator.
1155 ;; Set updn to T if the coefficient of IVAR in the
1156 ;; polynomial is known to be positive. Otherwise set to NIL.
1157 ;; (What does updn really mean?)
1168 term)
1203 ;; exponentialize sin and cos
1209 '$%i
1231 exp)
1247 ;; computes integral from minf to inf for expressions of the form
1248 ;; exp(%i*m*x)*r(x) by residues on either the upper half
1249 ;; plane or the lower half plane, depending on whether
1250 ;; m is positive or negative. [wang p. 77]
1251 ;;
1252 ;; exponentializes sin and cos before applying residue method.
1253 ;; can handle some expressions with poles on real line, such as
1254 ;; sin(x)*cos(x)/x.
1264 ivar
1265 nil)
1268 ;;;Above tests for applicability of this method.
1277 ;; call to $expand can create rational terms
1278 ;; with no exp(m*%i*x)
1282 ;;;Function RIB sets up the values of BPTU and BPTD
1289 ;;;If there is BPTU then CSEMIUP must succeed.
1290 ;;;Likewise for BPTD.
1294 ;; The original integrand was already
1295 ;; determined to have no poles by initial-analysis.
1296 ;; If individual terms of the expansion have poles, the poles
1297 ;; must cancel each other out, so we can ignore them.
1300 ;; if integral of ratterms is divergent, ratans is nil,
1301 ;; and mtosc returns nil
1318 t)
1320 t)
1328 t)
1330 t)
1335 s nc dc
1337 nn-var dn-var)
1358 sd new-sd))
1359 result))
1362 sn nil)
1364 sd nil))
1367 ;;;
1368 ;;;New section.
1386 s
1389 ivar)))
1399 findout
1403 on
1426 s)
1430 ;; It's very important here to bind VAR
1431 ;; because the PLOG simplifier checks
1432 ;; for VAR. Without this, the
1433 ;; simplifier converts plog(-x) to
1434 ;; log(x)+%i*%pi, which messes up the
1435 ;; keyhole routine.
1437 d
1438 ivar)))
1441 ;; Let's not signal an error here. Return nil so that we
1442 ;; eventually return a noun form if no other algorithm gives
1443 ;; a result.
1446 nil)))
1448 ;;(setq *dflag* nil)
1477 nn-var dn-var)
1504 sd new-sd))
1505 result))
1541 ;; Test to see if exp is of the form p(x)*f(exp(x)). If so, set pp to
1542 ;; be p(x) and set pe to f(exp(x)).
1555 s
1558 ivar)))
1565 ;; We have an integral of the form p(x)*F(exp(x)), where
1566 ;; p(x) is a polynomial.
1568 ;; No polynomial
1572 ;; These limits must exist for the integral to converge.
1575 ;; This only handles the case when the F(z) is a
1576 ;; rational function.
1579 ;; If we get here, F(z) is not a rational function.
1580 ;; We transform it using the substitution x=log(y)
1581 ;; which gives us an integral of the form
1582 ;; p(log(y))*F(y)/y, which maxima should be able to
1583 ;; handle.
1586 ;; Give up. We don't know how to handle this.
1588 en
1602 exp)))
1604 ;;; given (b*x+a)^n+c returns (a b n c)
1612 nil)
1614 ;;;get high degree of poly
1616 ;;;diff down to linear.
1618 ;;;all the way to constant.
1621 ;;;get rid of factorial from differentiation.
1624 ;;;Sees if can be expressed as (a*x+b)^n + part freeof x.
1634 s)
1659 n
1660 d)))))
1663 on
1718 e nil)
1732 e))))))
1738 ;; Compute the integral(n/d,x,0,inf) by computing the negative of the
1739 ;; sum of residues of log(-x)*n/d over the poles of n/d inside the
1740 ;; keyhole contour. This contour is basically an disk with a slit
1741 ;; along the positive real axis. n/d must be a rational function.
1746 ;; Ok if not on the positive real axis.
1751 t)
1762 ;; Look at an expression e of the form sin(r*x)^k, where k is an
1763 ;; integer. Return the list (1 r k). (Not sure if the first element
1764 ;; of the list is always 1 because I'm not sure what partition is
1765 ;; trying to do here.)
1779 ;; Look at an expression e of the form sin(r*x) and return r.
1790 ;; integrate(a*sc(r*x)^k/x^n,x,0,inf).
1793 ;; Get the argument of the involved trig function.
1796 ;; I don't think this needs to be special.
1797 #+nil
1799 ;; Replace (1-cos(arg)^2) with sin(arg)^2.
1801 ;(m+t 1. (m- (m^t `((%cos) ,ivar) 2.)))
1802 ;; The code from above generates expressions with
1803 ;; a missing simp flag. Furthermore, the
1804 ;; substitution has to be done for the complete
1805 ;; argument of the trig function. (DK 02/2010)
1808 exp))
1813 ;; n is the power of the denominator.
1815 ;; The simple case.
1821 ;; Do this for a sum of such terms.
1823 c))))))))))
1825 ;; We have an integral of the form sin(r*x)^k/x^n. C is the list (1 r k).
1826 ;;
1827 ;; The substitution y=r*x converts this integral to
1828 ;;
1829 ;; r^(n-1)*integral(sin(y)^k/y^n,y,0,inf)
1830 ;;
1831 ;; (If r is negative, we need to negate the result.)
1832 ;;
1833 ;; The recursion Wang gives on p. 87 has a typo. The second term
1834 ;; should be subtracted from the first. This formula is given in G&R,
1835 ;; 3.82, formula 12.
1836 ;;
1837 ;; integrate(sin(x)^r/x^s,x) =
1838 ;; r*(r-1)/(s-1)/(s-2)*integrate(sin(x)^(r-2)/x^(s-2),x)
1839 ;; - r^2/(s-1)/(s-2)*integrate(sin(x)^r/x^(s-2),x)
1840 ;;
1841 ;; (Limits are assumed to be 0 to inf.)
1842 ;;
1843 ;; This recursion ends up with integrals with s = 1 or 2 and
1844 ;;
1845 ;; integrate(sin(x)^p/x,x,0,inf) = integrate(sin(x)^(p-1),x,0,%pi/2)
1846 ;;
1847 ;; with p > 0 and odd. This latter integral is known to maxima, and
1848 ;; it's value is beta(p/2,1/2)/2.
1849 ;;
1850 ;; integrate(sin(x)^2/x^2,x,0,inf) = %pi/2*binomial(q-3/2,q-1)
1851 ;;
1852 ;; where q >= 2.
1853 ;;
1855 ;; Compute sign(r)*r^(n-1)*integrate(sin(y)^k/y^n,y,0,inf)
1857 c
1863 recursion)
1864 ;; Recursion failed. Return the integrand
1865 ;; The following code generates expressions with a missing simp flag
1866 ;; for the sin function. Use better simplifying code. (DK 02/2010)
1867 ; (let ((integrand (div (pow `((%sin) ,(m* r ivar))
1868 ; k)
1869 ; (pow ivar n))))
1871 k)
1876 ;; integrate(sin(x)^n/x^2,x,0,inf) = pi/2*binomial(n-3/2,n-1).
1877 ;; Express in terms of Gamma functions, though.
1883 ;; integrate(sin(x)^n/x,x,0,inf) = beta((n+1)/2,1/2)/2, for n odd and
1884 ;; n > 0.
1887 half%pi)
1890 ;; This implements the recursion for computing
1891 ;; integrate(sin(y)^l/y^k,y,0,inf). (Note the change in notation from
1892 ;; the above!)
1898 ;; Integral diverges if l-(k-1) < 0.
1901 ;; If l + k is not even, return NIL. (Is this the right
1902 ;; thing to do?)
1903 nil)
1905 ;; We have integrate(sin(y)^l/y^2). Use sevn to evaluate.
1908 ;; We have integrate(sin(y)^l/y,y)
1914 ;; j = k-2, i = (k-1)*(k-2)
1915 ;;
1916 ;;
1917 ;; The main recursion:
1918 ;;
1919 ;; i(sin(y)^l/y^k)
1920 ;; = l*(l-1)/(k-1)/(k-2)*i(sin(y)^(l-2)/y^k)
1921 ;; - l^2/(k-1)/(k-1)*i(sin(y)^l/y^(k-2))
1929 ;; Returns the fractional part of a?
1933 ;; Why do we return 0 if a is a number? Perhaps we really
1934 ;; mean integer?
1937 ;; If we're here, this basically assumes a is a rational.
1938 ;; Compute the remainder and return the result.
1946 ;; Doesn't appear to be used anywhere in Maxima. Not sure what this
1947 ;; was intended to do.
1948 #+nil
1959 ;;; THE FOLLOWING FUNCTION IS FOR TRIG FUNCTIONS OF THE FOLLOWING TYPE:
1960 ;;; LOWER LIMIT=0 B A MULTIPLE OF %PI SCA FUNCTION OF SIN (X) COS (X)
1961 ;;; B<=%PI2
1966 ;; means there was no error
1969 ; returns cons of (integer_part . fractional_part) of a
1971 ;; I think we really want to compute how many full periods are in a
1972 ;; and the remainder.
1977 ; returns cons of (integer_part . fractional_part) of a
1979 ;; I think we really want to compute how many full periods are in a
1980 ;; and the remainder.
1987 ;; Return the integer part of r.
1989 ;; r - fpart(r)
1993 ;;;Try making exp(%i*ivar) --> yy, if result is rational then do integral
1994 ;;;around unit circle. Make corrections for limits of integration if possible.
2014 rat-form
2015 ivar)))
2020 ;;; Do integrals of sin and cos. this routine makes sure lower limit
2021 ;;; is zero.
2023 ;; integrate(e,var,a,b)
2039 ivar e)
2040 cc))
2050 ;; Exit if b or a is not a constant or if the integrand
2051 ;; doesn't appear to have a period of 2 pi.
2054 ;; Multiples of 2*%pi in limits.
2061 ;; The integral is not over a full period (2*%pi) or multiple
2062 ;; of a full period.
2064 ;; Wang p. 111: The integral integrate(f(x),x,a,b) can be
2065 ;; written as:
2066 ;;
2067 ;; n * integrate(f,x,0,2*%pi) + integrate(f,x,0,c)
2068 ;; - integrate(f,x,0,d)
2069 ;;
2070 ;; for some integer n and d >= 0, c < 2*%pi because there exist
2071 ;; integers p and q such that a = 2 * p *%pi + d and b = 2 * q
2072 ;; * %pi + c. Then n = q - p.
2074 ;; Compute q and c for the upper limit b.
2081 ;; Compute p and d for the lower limit a.
2083 ;; avoid an extra trip around the circle - helps skip principal values
2088 ;; Compute -integrate(f,x,0,d)
2093 ;; Compute n = q - p (stored in nzp2)
2095 out
2096 ;; Compute n*integrate(f,x,0,2*%pi)
2098 ;; n = 0
2101 ;; n is not zero, so compute
2102 ;; integrate(f,x,0,2*%pi)
2104 ;; Unable to compute integrate(f,x,0,2*%pi)
2106 ;; Compute integrate(f,x,0,c)
2110 ;; All three pieces succeeded.
2113 ;; Less than 1 full period, so intsc can integrate it.
2114 ;; Apply the substitution to make the lower limit 0.
2115 ;; This is last resort because substitution often causes intsc to fail.
2117 limit-diff ivar))
2118 ;; nothing worked
2121 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)).
2122 ;; calls intsc with a wrapper to just return nil if integral is divergent,
2123 ;; rather than generating an error.
2128 nil
2129 ans)))
2131 ;; integrate(sc, ivar, 0, b), where sc is f(sin(x), cos(x)). I (rtoy)
2132 ;; think this expects b to be less than 2*%pi.
2135 0
2142 ;; Partition the integrand SC into the factors that do not
2143 ;; contain VAR (the car part) and the parts that do (the cdr
2144 ;; part).
2149 ;; integrate(sc, ivar, 0, b), where sc is f(sin(x), cos(x)).
2151 ;; Determine if sc is a product of sin's and cos's.
2155 ;; We have a product of sin's and cos's. We handle some
2156 ;; special cases here.
2158 ;; Wang p. 110, formula (1):
2159 ;; integrate(sin(x)^m*cos(x)^n, x, 0, %pi/2) =
2160 ;; gamma((m+1)/2)*gamma((n+1)/2)/2/gamma((n+m+2)/2)
2163 ;; Wang p. 110, near the bottom, says
2164 ;;
2165 ;; int(f(sin(x),cos(x)), x, 0, %pi) =
2166 ;; int(f(sin(x),cos(x)) + f(sin(x),-cos(x)),x,0,%pi/2)
2187 ;; Wang, p. 111 says
2188 ;;
2189 ;; int(f(sin(x),cos(x)),x,0,3*%pi/2) =
2190 ;; int(f(sin(x),cos(x)),x,0,%pi)
2191 ;; + int(f(-sin(x),-cos(x)),x,0,%pi/2)
2195 ;; We don't have a product of sin's and cos's.
2200 dn*)
2205 ;;;Is careful about substitution of limits where the denominator may be zero
2206 ;;;because of various assumptions made.
2227 nil
2228 ans)))
2234 nil
2235 ans)))
2237 ;; Determine whether E is of the form sin(x)^m*cos(x)^n and return the
2238 ;; list (m n).
2242 m n)
2261 ;; Says whether (m^t value exponent) has at least one real branch.
2262 ;; Only works for values of 1 and -1 now. Returns $yes $no
2263 ;; $unknown.
2274 ;; Compute beta((m+1)/2,(n+1)/2)/2.
2280 ;;Seems like Guys who call this don't agree on what it should return.
2293 ;; Check e for an expression of the form x^kk*(b*x^n+a)^l. If it
2294 ;; matches, Return the two values kk and (list l a n b).
2299 ;; We have f(x)^y. Look to see if f(x) has the desired
2300 ;; form. Then f(x)^y has the desired form too, with
2301 ;; suitably modified values.
2302 ;;
2303 ;; XXX: Should we ask for the sign of f(x) if y is not an
2304 ;; integer? This transformation we're going to do requires
2305 ;; that f(x)^y be real.
2307 e
2312 ;; Got a match. Adjust kk and cc appropriately.
2314 cc
2329 ;;(DEFUN BATAP (E)
2330 ;; (PROG (K C L)
2331 ;; (COND ((NOT (BATA0 E)) (RETURN NIL))
2332 ;; ((AND (EQUAL -1. (CADDDR C))
2333 ;; (EQ ($askSIGN (SETQ K (m+ 1. K)))
2334 ;; '$pos)
2335 ;; (EQ ($askSIGN (SETQ L (m+ 1. (CAR C))))
2336 ;; '$pos)
2337 ;; (ALIKE1 (CADR C)
2338 ;; (m^ UL (CADDR C)))
2339 ;; (SETQ E (CADR C))
2340 ;; (EQ ($askSIGN (SETQ C (CADDR C))) '$pos))
2341 ;; (RETURN (M// (m* (m^ UL (m+t K (m* C (m+t -1. L))))
2342 ;; `(($BETA) ,(SETQ NN* (M// K C))
2343 ;; ,(SETQ DN* L)))
2344 ;; C))))))
2347 ;; Integrals of the form i(log(x)^m*x^k*(a+b*x^n)^l,x,0,ul). There
2348 ;; are two cases to consider: One case has ul>0, b<0, a=-b*ul^n, k>-1,
2349 ;; l>-1, n>0, m a nonnegative integer. The second case has ul=inf, l < 0.
2350 ;;
2351 ;; These integrals are essentially partial derivatives of the Beta
2352 ;; function (i.e. the Eulerian integral of the first kind). Note
2353 ;; that, currently, with the default setting intanalysis:true, this
2354 ;; function might not even be called for some of these integrals.
2355 ;; However, this can be palliated by setting intanalysis:false.
2374 ;; We have i(x^kk/(d+cc*x^r)^s,x,0,inf) =
2375 ;; beta(aa,bb)/(cc^aa*d^bb*r). Compute this, and then
2376 ;; differentiate it m times to get the log term
2377 ;; incorporated.
2382 0
2383 m)))
2387 r))))
2396 ;; The result looks like B(k/n,l) ( ... ).
2397 ;; Perhaps, we should just $factor, instead of
2398 ;; pulling out beta like this.
2400 beta
2401 ($fullratsimp
2409 ivar m)
2411 beta))))))))))))
2414 ;;; If e is of the form given below, make the obvious change
2415 ;;; of variables (substituting ul*x^(1/n) for x) in order to reduce
2416 ;;; e to the usual form of the integrand in the Eulerian
2417 ;;; integral of the first kind.
2418 ;;; N. B: The old version of ZTO1 completely ignored this
2419 ;;; substitution; the log(x)s were just thrown in, which,
2420 ;;; of course would give wrong results.
2423 ;; Parse e
2427 ;; e=x^k*(a+b*x^n)^l
2429 c
2441 ;; Wang p. 71 gives the following formula for a beta function:
2442 ;;
2443 ;; integrate(x^(k-1)/(c*x^r+d)^s,x,0,inf)
2444 ;; = beta(a,b)/(c^a*d^b*r)
2445 ;;
2446 ;; where a = k/r > 0, b = s - a > 0, s > k > 0, r > 0, c*d > 0.
2447 ;;
2448 ;; This function matches this and returns k-1, d, r, c, a, b. And
2449 ;; also checks that all the conditions hold. If not, NIL is returned.
2450 ;;
2456 c
2473 ;; Handles beta integrals.
2481 nil)
2484 c
2485 ;; e = x^k*(d+c*x^al)^l.
2490 new-k)
2498 ;; Compute exp(d)*gamma((c+1)/b)/b/a^((c+1)/b). In essence, this is
2499 ;; the value of integrate(x^c*exp(d-a*x^b),x,0,inf).
2510 ;; Evaluates the contour integral of GRAND around the unit circle
2511 ;; using residues.
2518 ;; Is pt stricly inside the unit circle?
2524 ;; Is pt on the unit circle? (Use
2525 ;; the cached value computed
2526 ;; above.)
2547 ;; Wang 81-83. Unfortunately, the pdf version has page 82 as all
2548 ;; black, so here is, as best as I can tell, what Wang is doing.
2549 ;; Fortunately, p. 81 has the necessary hints.
2550 ;;
2551 ;; First consider integrate(exp(%i*k*x^n),x) around the closed contour
2552 ;; consisting of the real axis from 0 to R, the arc from the angle 0
2553 ;; to %pi/(2*n) and the ray from the arc back to the origin.
2554 ;;
2555 ;; There are no poles in this region, so the integral must be zero.
2556 ;; But consider the integral on the three parts. The real axis is the
2557 ;; integral we want. The return ray is
2558 ;;
2559 ;; exp(%i*%pi/2/n) * integrate(exp(%i*k*(t*exp(%i*%pi/2/n))^n),t,R,0)
2560 ;; = exp(%i*%pi/2/n) * integrate(exp(%i*k*t^n*exp(%i*%pi/2)),t,R,0)
2561 ;; = -exp(%i*%pi/2/n) * integrate(exp(-k*t^n),t,0,R)
2562 ;;
2563 ;; As R -> infinity, this last integral is gamma(1/n)/k^(1/n)/n.
2564 ;;
2565 ;; We assume the integral on the circular arc approaches 0 as R ->
2566 ;; infinity. (Need to prove this.)
2567 ;;
2568 ;; Thus, we have
2569 ;;
2570 ;; integrate(exp(%i*k*t^n),t,0,inf)
2571 ;; = exp(%i*%pi/2/n) * gamma(1/n)/k^(1/n)/n.
2572 ;;
2573 ;; Equating real and imaginary parts gives us the desired results:
2574 ;;
2575 ;; integrate(cos(k*t^n),t,0,inf) = G * cos(%pi/2/n)
2576 ;; integrate(sin(k*t^n),t,0,inf) = G * sin(%pi/2/n)
2577 ;;
2578 ;; where G = gamma(1/n)/k^(1/n)/n.
2579 ;;
2587 ;; Ok, we have cos(b*x^n) or sin(b*x^n), and we set e = (n
2588 ;; b)
2590 ;; n = 1. Give up. (Why not divergent?)
2591 nil)
2596 ;; s is the sign of b. Give up if it's zero.
2597 nil)
2599 ;; Give up if n-1 <= 0. (Why give up? Isn't the
2600 ;; integral divergent?)
2601 nil)
2603 ;; We can apply our formula now. g = gamma(1/n)/n/b^(1/n)
2610 e)))))))
2613 ;; this is the second part of the definite integral package
2618 ())
2620 ivar t)
2642 nil)
2644 t)))))
2654 n
2657 plm))))
2669 plm
2670 factors
2671 pl
2672 rl
2673 pl1
2674 rl1))))
2682 ans)
2684 ;; AFAICT, this call to PLOG doesn't need
2685 ;; to bind VAR.
2687 ans))))
2694 ;; AFAICT, this call to PLOG doesn't need
2695 ;; to bind VAR. An example where this is
2696 ;; used is
2697 ;; integrate(log(x)^2/(1+x^2),x,0,1) =
2698 ;; %pi^3/16.
2701 n)
2702 d
2703 plm)))))
2707 0
2710 ;; Handle integral(n(x)/d(x)*log(x)^m,x,0,inf). n and d are
2711 ;; polynomials.
2729 ans))))
2753 n
2757 ;;Makes a new poly such that np(x)-np(x+2*%i*%pi)=p(x).
2758 ;;Constructs general POLY of degree one higher than P with
2759 ;;arbitrary coeff. and then solves for coeffs by equating like powers
2760 ;;of the varibale of integration.
2761 ;;Can probably be made simpler now.
2769 ;;;Now, poly(x)-poly(x+2*%i*%pi)=p(x), P is the original poly.
2778 -1
2787 ;; Integrate(p(x)*f(exp(x))/g(exp(x)),x,minf,inf) by applying the
2788 ;; transformation y = exp(x) to get
2789 ;; integrate(p(log(y))*f(y)/g(y)/y,y,0,inf). This should be handled
2790 ;; by dintlog.
2797 ;; This implements Wang's algorithm in Chapter 5.2, pp. 98-100.
2798 ;;
2799 ;; This is a very brief description of the algorithm. Basically, we
2800 ;; have integrate(R(exp(x))*p(x),x,minf,inf), where R(x) is a rational
2801 ;; function and p(x) is a polynomial.
2802 ;;
2803 ;; We find a polynomial q(x) such that q(x) - q(x+2*%i*%pi) = p(x).
2804 ;; Then consider a contour integral of R(exp(z))*q(z) over a
2805 ;; rectangular contour. Opposite corners of the rectangle are (-R,
2806 ;; 2*%i*%pi) and (R, 0).
2807 ;;
2808 ;; Wang shows that this contour integral, in the limit, is the
2809 ;; integral of R(exp(x))*q(x)-R(exp(x))*q(x+2*%i*%pi), which is
2810 ;; exactly the integral we're looking for.
2811 ;;
2812 ;; Thus, to find the value of the contour integral, we just need the
2813 ;; residues of R(exp(z))*q(z). The only tricky part is that we want
2814 ;; the log function to have an imaginary part between 0 and 2*%pi
2815 ;; instead of -%pi to %pi.
2817 ;; We have R(exp(x))*p(x) represented as p(x)*pe(exp(x))/d(exp(x)).
2823 ;; At this point denom-exponential has converted d(exp(x)) to the
2824 ;; polynomial d(z), where z = exp(x).
2827 pe))
2829 ;; Find the poles of the denominator. denom-exponential is the
2830 ;; denominator of R(x).
2831 ;;
2832 ;; It seems as if polelist returns a list of several items.
2833 ;; The first element is a list consisting of the pole and (z -
2834 ;; pole). We don't care about this, so we take the rest of the
2835 ;; result.
2838 ;; The imaginary part is nonzero,
2839 ;; or the realpart is negative.
2843 ;; The realpart is not zero.
2845 ;; Not sure what this does.
2847 ;; No roots at all, so return
2851 ;; We have simple roots or roots in REGION1
2854 ;; The cadr of pl is the list of the simple poles of
2855 ;; denom-exponential. Take the log of them to find the
2856 ;; poles of the original expression. Then compute the
2857 ;; residues at each of these poles and sum them up and put
2858 ;; the result in B. (If no simple poles set B to 0.)
2864 ;; I think this handles the case of poles outside the
2865 ;; regions. The sum of these residues are placed in C.
2869 temp)))
2874 ;; We have the repeated poles of deonom-exponential, so we
2875 ;; need to convert them to the actual pole values for
2876 ;; R(exp(x)), by taking the log of the value of poles.
2881 ;; Compute the residues at all of these poles and sum
2882 ;; them up.
2885 poles))
2900 ;; Check to see if each term in exp that is of the form exp(k*x) has
2901 ;; an integer value for k.
2917 ;; Test to see if exp is of the form f(exp(x)), and if so, replace
2918 ;; exp(x) with 'z*. That is, basically return f(z*).
2921 exp)
2925 ivar))
2928 ;; Does e really need to be special here? Seems to be ok without
2929 ;; it; testsuite works.
2930 #+nil
2933 e)
2964 ;; I think this is looking at f(exp(x)) and tries to find some
2965 ;; rational function R and some number k such that f(exp(x)) =
2966 ;; R(exp(k*x)).
2983 ;; Integration by parts.
2984 ;;
2985 ;; integrate(u(x)*diff(v(x),x),x,a,b)
2986 ;; |b
2987 ;; = u(x)*v(x)| - integrate(v(x)*diff(u(x),x))
2988 ;; |a
2989 ;;
2991 ;;;SINCE ONLY CALLED FROM DINTLOG TO get RID OF LOGS - IF LOG REMAINS, QUIT
2995 nil)
3002 nil)
3009 ;; integrate(f(exp(k*x)),x,a,b), where f(z) is rational.
3010 ;;
3011 ;; See Wang p. 96-97.
3012 ;;
3013 ;; If the limits are minf to inf, we use the substitution y=exp(k*x)
3014 ;; to get integrate(f(y)/y,y,0,inf)/k. If the limits are 0 to inf,
3015 ;; use the substitution s+1=exp(k*x) to get
3016 ;; integrate(f(s+1)/(s+1),s,0,inf).
3022 ;; If we can integrate it directly, do so and take the
3023 ;; appropriate limits.
3024 )
3026 ;; ans is the list (f(x) exp(k*x)).
3029 ;; Use the substitution s + 1 = exp(k*x). The
3030 ;; integral becomes integrate(f(s+1)/(s+1),s,0,inf)
3033 ;; Use the substitution y=exp(k*x) because the
3034 ;; limits are minf to inf.
3036 ;; Apply the substitution and integrate it.
3039 ;; integrate(log(g(x))*f(x),x,0,inf)
3047 exp
3048 ivar))
3050 ;; Make the substitution y=1/x. If the integrand has
3051 ;; exactly the same form, the answer has to be 0.
3057 ;; It's easy if we have the antiderivative.
3058 ;; but intsubs sometimes gives results containing %limit
3060 ;; Ok, the easy cases didn't work. We now try integration by
3061 ;; parts. Set ANS to f(x).
3064 ;; Bad. f(x) contains a log term, so we give up.
3070 ;; The arg of the log function is the same as the
3071 ;; integration variable. We can do something a little
3072 ;; simpler than integration by parts. We have something
3073 ;; like f(x)*log(x). Consider f(x)*x^z. If we
3074 ;; differentiate this wrt to z, the integrand becomes
3075 ;; f(x)*log(x)*x^z. When we evaluate this at z = 0, we
3076 ;; get the desired integrand.
3077 ;;
3078 ;; So we need f(0) to be 0 at 0. If we can integrate
3079 ;; f(x)*x^z, then we differentiate the result and
3080 ;; evaluate it at z = 0.
3083 ;; Try integration by parts.
3086 ;; Compute diff(e,ivar,n) at the point pt.
3090 ;;; GGR and friends
3092 ;; MAYBPC returns (COEF EXPO CONST)
3093 ;;
3094 ;; This basically picks off b*x^n+a and returns the list
3095 ;; (b n a).
3102 ;; At this point, e is of the form (a n b)
3107 ;; If we're here, b is complex, and n > 0. zn = imagpart(b).
3108 ;;
3109 ;; Set ivar to the same sign as zn.
3114 ;; zd = exp(ivar*%i*%pi*(1+nd)/(2*n). (ZD is special!)
3117 ;; Return zn, n, a, zd.
3123 ;; We're here if realpart(b) <= 0, and n >= 0. Then return -b, n, a.
3126 ;; Integrate x^m*exp(b*x^n+a), with realpart(m) > -1.
3127 ;;
3128 ;; See Wang, pp. 84-85.
3129 ;;
3130 ;; I believe the formula Wang gives is incorrect. The derivation is
3131 ;; correct except for the last step.
3132 ;;
3133 ;; Let J = integrate(x^m*exp(%i*k*x^n),x,0,inf), with real k.
3134 ;;
3135 ;; Consider the case for k < 0. Take a sector of a circle bounded by
3136 ;; the real line and the angle -%pi/(2*n), and by the radii, r and R.
3137 ;; Since there are no poles inside this contour, the integral
3138 ;;
3139 ;; integrate(z^m*exp(%i*k*z^n),z) = 0
3140 ;;
3141 ;; Then J = exp(-%pi*%i*(m+1)/(2*n))*integrate(R^m*exp(k*R^n),R,0,inf)
3142 ;;
3143 ;; because the integral along the boundary is zero except for the part
3144 ;; on the real axis. (Proof?)
3145 ;;
3146 ;; Wang seems to say this last integral is gamma(s/n/(-k)^s) where s =
3147 ;; (m+1)/n. But that seems wrong. If we use the substitution R =
3148 ;; (y/(-k))^(1/n), we end up with the result:
3149 ;;
3150 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n).
3151 ;;
3152 ;; or gamma((m+1)/n)/k^((m+1)/n)/n.
3153 ;;
3154 ;; Note that this also handles the case of
3155 ;;
3156 ;; integrate(x^m*exp(-k*x^n),x,0,inf);
3157 ;;
3158 ;; where k is positive real number. A simple change of variables,
3159 ;; y=k*x^n, gives
3160 ;;
3161 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n))
3162 ;;
3163 ;; which is the same form above.
3182 ;; e = (m b n a). That is, the integral is of the form
3183 ;; x^m*exp(b*x^n+a). I think we want to compute
3184 ;; gamma((m+1)/n)/b^((m+1)/n)/n.
3185 ;;
3186 ;; FIXME: If n > m + 1, the integral converges. We need
3187 ;; to check for this.
3189 e
3192 ;; The derivation only holds if b is purely real or
3193 ;; purely imaginary. Give up if it's not.
3195 ;; Check for convergence. If b is complex, we need n -
3196 ;; m > 1. If b is real, we need b < 0.
3205 ;; If we're here, b must be negative.
3208 ;; Complex b. Take the imaginary part
3210 n a))
3212 ;; FIXME: Why do we set %emode here? Shouldn't we just
3213 ;; bind it? And why do we want it bound to T anyway?
3214 ;; Shouldn't the user control that? The same goes for
3215 ;; dosimp.
3216 ;;(setq $%emode t)
3222 ;; Match x^m*exp(b*x^n+a). If it does, return (list m b n a).
3228 ;; We're looking at something like exp(f(ivar)). See if it's
3229 ;; of the form b*x^n+a, and return (list 0 b n a). (The 0 is
3230 ;; so we can graft something onto it if needed.)
3235 ;; E should be the product of exactly 2 terms
3237 ;; Check to see if one of the terms is of the form
3238 ;; ivar^p. If so, make sure the realpart of p > -1. If
3239 ;; so, check the other term has the right form via
3240 ;; another call to ggr1.
3251 ;; Both terms have the right form and nn* contains the ivar of
3252 ;; the exponential term. Put dn* as the car of nn*. The
3253 ;; result is something like (m b n a) when we have the
3254 ;; expression x^m*exp(b*x^n+a).
3258 ;; Match b*x^n+a. If a match is found, return the list (a n b).
3259 ;; Otherwise, return NIL
3272 ;; Match b*x^n. Return the list (n b) if found or NIL if not.
3282 ;; nn* should be the value of var. This is only called by bx**n with
3283 ;; the second arg of var.
3294 ;;; given (b*x^n+a)^m returns (m a n b)
3301 nil)
3308 ;;;Catches Unfactored forms.
3311 ivar)
3324 ;;;Is E = VAR raised to some power? If so return power or 0.
3330 ;;;Is E = VAR1 raised to some power? If so return power.
3351 ans)))))
3379 ;; Note: This doesn't seem be called from anywhere.
3404 ivar)
3432 ;;;Looks at the IMAGINARY part of a log and puts it in the interval 0 2*%pi.
3435 ;; We take the $rectform above to make sure that the log is
3436 ;; expanded out for the situations where simplifying plog itself
3437 ;; doesn't do it. This should probably be considered a bug in the
3438 ;; plog simplifier and should be fixed there.
3451 ;;; Temporary fix for a lacking in taylor, which loses with %i in denom.
3452 ;;; Besides doesn't seem like a bad thing to do in general.
3456 ;; Multiply the denominator by it's conjugate to get rid of
3457 ;; %i.
3462 ;; If the new denominator still contains %i, just give up.
3464 exp
3465 new-exp)))
3468 ;;; LL and UL must be real otherwise this routine return $UNKNOWN.
3469 ;;; Returns $no $unknown or a list of poles in the interval (*ll* *ul*)
3470 ;;; for exp w.r.t. ivar.
3471 ;;; Form of list ((pole . multiplicity) (pole1 . multiplicity) ....)
3488 ;; this clause handles cases where we can't find the exact roots,
3489 ;; but we know that they occur outside the interval of integration.
3490 ;; example: integrate ((1+exp(t))/sqrt(t+exp(t)), t, 0, 1);
3506 ;; (multiplicity (cdar dummy)) (not used? -- cwh)
3516 pole-list))))))))))))
3519 ;;;Returns $YES if there is no pole and $NO if there is one.
3532 ;;;Takes care of forms that the ratio test fails on.
3548 (%einvolve-var
3551 ivar))
3585 ;; real values for *ll* and *ul*; place can be imaginary.
3592 '$no
3597 ;; returns true or nil
3599 ;; real values for *ll* and *ul*; place can be imaginary.
3615 ;; If *failures is set, we may have missed some roots.
3616 ;; We still return the roots that we have found.
3621 rootlist)
3628 rootlist)))))))))
3631 ;;; Is x > y. X or Y can be $MINF or $INF, zeroA or zeroB.
3671 nil)
3674 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3675 ;;;
3676 ;;; Integrate Definite Integrals involving log and exp functions. The algorithm
3677 ;;; are taken from the paper "Evaluation of CLasses of Definite Integrals ..."
3678 ;;; by K.O.Geddes et. al.
3679 ;;;
3680 ;;; 1. CASE: Integrals generated by the Gamma function.
3681 ;;;
3682 ;;; inf
3683 ;;; /
3684 ;;; [ w m s - m - 1
3685 ;;; I t log (t) expt(- t ) dt = s signum(s)
3686 ;;; ]
3687 ;;; /
3688 ;;; 0
3689 ;;; !
3690 ;;; m !
3691 ;;; d !
3692 ;;; (--- (gamma(z))! )
3693 ;;; m !
3694 ;;; dz ! w + 1
3695 ;;; !z = -----
3696 ;;; s
3697 ;;;
3698 ;;; The integral converges for:
3699 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3700 ;;;
3701 ;;; 2. CASE: Integrals generated by the Incomplete Gamma function.
3702 ;;;
3703 ;;; inf !
3704 ;;; / m !
3705 ;;; [ w m s d s !
3706 ;;; I t log (t) exp(- t ) dt = (--- (gamma_incomplete(a, x ))! )
3707 ;;; ] m !
3708 ;;; / da ! w + 1
3709 ;;; x !z = -----
3710 ;;; s
3711 ;;; - m - 1
3712 ;;; s signum(s)
3713 ;;;
3714 ;;; The integral converges for:
3715 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3716 ;;; The shown solution is valid for s>0. For s<0 gamma_incomplete has to be
3717 ;;; replaced by gamma(a) - gamma_incomplete(a,x^s).
3718 ;;;
3719 ;;; 3. CASE: Integrals generated by the beta function.
3720 ;;;
3721 ;;; 1
3722 ;;; /
3723 ;;; [ m s r n
3724 ;;; I log (1 - t) (1 - t) t log (t) dt =
3725 ;;; ]
3726 ;;; /
3727 ;;; 0
3728 ;;; !
3729 ;;; ! !
3730 ;;; n m ! !
3731 ;;; d d ! !
3732 ;;; --- (--- (beta(z, w))! )!
3733 ;;; n m ! !
3734 ;;; dz dw ! !
3735 ;;; !w = s + 1 !
3736 ;;; !z = r + 1
3737 ;;;
3738 ;;; The integral converges for:
3739 ;;; n, m = 0, 1, 2, ..., s > -1 and r > -1.
3740 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3744 ;;; Recognize c*z^w*log(z)^m*exp(-t^s)
3756 ;;; Recognize c*z^r*log(z)^n*(1-z)^s*log(1-z)^m
3769 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3776 ;; var1 is used as a parameter for differentiation. Add var1>0 to the
3777 ;; database, to get the desired simplification of the differentiation of
3778 ;; the gamma_incomplete function.
3786 ;; The integrand matches the cases 1 and 2.
3805 ;; Case 1: Generated by the Gamma function.
3812 var1
3817 ; derivate the involved hypergeometric
3818 ; functions.
3823 ;; Case 2: Generated by the Incomplete Gamma function.
3839 ;; Case 3: Generated by the Beta function.
3865 var2 m)
3867 var1 n)
3873 ;; Simplify result and set $gamma_expand to global value
3876 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;