| blob | 1d2f09fc015bec1c2d53909971f404b4ad767a16 |
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.
128 *nodiverg exp1
131 nd*
137 ;;;rsn* is in comdenom. does a ratsimp of numerator.
138 ;expvar
140 ;impvar
142 context
143 ;;LIMITP T Causes $ASKSIGN to do special things
144 ;;For DEFINT like eliminate epsilon look for prin-inf
145 ;;take realpart and imagpart.
146 integer-info
147 ;;If LIMITP is non-null ask-integer conses
148 ;;its assumptions onto this list.
149 ))
154 "When @code{true}, definite integration tries to find poles in the integrand
161 ;; Distribute $defint over equations, lists, and matrices.
193 (merror (intl:gettext "defint: variable of integration must be a simple or subscripted variable.~%defint: found ~M") ivar))
222 ;; If antideriv can't do it, returns nil
223 ;; use limit to evaluate every answer returned by antideriv.
227 ;;;Hack the expression up for exponentials.
230 ;; Is this expr a candidate for SININT ?
239 t)))
240 ;;ERF type things go here.
245 t))))))
265 ;; We have some subscripted variable that's not our variable
266 ;; (I think), so it's deg-lessp.
267 ;;
268 ;; FIXME: Is this the best way to handle this? Are there
269 ;; other cases we're mising here?
270 t)))
280 ;; This routine tries to take a limit a couple of ways.
286 ans
289 ivar
290 0
299 ;; Test whether fun2 is inverse of fun1 at val.
305 ;; integration change of variable
313 ())
315 ;; This is a hack! If nv is of the form b*x^n+a, we can
316 ;; solve the equation manually instead of using solve.
317 ;; Why? Because solve asks us for the sign of yx and
318 ;; that's bogus.
320 ;; Solve yx = b*x^n+a, for x. Any root will do. So we
321 ;; have x = ((yx-a)/b)^(1/n).
323 d
329 ))))
346 ;; d: original variable (ivar) as a function of 'yx
347 ;; ind: boolean flag
348 ;; nv: new variable ('yx) as a function of original variable (ivar)
357 ;; converts limits of integration to values for new variable 'yx
367 ;; wrapper around limit, returns nil if
368 ;; limit not found (nounform returned), or undefined ($und or $ind)
375 ans))))
377 ;; rewrites exp, the integrand in terms of ivar,
378 ;; into exp1, the integrand in terms of 'yx.
399 ;; D (not used? -- cwh)
412 ;; Look for integrals which involve log and exp functions.
413 ;; Maxima has a special algorithm to get general results.
424 %tan %sinh %cosh %tanh
425 %log %asin %acos %atan
426 %cot %acot %sec
427 %asec %csc %acsc
429 ;; Call ANTIDERIV with logabs disabled,
430 ;; because the Risch algorithm assumes
431 ;; the integral of 1/x is log(x), not log(abs(x)).
432 ;; Why not just assume logabs = false within RISCHINT itself?
433 ;; Well, there's at least one existing result which requires
434 ;; logabs = true in RISCHINT, so try to make a minimal change here instead.
475 ans)
478 ans)
491 ;;Should be a PROG inside of unwind-protect, but Multics has a compiler
492 ;;bug wrt. and I want to test this code now.
500 ;;;This seems((and (and (eq *ul* '$inf)
501 ;;;fairly losing (setq exp (subin (m+ *ll* ivar) exp))
502 ;;; (setq *ll* 0.))
503 ;;; (ztoinf exp ivar)))
508 ;; ((and (and (equal *ul* 1.)
509 ;; (equal *ll* 0.)) (zto1 exp)))
517 ;;;NOT COMPLETE for sin's and cos's.
539 ;; Call ANTIDERIV with logabs disabled,
540 ;; because the Risch algorithm assumes
541 ;; the integral of 1/x is log(x), not log(abs(x)).
542 ;; Why not just assume logabs = false within RISCHINT itself?
543 ;; Well, there's at least one existing result which requires
544 ;; logabs = true in RISCHINT, so try to make a minimal change here instead.
550 ;;;Recursion stopper
573 ;; adds up integrals of ranges between each pair of poles.
574 ;; checks if whole thing is divergent as limits of integration approach poles.
576 ;;; calling $logcontract causes antiderivative of 1/(1-x^5) to blow up
577 ;; (setq anti-deriv (cond ((involve anti-deriv '(%log))
578 ;; ($logcontract anti-deriv))
579 ;; (t anti-deriv)))
588 ivar))))
591 ;;Return section.
616 pole-list)
618 ;; Assumes EXP is a rational expression with no polynomial part and
619 ;; converts the finite integration to integration over a half-infinite
620 ;; interval. The substitution y = (x-a)/(b-x) is used. Equivalently,
621 ;; x = (b*y+a)/(y+1).
622 ;;
623 ;; (I'm guessing CV means Change Variable here.)
626 ;; FIXME! This is a hack. We apply the transformation with
627 ;; symbolic limits and then substitute the actual limits later.
628 ;; That way method-by-limits (usually?) sees a simpler
629 ;; integrand.
630 ;;
631 ;; See Bugs 938235 and 941457. These fail because $FACTOR is
632 ;; unable to factor the transformed result. This needs more
633 ;; work (in other places).
637 ;; If the limit is a number, use $substitute so we simplify
638 ;; the result. Do we really want to do this?
646 ()))
648 ;; Integrate rational functions over a finite interval by doing the
649 ;; polynomial part directly, and converting the rational part to an
650 ;; integral from 0 to inf. This is evaluated via residues.
653 ;; PQR divides the rational expression and returns the quotient
654 ;; and remainder
661 ;; No polynomial part
664 ans
667 ;; Only polynomial part
670 ;; A non-zero quotient and remainder. Combine the results
671 ;; together.
675 ans))
680 ;; I think this takes a rational expression E, and finds the
681 ;; polynomial part. A cons is returned. The car is the quotient and
682 ;; the cdr is the remainder.
707 ;;;If leadop isn't plus don't do anything.
717 t)
723 t)
778 ;; *Z* is a "zero parameter" for this package.
779 ;; Its also used to transform.
780 ;; limit(exp,var,val,dir) -- limit(exp,tvar,0,dir)
787 ;; EPSILON is used in principal value code to denote the familiar
788 ;; mathematical entity.
793 ;;; PRIN-INF Is a special symbol in the principal value code used to
794 ;;; denote an end-point which is proceeding to infinity.
817 ; We have minf <= *ll* = *ul* <= inf
818 )
820 ; We have minf <= *ll* < *ul* = inf
821 )
823 ; We have minf = *ll* < *ul* < inf
824 ;
825 ; Now substitute
826 ;
827 ; ivar -> -ivar
828 ; *ll* -> -*ul*
829 ; *ul* -> inf
830 ;
831 ; so that minf < *ll* < *ul* = inf
837 ; We have minf <= *ul* < *ll*
838 ;
839 ; Now substitute
840 ;
841 ; exp -> -exp
842 ; *ll* <-> *ul*
843 ;
844 ; so that minf <= *ll* < *ul*
847 t)))
861 ;; Substitute a and b into integral e
862 ;;
863 ;; Looks for discontinuties in integral, and works around them.
864 ;; For example, in
865 ;;
866 ;; integrate(x^(2*n)*exp(-(x)^2),x) ==>
867 ;; -gamma_incomplete((2*n+1)/2,x^2)*x^(2*n+1)*abs(x)^(-2*n-1)/2
868 ;;
869 ;; the integral has a discontinuity at x=0.
870 ;;
877 ivar a b)))))
891 total))))))
893 ;; look for terms with a negative exponent
894 ;;
895 ;; recursively traverses exp in order to find discontinuities such as
896 ;; erfc(1/x-x) at x=0
908 ;; returns list of places where exp might be discontinuous in ivar.
909 ;; list begins with *ll* and ends with *ul*, and include any values between
910 ;; *ll* and *ul*.
911 ;; return '$no or '$unknown if no discontinuities found.
928 ;; (multiplicity (cdar dummy)) ;; not used
933 ;; Carefully substitute the integration limits A and B into the
934 ;; expression E.
944 ;; Try easy substitutions. Return NIL if we can't.
947 ;; Infinite limits aren't easy
948 nil)
952 %gamma_incomplete %expintegral_ei)))
954 ;; It's easy if we have a polynomial. I (rtoy) think
955 ;; it's also easy if the denominator is free of the
956 ;; integration variable and also if the expression
957 ;; doesn't involve inverse functions.
958 ;;
959 ;; gamma_incomplete and expintegral_ie
960 ;; included because of discontinuity in
961 ;; gamma_incomplete(0, exp(%i*x)) and
962 ;; expintegral_ei(exp(%i*x))
963 ;;
964 ;; XXX: Are there other cases we've forgotten about?
965 ;;
966 ;; So just try to substitute the limits into the
967 ;; expression. If no errors are produced, we're done.
972 ;; no-err-sub has returned T. An error was catched.
973 nil)
987 ;; check for divergent integral
999 ;;;This function works only on things with ATAN's in them now.
1001 ;; POLES-IN-INTERVAL doesn't know about the poles of tan(x). Call
1002 ;; trigsimp to convert tan into sin/cos, which POLES-IN-INTERVAL
1003 ;; knows how to handle.
1004 ;;
1005 ;; XXX Should we fix POLES-IN-INTERVAL instead?
1006 ;;
1007 ;; XXX Is calling trigsimp too much? Should we just only try to
1008 ;; substitute sin/cos for tan?
1009 ;;
1010 ;; XXX Should the result try to convert sin/cos back into tan? (A
1011 ;; call to trigreduce would do it, among other things.)
1014 ;;POLES -> ((mlist) ((mequal) ((%atan) foo) replacement) ......)
1015 ;;We can then use $SUBSTITUTE
1020 ul-ans)
1022 nil)))
1053 llist)))))))))))))))
1068 ;; There are no calls where fn1 is ever equal to
1069 ;; 'mtorat. Hence this case is never true. What is
1070 ;; this testing for?
1077 nil)
1096 ;; e is of form poly(x)*exp(m*%i*x)
1097 ;; s is degree of denominator
1098 ;; adds e to bptu or bptd according to sign of m
1114 ;; Check term is of form poly(x)*exp(m*%i*x)
1115 ;; n is degree of denominator.
1125 ;; Set updn to T if the coefficient of IVAR in the
1126 ;; polynomial is known to be positive. Otherwise set to NIL.
1127 ;; (What does updn really mean?)
1138 term)
1173 ;; exponentialize sin and cos
1179 '$%i
1201 exp)
1217 ;; computes integral from minf to inf for expressions of the form
1218 ;; exp(%i*m*x)*r(x) by residues on either the upper half
1219 ;; plane or the lower half plane, depending on whether
1220 ;; m is positive or negative. [wang p. 77]
1221 ;;
1222 ;; exponentializes sin and cos before applying residue method.
1223 ;; can handle some expressions with poles on real line, such as
1224 ;; sin(x)*cos(x)/x.
1229 ratterms ratans
1230 plf bptu bptd s upans downans)
1235 ivar
1236 nil)
1239 ;;;Above tests for applicability of this method.
1248 ;; call to $expand can create rational terms
1249 ;; with no exp(m*%i*x)
1253 ;;;Function RIB sets up the values of BPTU and BPTD
1260 ;;;If there is BPTU then CSEMIUP must succeed.
1261 ;;;Likewise for BPTD.
1265 ;; The original integrand was already
1266 ;; determined to have no poles by initial-analysis.
1267 ;; If individual terms of the expansion have poles, the poles
1268 ;; must cancel each other out, so we can ignore them.
1271 ;; if integral of ratterms is divergent, ratans is nil,
1272 ;; and mtosc returns nil
1289 t)
1291 t)
1299 t)
1301 t)
1306 s nc dc
1328 sd new-sd))
1329 result))
1332 sn nil)
1334 sd nil))
1336 ;;;
1337 ;;;New section.
1355 s
1358 ivar)))
1368 findout
1372 on
1395 s)
1399 ;; It's very important here to bind VAR
1400 ;; because the PLOG simplifier checks
1401 ;; for VAR. Without this, the
1402 ;; simplifier converts plog(-x) to
1403 ;; log(x)+%i*%pi, which messes up the
1404 ;; keyhole routine.
1406 d
1407 ivar)))
1410 ;; Let's not signal an error here. Return nil so that we
1411 ;; eventually return a noun form if no other algorithm gives
1412 ;; a result.
1415 nil)))
1472 sd new-sd))
1473 result))
1508 ;; Test to see if exp is of the form p(x)*f(exp(x)). If so, set pp to
1509 ;; be p(x) and set pe to f(exp(x)).
1522 s
1525 ivar)))
1532 ;; We have an integral of the form p(x)*F(exp(x)), where
1533 ;; p(x) is a polynomial.
1535 ;; No polynomial
1539 ;; These limits must exist for the integral to converge.
1542 ;; This only handles the case when the F(z) is a
1543 ;; rational function.
1546 ;; If we get here, F(z) is not a rational function.
1547 ;; We transform it using the substitution x=log(y)
1548 ;; which gives us an integral of the form
1549 ;; p(log(y))*F(y)/y, which maxima should be able to
1550 ;; handle.
1553 ;; Give up. We don't know how to handle this.
1555 en
1569 exp)))
1571 ;;; given (b*x+a)^n+c returns (a b n c)
1579 nil)
1581 ;;;get high degree of poly
1583 ;;;diff down to linear.
1585 ;;;all the way to constant.
1588 ;;;get rid of factorial from differentiation.
1591 ;;;Sees if can be expressed as (a*x+b)^n + part freeof x.
1601 s)
1626 n
1627 d)))))
1630 on
1685 e nil)
1699 e))))))
1705 ;; Compute the integral(n/d,x,0,inf) by computing the negative of the
1706 ;; sum of residues of log(-x)*n/d over the poles of n/d inside the
1707 ;; keyhole contour. This contour is basically an disk with a slit
1708 ;; along the positive real axis. n/d must be a rational function.
1713 ;; Ok if not on the positive real axis.
1718 t)
1729 ;; Look at an expression e of the form sin(r*x)^k, where k is an
1730 ;; integer. Return the list (1 r k). (Not sure if the first element
1731 ;; of the list is always 1 because I'm not sure what partition is
1732 ;; trying to do here.)
1746 ;; Look at an expression e of the form sin(r*x) and return r.
1757 ;; integrate(a*sc(r*x)^k/x^n,x,0,inf).
1760 ;; Get the argument of the involved trig function.
1763 ;; I don't think this needs to be special.
1764 #+nil
1766 ;; Replace (1-cos(arg)^2) with sin(arg)^2.
1768 ;(m+t 1. (m- (m^t `((%cos) ,ivar) 2.)))
1769 ;; The code from above generates expressions with
1770 ;; a missing simp flag. Furthermore, the
1771 ;; substitution has to be done for the complete
1772 ;; argument of the trig function. (DK 02/2010)
1775 exp))
1780 ;; n is the power of the denominator.
1782 ;; The simple case.
1788 ;; Do this for a sum of such terms.
1790 c)))))))))
1792 ;; We have an integral of the form sin(r*x)^k/x^n. C is the list (1 r k).
1793 ;;
1794 ;; The substitution y=r*x converts this integral to
1795 ;;
1796 ;; r^(n-1)*integral(sin(y)^k/y^n,y,0,inf)
1797 ;;
1798 ;; (If r is negative, we need to negate the result.)
1799 ;;
1800 ;; The recursion Wang gives on p. 87 has a typo. The second term
1801 ;; should be subtracted from the first. This formula is given in G&R,
1802 ;; 3.82, formula 12.
1803 ;;
1804 ;; integrate(sin(x)^r/x^s,x) =
1805 ;; r*(r-1)/(s-1)/(s-2)*integrate(sin(x)^(r-2)/x^(s-2),x)
1806 ;; - r^2/(s-1)/(s-2)*integrate(sin(x)^r/x^(s-2),x)
1807 ;;
1808 ;; (Limits are assumed to be 0 to inf.)
1809 ;;
1810 ;; This recursion ends up with integrals with s = 1 or 2 and
1811 ;;
1812 ;; integrate(sin(x)^p/x,x,0,inf) = integrate(sin(x)^(p-1),x,0,%pi/2)
1813 ;;
1814 ;; with p > 0 and odd. This latter integral is known to maxima, and
1815 ;; it's value is beta(p/2,1/2)/2.
1816 ;;
1817 ;; integrate(sin(x)^2/x^2,x,0,inf) = %pi/2*binomial(q-3/2,q-1)
1818 ;;
1819 ;; where q >= 2.
1820 ;;
1822 ;; Compute sign(r)*r^(n-1)*integrate(sin(y)^k/y^n,y,0,inf)
1824 c
1830 recursion)
1831 ;; Recursion failed. Return the integrand
1832 ;; The following code generates expressions with a missing simp flag
1833 ;; for the sin function. Use better simplifying code. (DK 02/2010)
1834 ; (let ((integrand (div (pow `((%sin) ,(m* r ivar))
1835 ; k)
1836 ; (pow ivar n))))
1838 k)
1843 ;; integrate(sin(x)^n/x^2,x,0,inf) = pi/2*binomial(n-3/2,n-1).
1844 ;; Express in terms of Gamma functions, though.
1850 ;; integrate(sin(x)^n/x,x,0,inf) = beta((n+1)/2,1/2)/2, for n odd and
1851 ;; n > 0.
1854 half%pi)
1857 ;; This implements the recursion for computing
1858 ;; integrate(sin(y)^l/y^k,y,0,inf). (Note the change in notation from
1859 ;; the above!)
1865 ;; Integral diverges if l-(k-1) < 0.
1868 ;; If l + k is not even, return NIL. (Is this the right
1869 ;; thing to do?)
1870 nil)
1872 ;; We have integrate(sin(y)^l/y^2). Use sevn to evaluate.
1875 ;; We have integrate(sin(y)^l/y,y)
1881 ;; j = k-2, i = (k-1)*(k-2)
1882 ;;
1883 ;;
1884 ;; The main recursion:
1885 ;;
1886 ;; i(sin(y)^l/y^k)
1887 ;; = l*(l-1)/(k-1)/(k-2)*i(sin(y)^(l-2)/y^k)
1888 ;; - l^2/(k-1)/(k-1)*i(sin(y)^l/y^(k-2))
1896 ;; Returns the fractional part of a?
1900 ;; Why do we return 0 if a is a number? Perhaps we really
1901 ;; mean integer?
1904 ;; If we're here, this basically assumes a is a rational.
1905 ;; Compute the remainder and return the result.
1913 ;; Doesn't appear to be used anywhere in Maxima. Not sure what this
1914 ;; was intended to do.
1915 #+nil
1926 ;;; THE FOLLOWING FUNCTION IS FOR TRIG FUNCTIONS OF THE FOLLOWING TYPE:
1927 ;;; LOWER LIMIT=0 B A MULTIPLE OF %PI SCA FUNCTION OF SIN (X) COS (X)
1928 ;;; B<=%PI2
1933 ;; means there was no error
1936 ; returns cons of (integer_part . fractional_part) of a
1938 ;; I think we really want to compute how many full periods are in a
1939 ;; and the remainder.
1944 ; returns cons of (integer_part . fractional_part) of a
1946 ;; I think we really want to compute how many full periods are in a
1947 ;; and the remainder.
1954 ;; Return the integer part of r.
1956 ;; r - fpart(r)
1960 ;;;Try making exp(%i*ivar) --> yy, if result is rational then do integral
1961 ;;;around unit circle. Make corrections for limits of integration if possible.
1981 rat-form
1982 ivar)))
1987 ;;; Do integrals of sin and cos. this routine makes sure lower limit
1988 ;;; is zero.
1990 ;; integrate(e,var,a,b)
2006 ivar e)
2007 cc))
2017 ;; Exit if b or a is not a constant or if the integrand
2018 ;; doesn't appear to have a period of 2 pi.
2021 ;; Multiples of 2*%pi in limits.
2028 ;; The integral is not over a full period (2*%pi) or multiple
2029 ;; of a full period.
2031 ;; Wang p. 111: The integral integrate(f(x),x,a,b) can be
2032 ;; written as:
2033 ;;
2034 ;; n * integrate(f,x,0,2*%pi) + integrate(f,x,0,c)
2035 ;; - integrate(f,x,0,d)
2036 ;;
2037 ;; for some integer n and d >= 0, c < 2*%pi because there exist
2038 ;; integers p and q such that a = 2 * p *%pi + d and b = 2 * q
2039 ;; * %pi + c. Then n = q - p.
2041 ;; Compute q and c for the upper limit b.
2048 ;; Compute p and d for the lower limit a.
2050 ;; avoid an extra trip around the circle - helps skip principal values
2055 ;; Compute -integrate(f,x,0,d)
2060 ;; Compute n = q - p (stored in nzp2)
2062 out
2063 ;; Compute n*integrate(f,x,0,2*%pi)
2065 ;; n = 0
2068 ;; n is not zero, so compute
2069 ;; integrate(f,x,0,2*%pi)
2071 ;; Unable to compute integrate(f,x,0,2*%pi)
2073 ;; Compute integrate(f,x,0,c)
2077 ;; All three pieces succeeded.
2080 ;; Less than 1 full period, so intsc can integrate it.
2081 ;; Apply the substitution to make the lower limit 0.
2082 ;; This is last resort because substitution often causes intsc to fail.
2084 limit-diff ivar))
2085 ;; nothing worked
2088 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)).
2089 ;; calls intsc with a wrapper to just return nil if integral is divergent,
2090 ;; rather than generating an error.
2095 nil
2096 ans)))
2098 ;; integrate(sc, ivar, 0, b), where sc is f(sin(x), cos(x)). I (rtoy)
2099 ;; think this expects b to be less than 2*%pi.
2102 0
2109 ;; Partition the integrand SC into the factors that do not
2110 ;; contain VAR (the car part) and the parts that do (the cdr
2111 ;; part).
2116 ;; integrate(sc, ivar, 0, b), where sc is f(sin(x), cos(x)).
2118 ;; Determine if sc is a product of sin's and cos's.
2122 ;; We have a product of sin's and cos's. We handle some
2123 ;; special cases here.
2125 ;; Wang p. 110, formula (1):
2126 ;; integrate(sin(x)^m*cos(x)^n, x, 0, %pi/2) =
2127 ;; gamma((m+1)/2)*gamma((n+1)/2)/2/gamma((n+m+2)/2)
2130 ;; Wang p. 110, near the bottom, says
2131 ;;
2132 ;; int(f(sin(x),cos(x)), x, 0, %pi) =
2133 ;; int(f(sin(x),cos(x)) + f(sin(x),-cos(x)),x,0,%pi/2)
2154 ;; Wang, p. 111 says
2155 ;;
2156 ;; int(f(sin(x),cos(x)),x,0,3*%pi/2) =
2157 ;; int(f(sin(x),cos(x)),x,0,%pi)
2158 ;; + int(f(-sin(x),-cos(x)),x,0,%pi/2)
2162 ;; We don't have a product of sin's and cos's.
2167 dn*)
2172 ;;;Is careful about substitution of limits where the denominator may be zero
2173 ;;;because of various assumptions made.
2194 nil
2195 ans)))
2201 nil
2202 ans)))
2204 ;; Determine whether E is of the form sin(x)^m*cos(x)^n and return the
2205 ;; list (m n).
2209 m n)
2228 ;; Says whether (m^t value exponent) has at least one real branch.
2229 ;; Only works for values of 1 and -1 now. Returns $yes $no
2230 ;; $unknown.
2241 ;; Compute beta((m+1)/2,(n+1)/2)/2.
2247 ;;Seems like Guys who call this don't agree on what it should return.
2260 ;; Check e for an expression of the form x^kk*(b*x^n+a)^l. If it
2261 ;; matches, Return the two values kk and (list l a n b).
2266 ;; We have f(x)^y. Look to see if f(x) has the desired
2267 ;; form. Then f(x)^y has the desired form too, with
2268 ;; suitably modified values.
2269 ;;
2270 ;; XXX: Should we ask for the sign of f(x) if y is not an
2271 ;; integer? This transformation we're going to do requires
2272 ;; that f(x)^y be real.
2274 e
2279 ;; Got a match. Adjust kk and cc appropriately.
2281 cc
2296 ;;(DEFUN BATAP (E)
2297 ;; (PROG (K C L)
2298 ;; (COND ((NOT (BATA0 E)) (RETURN NIL))
2299 ;; ((AND (EQUAL -1. (CADDDR C))
2300 ;; (EQ ($askSIGN (SETQ K (m+ 1. K)))
2301 ;; '$pos)
2302 ;; (EQ ($askSIGN (SETQ L (m+ 1. (CAR C))))
2303 ;; '$pos)
2304 ;; (ALIKE1 (CADR C)
2305 ;; (m^ UL (CADDR C)))
2306 ;; (SETQ E (CADR C))
2307 ;; (EQ ($askSIGN (SETQ C (CADDR C))) '$pos))
2308 ;; (RETURN (M// (m* (m^ UL (m+t K (m* C (m+t -1. L))))
2309 ;; `(($BETA) ,(SETQ NN* (M// K C))
2310 ;; ,(SETQ DN* L)))
2311 ;; C))))))
2314 ;; Integrals of the form i(log(x)^m*x^k*(a+b*x^n)^l,x,0,ul). There
2315 ;; are two cases to consider: One case has ul>0, b<0, a=-b*ul^n, k>-1,
2316 ;; l>-1, n>0, m a nonnegative integer. The second case has ul=inf, l < 0.
2317 ;;
2318 ;; These integrals are essentially partial derivatives of the Beta
2319 ;; function (i.e. the Eulerian integral of the first kind). Note
2320 ;; that, currently, with the default setting intanalysis:true, this
2321 ;; function might not even be called for some of these integrals.
2322 ;; However, this can be palliated by setting intanalysis:false.
2341 ;; We have i(x^kk/(d+cc*x^r)^s,x,0,inf) =
2342 ;; beta(aa,bb)/(cc^aa*d^bb*r). Compute this, and then
2343 ;; differentiate it m times to get the log term
2344 ;; incorporated.
2349 0
2350 m)))
2354 r))))
2363 ;; The result looks like B(k/n,l) ( ... ).
2364 ;; Perhaps, we should just $factor, instead of
2365 ;; pulling out beta like this.
2367 beta
2368 ($fullratsimp
2376 ivar m)
2378 beta))))))))))))
2381 ;;; If e is of the form given below, make the obvious change
2382 ;;; of variables (substituting ul*x^(1/n) for x) in order to reduce
2383 ;;; e to the usual form of the integrand in the Eulerian
2384 ;;; integral of the first kind.
2385 ;;; N. B: The old version of ZTO1 completely ignored this
2386 ;;; substitution; the log(x)s were just thrown in, which,
2387 ;;; of course would give wrong results.
2390 ;; Parse e
2394 ;; e=x^k*(a+b*x^n)^l
2396 c
2408 ;; Wang p. 71 gives the following formula for a beta function:
2409 ;;
2410 ;; integrate(x^(k-1)/(c*x^r+d)^s,x,0,inf)
2411 ;; = beta(a,b)/(c^a*d^b*r)
2412 ;;
2413 ;; where a = k/r > 0, b = s - a > 0, s > k > 0, r > 0, c*d > 0.
2414 ;;
2415 ;; This function matches this and returns k-1, d, r, c, a, b. And
2416 ;; also checks that all the conditions hold. If not, NIL is returned.
2417 ;;
2423 c
2440 ;; Handles beta integrals.
2448 nil)
2451 c
2452 ;; e = x^k*(d+c*x^al)^l.
2457 new-k)
2465 ;; Compute exp(d)*gamma((c+1)/b)/b/a^((c+1)/b). In essence, this is
2466 ;; the value of integrate(x^c*exp(d-a*x^b),x,0,inf).
2477 ;; Evaluates the contour integral of GRAND around the unit circle
2478 ;; using residues.
2484 ;; Is pt stricly inside the unit circle?
2490 ;; Is pt on the unit circle? (Use
2491 ;; the cached value computed
2492 ;; above.)
2513 ;; Wang 81-83. Unfortunately, the pdf version has page 82 as all
2514 ;; black, so here is, as best as I can tell, what Wang is doing.
2515 ;; Fortunately, p. 81 has the necessary hints.
2516 ;;
2517 ;; First consider integrate(exp(%i*k*x^n),x) around the closed contour
2518 ;; consisting of the real axis from 0 to R, the arc from the angle 0
2519 ;; to %pi/(2*n) and the ray from the arc back to the origin.
2520 ;;
2521 ;; There are no poles in this region, so the integral must be zero.
2522 ;; But consider the integral on the three parts. The real axis is the
2523 ;; integral we want. The return ray is
2524 ;;
2525 ;; exp(%i*%pi/2/n) * integrate(exp(%i*k*(t*exp(%i*%pi/2/n))^n),t,R,0)
2526 ;; = exp(%i*%pi/2/n) * integrate(exp(%i*k*t^n*exp(%i*%pi/2)),t,R,0)
2527 ;; = -exp(%i*%pi/2/n) * integrate(exp(-k*t^n),t,0,R)
2528 ;;
2529 ;; As R -> infinity, this last integral is gamma(1/n)/k^(1/n)/n.
2530 ;;
2531 ;; We assume the integral on the circular arc approaches 0 as R ->
2532 ;; infinity. (Need to prove this.)
2533 ;;
2534 ;; Thus, we have
2535 ;;
2536 ;; integrate(exp(%i*k*t^n),t,0,inf)
2537 ;; = exp(%i*%pi/2/n) * gamma(1/n)/k^(1/n)/n.
2538 ;;
2539 ;; Equating real and imaginary parts gives us the desired results:
2540 ;;
2541 ;; integrate(cos(k*t^n),t,0,inf) = G * cos(%pi/2/n)
2542 ;; integrate(sin(k*t^n),t,0,inf) = G * sin(%pi/2/n)
2543 ;;
2544 ;; where G = gamma(1/n)/k^(1/n)/n.
2545 ;;
2553 ;; Ok, we have cos(b*x^n) or sin(b*x^n), and we set e = (n
2554 ;; b)
2556 ;; n = 1. Give up. (Why not divergent?)
2557 nil)
2562 ;; s is the sign of b. Give up if it's zero.
2563 nil)
2565 ;; Give up if n-1 <= 0. (Why give up? Isn't the
2566 ;; integral divergent?)
2567 nil)
2569 ;; We can apply our formula now. g = gamma(1/n)/n/b^(1/n)
2576 e)))))))
2579 ;; this is the second part of the definite integral package
2584 ())
2586 ivar t)
2608 nil)
2610 t)))))
2620 n
2623 plm))))
2635 plm
2636 factors
2637 pl
2638 rl
2639 pl1
2640 rl1))))
2648 ans)
2650 ;; AFAICT, this call to PLOG doesn't need
2651 ;; to bind VAR.
2653 ans))))
2660 ;; AFAICT, this call to PLOG doesn't need
2661 ;; to bind VAR. An example where this is
2662 ;; used is
2663 ;; integrate(log(x)^2/(1+x^2),x,0,1) =
2664 ;; %pi^3/16.
2667 n)
2668 d
2669 plm)))))
2673 0
2676 ;; Handle integral(n(x)/d(x)*log(x)^m,x,0,inf). n and d are
2677 ;; polynomials.
2695 ans))))
2719 n
2723 ;;Makes a new poly such that np(x)-np(x+2*%i*%pi)=p(x).
2724 ;;Constructs general POLY of degree one higher than P with
2725 ;;arbitrary coeff. and then solves for coeffs by equating like powers
2726 ;;of the varibale of integration.
2727 ;;Can probably be made simpler now.
2735 ;;;Now, poly(x)-poly(x+2*%i*%pi)=p(x), P is the original poly.
2744 -1
2753 ;; Integrate(p(x)*f(exp(x))/g(exp(x)),x,minf,inf) by applying the
2754 ;; transformation y = exp(x) to get
2755 ;; integrate(p(log(y))*f(y)/g(y)/y,y,0,inf). This should be handled
2756 ;; by dintlog.
2763 ;; This implements Wang's algorithm in Chapter 5.2, pp. 98-100.
2764 ;;
2765 ;; This is a very brief description of the algorithm. Basically, we
2766 ;; have integrate(R(exp(x))*p(x),x,minf,inf), where R(x) is a rational
2767 ;; function and p(x) is a polynomial.
2768 ;;
2769 ;; We find a polynomial q(x) such that q(x) - q(x+2*%i*%pi) = p(x).
2770 ;; Then consider a contour integral of R(exp(z))*q(z) over a
2771 ;; rectangular contour. Opposite corners of the rectangle are (-R,
2772 ;; 2*%i*%pi) and (R, 0).
2773 ;;
2774 ;; Wang shows that this contour integral, in the limit, is the
2775 ;; integral of R(exp(x))*q(x)-R(exp(x))*q(x+2*%i*%pi), which is
2776 ;; exactly the integral we're looking for.
2777 ;;
2778 ;; Thus, to find the value of the contour integral, we just need the
2779 ;; residues of R(exp(z))*q(z). The only tricky part is that we want
2780 ;; the log function to have an imaginary part between 0 and 2*%pi
2781 ;; instead of -%pi to %pi.
2783 ;; We have R(exp(x))*p(x) represented as p(x)*pe(exp(x))/d(exp(x)).
2789 ;; At this point denom-exponential has converted d(exp(x)) to the
2790 ;; polynomial d(z), where z = exp(x).
2793 pe))
2795 ;; Find the poles of the denominator. denom-exponential is the
2796 ;; denominator of R(x).
2797 ;;
2798 ;; It seems as if polelist returns a list of several items.
2799 ;; The first element is a list consisting of the pole and (z -
2800 ;; pole). We don't care about this, so we take the rest of the
2801 ;; result.
2804 ;; The imaginary part is nonzero,
2805 ;; or the realpart is negative.
2809 ;; The realpart is not zero.
2811 ;; Not sure what this does.
2813 ;; No roots at all, so return
2817 ;; We have simple roots or roots in REGION1
2820 ;; The cadr of pl is the list of the simple poles of
2821 ;; denom-exponential. Take the log of them to find the
2822 ;; poles of the original expression. Then compute the
2823 ;; residues at each of these poles and sum them up and put
2824 ;; the result in B. (If no simple poles set B to 0.)
2830 ;; I think this handles the case of poles outside the
2831 ;; regions. The sum of these residues are placed in C.
2835 temp)))
2840 ;; We have the repeated poles of deonom-exponential, so we
2841 ;; need to convert them to the actual pole values for
2842 ;; R(exp(x)), by taking the log of the value of poles.
2847 ;; Compute the residues at all of these poles and sum
2848 ;; them up.
2851 poles))
2866 ;; Check to see if each term in exp that is of the form exp(k*x) has
2867 ;; an integer value for k.
2883 ;; Test to see if exp is of the form f(exp(x)), and if so, replace
2884 ;; exp(x) with 'z*. That is, basically return f(z*).
2887 exp)
2891 ivar))
2894 ;; Does e really need to be special here? Seems to be ok without
2895 ;; it; testsuite works.
2896 #+nil
2899 e)
2929 ;; I think this is looking at f(exp(x)) and tries to find some
2930 ;; rational function R and some number k such that f(exp(x)) =
2931 ;; R(exp(k*x)).
2948 ;; Integration by parts.
2949 ;;
2950 ;; integrate(u(x)*diff(v(x),x),x,a,b)
2951 ;; |b
2952 ;; = u(x)*v(x)| - integrate(v(x)*diff(u(x),x))
2953 ;; |a
2954 ;;
2956 ;;;SINCE ONLY CALLED FROM DINTLOG TO get RID OF LOGS - IF LOG REMAINS, QUIT
2960 nil)
2967 nil)
2975 ;; integrate(f(exp(k*x)),x,a,b), where f(z) is rational.
2976 ;;
2977 ;; See Wang p. 96-97.
2978 ;;
2979 ;; If the limits are minf to inf, we use the substitution y=exp(k*x)
2980 ;; to get integrate(f(y)/y,y,0,inf)/k. If the limits are 0 to inf,
2981 ;; use the substitution s+1=exp(k*x) to get
2982 ;; integrate(f(s+1)/(s+1),s,0,inf).
2988 ;; If we can integrate it directly, do so and take the
2989 ;; appropriate limits.
2990 )
2992 ;; ans is the list (f(x) exp(k*x)).
2995 ;; Use the substitution s + 1 = exp(k*x). The
2996 ;; integral becomes integrate(f(s+1)/(s+1),s,0,inf)
2999 ;; Use the substitution y=exp(k*x) because the
3000 ;; limits are minf to inf.
3002 ;; Apply the substitution and integrate it.
3005 ;; integrate(log(g(x))*f(x),x,0,inf)
3013 exp
3014 ivar))
3016 ;; Make the substitution y=1/x. If the integrand has
3017 ;; exactly the same form, the answer has to be 0.
3023 ;; It's easy if we have the antiderivative.
3024 ;; but intsubs sometimes gives results containing %limit
3026 ;; Ok, the easy cases didn't work. We now try integration by
3027 ;; parts. Set ANS to f(x).
3030 ;; Bad. f(x) contains a log term, so we give up.
3037 ;; The arg of the log function is the same as the
3038 ;; integration variable. We can do something a little
3039 ;; simpler than integration by parts. We have something
3040 ;; like f(x)*log(x). Consider f(x)*x^z. If we
3041 ;; differentiate this wrt to z, the integrand becomes
3042 ;; f(x)*log(x)*x^z. When we evaluate this at z = 0, we
3043 ;; get the desired integrand.
3044 ;;
3045 ;; So we need f(0) to be 0 at 0. If we can integrate
3046 ;; f(x)*x^z, then we differentiate the result and
3047 ;; evaluate it at z = 0.
3050 ;; Try integration by parts.
3053 ;; Compute diff(e,ivar,n) at the point pt.
3057 ;;; GGR and friends
3059 ;; MAYBPC returns (COEF EXPO CONST)
3060 ;;
3061 ;; This basically picks off b*x^n+a and returns the list
3062 ;; (b n a). It may also set the global *zd*.
3069 ;; At this point, e is of the form (a n b)
3074 ;; If we're here, b is complex, and n > 0. zn = imagpart(b).
3075 ;;
3076 ;; Set ivar to the same sign as zn.
3081 ;; zd = exp(ivar*%i*%pi*(1+nd)/(2*n). (ZD is special!)
3084 ;; Return zn, n, a, zd.
3090 ;; We're here if realpart(b) <= 0, and n >= 0. Then return -b, n, a.
3093 ;; Integrate x^m*exp(b*x^n+a), with realpart(m) > -1.
3094 ;;
3095 ;; See Wang, pp. 84-85.
3096 ;;
3097 ;; I believe the formula Wang gives is incorrect. The derivation is
3098 ;; correct except for the last step.
3099 ;;
3100 ;; Let J = integrate(x^m*exp(%i*k*x^n),x,0,inf), with real k.
3101 ;;
3102 ;; Consider the case for k < 0. Take a sector of a circle bounded by
3103 ;; the real line and the angle -%pi/(2*n), and by the radii, r and R.
3104 ;; Since there are no poles inside this contour, the integral
3105 ;;
3106 ;; integrate(z^m*exp(%i*k*z^n),z) = 0
3107 ;;
3108 ;; Then J = exp(-%pi*%i*(m+1)/(2*n))*integrate(R^m*exp(k*R^n),R,0,inf)
3109 ;;
3110 ;; because the integral along the boundary is zero except for the part
3111 ;; on the real axis. (Proof?)
3112 ;;
3113 ;; Wang seems to say this last integral is gamma(s/n/(-k)^s) where s =
3114 ;; (m+1)/n. But that seems wrong. If we use the substitution R =
3115 ;; (y/(-k))^(1/n), we end up with the result:
3116 ;;
3117 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n).
3118 ;;
3119 ;; or gamma((m+1)/n)/k^((m+1)/n)/n.
3120 ;;
3121 ;; Note that this also handles the case of
3122 ;;
3123 ;; integrate(x^m*exp(-k*x^n),x,0,inf);
3124 ;;
3125 ;; where k is positive real number. A simple change of variables,
3126 ;; y=k*x^n, gives
3127 ;;
3128 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n))
3129 ;;
3130 ;; which is the same form above.
3149 ;; e = (m b n a). That is, the integral is of the form
3150 ;; x^m*exp(b*x^n+a). I think we want to compute
3151 ;; gamma((m+1)/n)/b^((m+1)/n)/n.
3152 ;;
3153 ;; FIXME: If n > m + 1, the integral converges. We need
3154 ;; to check for this.
3156 e
3159 ;; The derivation only holds if b is purely real or
3160 ;; purely imaginary. Give up if it's not.
3162 ;; Check for convergence. If b is complex, we need n -
3163 ;; m > 1. If b is real, we need b < 0.
3172 ;; If we're here, b must be negative.
3175 ;; Complex b. Take the imaginary part
3177 n a))
3179 ;; FIXME: Why do we set %emode here? Shouldn't we just
3180 ;; bind it? And why do we want it bound to T anyway?
3181 ;; Shouldn't the user control that? The same goes for
3182 ;; dosimp.
3183 ;;(setq $%emode t)
3189 ;; Match x^m*exp(b*x^n+a). If it does, return (list m b n a).
3195 ;; We're looking at something like exp(f(ivar)). See if it's
3196 ;; of the form b*x^n+a, and return (list 0 b n a). (The 0 is
3197 ;; so we can graft something onto it if needed.)
3202 ;; E should be the product of exactly 2 terms
3204 ;; Check to see if one of the terms is of the form
3205 ;; ivar^p. If so, make sure the realpart of p > -1. If
3206 ;; so, check the other term has the right form via
3207 ;; another call to ggr1.
3218 ;; Both terms have the right form and nn* contains the ivar of
3219 ;; the exponential term. Put dn* as the car of nn*. The
3220 ;; result is something like (m b n a) when we have the
3221 ;; expression x^m*exp(b*x^n+a).
3225 ;; Match b*x^n+a. If a match is found, return the list (a n b).
3226 ;; Otherwise, return NIL
3239 ;; Match b*x^n. Return the list (n b) if found or NIL if not.
3249 ;; nn* should be the value of var. This is only called by bx**n with
3250 ;; the second arg of var.
3261 ;;; given (b*x^n+a)^m returns (m a n b)
3268 nil)
3275 ;;;Catches Unfactored forms.
3292 ;;;Is E = VAR raised to some power? If so return power or 0.
3298 ;;;Is E = VAR1 raised to some power? If so return power.
3319 ans)))))
3347 ;; Note: This doesn't seem be called from anywhere.
3372 ivar)
3400 ;;;Looks at the IMAGINARY part of a log and puts it in the interval 0 2*%pi.
3403 ;; We take the $rectform above to make sure that the log is
3404 ;; expanded out for the situations where simplifying plog itself
3405 ;; doesn't do it. This should probably be considered a bug in the
3406 ;; plog simplifier and should be fixed there.
3419 ;;; Temporary fix for a lacking in taylor, which loses with %i in denom.
3420 ;;; Besides doesn't seem like a bad thing to do in general.
3424 ;; Multiply the denominator by it's conjugate to get rid of
3425 ;; %i.
3430 ;; If the new denominator still contains %i, just give up.
3432 exp
3433 new-exp)))
3436 ;;; LL and UL must be real otherwise this routine return $UNKNOWN.
3437 ;;; Returns $no $unknown or a list of poles in the interval (*ll* *ul*)
3438 ;;; for exp w.r.t. ivar.
3439 ;;; Form of list ((pole . multiplicity) (pole1 . multiplicity) ....)
3456 ;; this clause handles cases where we can't find the exact roots,
3457 ;; but we know that they occur outside the interval of integration.
3458 ;; example: integrate ((1+exp(t))/sqrt(t+exp(t)), t, 0, 1);
3474 ;; (multiplicity (cdar dummy)) (not used? -- cwh)
3484 pole-list))))))))))))
3487 ;;;Returns $YES if there is no pole and $NO if there is one.
3500 ;;;Takes care of forms that the ratio test fails on.
3516 (%einvolve-var
3519 ivar))
3553 ;; real values for *ll* and *ul*; place can be imaginary.
3560 '$no
3565 ;; returns true or nil
3567 ;; real values for *ll* and *ul*; place can be imaginary.
3583 ;; If *failures is set, we may have missed some roots.
3584 ;; We still return the roots that we have found.
3589 rootlist)
3596 rootlist)))))))))
3599 ;;; Is x > y. X or Y can be $MINF or $INF, zeroA or zeroB.
3639 nil)
3642 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3643 ;;;
3644 ;;; Integrate Definite Integrals involving log and exp functions. The algorithm
3645 ;;; are taken from the paper "Evaluation of CLasses of Definite Integrals ..."
3646 ;;; by K.O.Geddes et. al.
3647 ;;;
3648 ;;; 1. CASE: Integrals generated by the Gamma function.
3649 ;;;
3650 ;;; inf
3651 ;;; /
3652 ;;; [ w m s - m - 1
3653 ;;; I t log (t) expt(- t ) dt = s signum(s)
3654 ;;; ]
3655 ;;; /
3656 ;;; 0
3657 ;;; !
3658 ;;; m !
3659 ;;; d !
3660 ;;; (--- (gamma(z))! )
3661 ;;; m !
3662 ;;; dz ! w + 1
3663 ;;; !z = -----
3664 ;;; s
3665 ;;;
3666 ;;; The integral converges for:
3667 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3668 ;;;
3669 ;;; 2. CASE: Integrals generated by the Incomplete Gamma function.
3670 ;;;
3671 ;;; inf !
3672 ;;; / m !
3673 ;;; [ w m s d s !
3674 ;;; I t log (t) exp(- t ) dt = (--- (gamma_incomplete(a, x ))! )
3675 ;;; ] m !
3676 ;;; / da ! w + 1
3677 ;;; x !z = -----
3678 ;;; s
3679 ;;; - m - 1
3680 ;;; s signum(s)
3681 ;;;
3682 ;;; The integral converges for:
3683 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3684 ;;; The shown solution is valid for s>0. For s<0 gamma_incomplete has to be
3685 ;;; replaced by gamma(a) - gamma_incomplete(a,x^s).
3686 ;;;
3687 ;;; 3. CASE: Integrals generated by the beta function.
3688 ;;;
3689 ;;; 1
3690 ;;; /
3691 ;;; [ m s r n
3692 ;;; I log (1 - t) (1 - t) t log (t) dt =
3693 ;;; ]
3694 ;;; /
3695 ;;; 0
3696 ;;; !
3697 ;;; ! !
3698 ;;; n m ! !
3699 ;;; d d ! !
3700 ;;; --- (--- (beta(z, w))! )!
3701 ;;; n m ! !
3702 ;;; dz dw ! !
3703 ;;; !w = s + 1 !
3704 ;;; !z = r + 1
3705 ;;;
3706 ;;; The integral converges for:
3707 ;;; n, m = 0, 1, 2, ..., s > -1 and r > -1.
3708 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3712 ;;; Recognize c*z^w*log(z)^m*exp(-t^s)
3724 ;;; Recognize c*z^r*log(z)^n*(1-z)^s*log(1-z)^m
3737 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3744 ;; var1 is used as a parameter for differentiation. Add var1>0 to the
3745 ;; database, to get the desired simplification of the differentiation of
3746 ;; the gamma_incomplete function.
3754 ;; The integrand matches the cases 1 and 2.
3773 ;; Case 1: Generated by the Gamma function.
3780 var1
3785 ; derivate the involved hypergeometric
3786 ; functions.
3791 ;; Case 2: Generated by the Incomplete Gamma function.
3807 ;; Case 3: Generated by the Beta function.
3833 var2 m)
3835 var1 n)
3841 ;; Simplify result and set $gamma_expand to global value
3844 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;