| blob | 850196102a51fb0f408dab9d6e5c0bbe6957d41b |
1 ;; -*- Mode: Lisp; Package: Maxima; Syntax: Common-Lisp; Base: 10 -*- ;;;;
2 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3 ;;; The data in this file contains enhancments. ;;;;;
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.
129 *nodiverg exp1
133 factors rlm*
134 $trigexpandplus $trigexpandtimes
140 ;;;rsn* is in comdenom. does a ratsimp of numerator.
141 ;expvar
143 ;impvar
145 $logabs $tlimswitch $maxposex $maxnegex
146 $trigsign $savefactors $radexpand $breakup $%emode
147 $float $exptsubst dosimp context rp-polylogp
148 %p%i half%pi %pi2 half%pi3 varlist genvar
149 $domain $m1pbranch errorsw
150 limitp $algebraic
151 ;;LIMITP T Causes $ASKSIGN to do special things
152 ;;For DEFINT like eliminate epsilon look for prin-inf
153 ;;take realpart and imagpart.
154 integer-info
155 ;;If LIMITP is non-null ask-integer conses
156 ;;its assumptions onto this list.
157 generate-atan2))
158 ;If this switch is () then RPART returns ATAN's
159 ;instead of ATAN2's
163 ;;These are really defined in LIMIT but DEFINT uses them also.
170 "When @code{true}, definite integration tries to find poles in the integrand
181 ;; Not really sure what this is meant to do, but it's used by MTORAT,
182 ;; KEYHOLE, and POLELIST.
187 ;; Distribute $defint over equations, lists, and matrices.
219 (merror (intl:gettext "defint: variable of integration must be a simple or subscripted variable.~%defint: found ~M") var))
248 ;; If antideriv can't do it, returns nil
249 ;; use limit to evaluate every answer returned by antideriv.
252 ;;;Hack the expression up for exponentials.
255 ;; Is this expr a candidate for SININT ?
264 t)))
265 ;;ERF type things go here.
270 t))))))
290 ;; We have some subscripted variable that's not our variable
291 ;; (I think), so it's deg-lessp.
292 ;;
293 ;; FIXME: Is this the best way to handle this? Are there
294 ;; other cases we're mising here?
295 t)))
305 ;; This routine tries to take a limit a couple of ways.
311 ans
314 var
315 0
325 ;; test whether fun2 is inverse of fun1 at val
333 ;; integration change of variable
341 ())
343 ;; This is a hack! If nv is of the form b*x^n+a, we can
344 ;; solve the equation manually instead of using solve.
345 ;; Why? Because solve asks us for the sign of yx and
346 ;; that's bogus.
348 ;; Solve yx = b*x^n+a, for x. Any root will do. So we
349 ;; have x = ((yx-a)/b)^(1/n).
351 d
357 ))))
374 ;; d: original variable (var) as a function of 'yx
375 ;; ind: boolean flag
376 ;; nv: new variable ('yx) as a function of original variable (var)
385 ;; converts limits of integration to values for new variable 'yx
395 ;; wrapper around limit, returns nil if
396 ;; limit not found (nounform returned), or undefined ($und or $ind)
403 ans))))
405 ;; rewrites exp, the integrand in terms of var,
406 ;; into exp1, the integrand in terms of 'yx.
427 ;; D (not used? -- cwh)
440 ;; Look for integrals which involve log and exp functions.
441 ;; Maxima has a special algorithm to get general results.
452 %tan %sinh %cosh %tanh
453 %log %asin %acos %atan
454 %cot %acot %sec
455 %asec %csc %acsc
457 ;; Call ANTIDERIV with logabs disabled,
458 ;; because the Risch algorithm assumes
459 ;; the integral of 1/x is log(x), not log(abs(x)).
460 ;; Why not just assume logabs = false within RISCHINT itself?
461 ;; Well, there's at least one existing result which requires
462 ;; logabs = true in RISCHINT, so try to make a minimal change here instead.
517 ;;Should be a PROG inside of unwind-protect, but Multics has a compiler
518 ;;bug wrt. and I want to test this code now.
526 ;;;This seems((and (and (eq ul '$inf)
527 ;;;fairly losing (setq exp (subin (m+ ll var) exp))
528 ;;; (setq ll 0.))
529 ;;; (ztoinf exp var)))
534 ;; ((and (and (equal ul 1.)
535 ;; (equal ll 0.)) (zto1 exp)))
543 ;;;NOT COMPLETE for sin's and cos's.
565 ;; Call ANTIDERIV with logabs disabled,
566 ;; because the Risch algorithm assumes
567 ;; the integral of 1/x is log(x), not log(abs(x)).
568 ;; Why not just assume logabs = false within RISCHINT itself?
569 ;; Well, there's at least one existing result which requires
570 ;; logabs = true in RISCHINT, so try to make a minimal change here instead.
576 ;;;Recursion stopper
599 ;; adds up integrals of ranges between each pair of poles.
600 ;; checks if whole thing is divergent as limits of integration approach poles.
602 ;;; calling $logcontract causes antiderivative of 1/(1-x^5) to blow up
603 ;; (setq anti-deriv (cond ((involve anti-deriv '(%log))
604 ;; ($logcontract anti-deriv))
605 ;; (t anti-deriv)))
616 ;;Return section.
641 pole-list)
643 ;; Assumes EXP is a rational expression with no polynomial part and
644 ;; converts the finite integration to integration over a half-infinite
645 ;; interval. The substitution y = (x-a)/(b-x) is used. Equivalently,
646 ;; x = (b*y+a)/(y+1).
647 ;;
648 ;; (I'm guessing CV means Change Variable here.)
651 ;; FIXME! This is a hack. We apply the transformation with
652 ;; symbolic limits and then substitute the actual limits later.
653 ;; That way method-by-limits (usually?) sees a simpler
654 ;; integrand.
655 ;;
656 ;; See Bugs 938235 and 941457. These fail because $FACTOR is
657 ;; unable to factor the transformed result. This needs more
658 ;; work (in other places).
662 ;; If the limit is a number, use $substitute so we simplify
663 ;; the result. Do we really want to do this?
671 ()))
673 ;; Integrate rational functions over a finite interval by doing the
674 ;; polynomial part directly, and converting the rational part to an
675 ;; integral from 0 to inf. This is evaluated via residues.
678 ;; PQR divides the rational expression and returns the quotient
679 ;; and remainder
686 ;; No polynomial part
689 ans
692 ;; Only polynomial part
695 ;; A non-zero quotient and remainder. Combine the results
696 ;; together.
700 ans))
705 ;; I think this takes a rational expression E, and finds the
706 ;; polynomial part. A cons is returned. The car is the quotient and
707 ;; the cdr is the remainder.
732 ;;;If leadop isn't plus don't do anything.
742 t)
748 t)
803 ;; *Z* is a "zero parameter" for this package.
804 ;; Its also used to transform.
805 ;; limit(exp,var,val,dir) -- limit(exp,tvar,0,dir)
812 ;; EPSILON is used in principal value code to denote the familiar
813 ;; mathematical entity.
818 ;;; PRIN-INF Is a special symbol in the principal value code used to
819 ;;; denote an end-point which is proceeding to infinity.
842 ; We have minf <= ll = ul <= inf
843 )
845 ; We have minf <= ll < ul = inf
846 )
848 ; We have minf = ll < ul < inf
849 ;
850 ; Now substitute
851 ;
852 ; var -> -var
853 ; ll -> -ul
854 ; ul -> inf
855 ;
856 ; so that minf < ll < ul = inf
862 ; We have minf <= ul < ll
863 ;
864 ; Now substitute
865 ;
866 ; exp -> -exp
867 ; ll <-> ul
868 ;
869 ; so that minf <= ll < ul
872 t)))
886 ;; Substitute a and b into integral e
887 ;;
888 ;; Looks for discontinuties in integral, and works around them.
889 ;; For example, in
890 ;;
891 ;; integrate(x^(2*n)*exp(-(x)^2),x) ==>
892 ;; -gamma_incomplete((2*n+1)/2,x^2)*x^(2*n+1)*abs(x)^(-2*n-1)/2
893 ;;
894 ;; the integral has a discontinuity at x=0.
895 ;;
902 var a b)))))
916 total))))))
918 ;; look for terms with a negative exponent
919 ;;
920 ;; recursively traverses exp in order to find discontinuities such as
921 ;; erfc(1/x-x) at x=0
931 ;; returns list of places where exp might be discontinuous in var.
932 ;; list begins with ll and ends with ul, and include any values between
933 ;; ll and ul.
934 ;; return '$no or '$unknown if no discontinuities found.
951 ;; (multiplicity (cdar dummy)) ;; not used
956 ;; Carefully substitute the integration limits A and B into the
957 ;; expression E.
967 ;; Try easy substitutions. Return NIL if we can't.
970 ;; Infinite limits aren't easy
971 nil)
975 %gamma_incomplete %expintegral_ei)))
977 ;; It's easy if we have a polynomial. I (rtoy) think
978 ;; it's also easy if the denominator is free of the
979 ;; integration variable and also if the expression
980 ;; doesn't involve inverse functions.
981 ;;
982 ;; gamma_incomplete and expintegral_ie
983 ;; included because of discontinuity in
984 ;; gamma_incomplete(0, exp(%i*x)) and
985 ;; expintegral_ei(exp(%i*x))
986 ;;
987 ;; XXX: Are there other cases we've forgotten about?
988 ;;
989 ;; So just try to substitute the limits into the
990 ;; expression. If no errors are produced, we're done.
995 ;; no-err-sub has returned T. An error was catched.
996 nil)
1010 ;; check for divergent integral
1022 ;;;This function works only on things with ATAN's in them now.
1024 ;; POLES-IN-INTERVAL doesn't know about the poles of tan(x). Call
1025 ;; trigsimp to convert tan into sin/cos, which POLES-IN-INTERVAL
1026 ;; knows how to handle.
1027 ;;
1028 ;; XXX Should we fix POLES-IN-INTERVAL instead?
1029 ;;
1030 ;; XXX Is calling trigsimp too much? Should we just only try to
1031 ;; substitute sin/cos for tan?
1032 ;;
1033 ;; XXX Should the result try to convert sin/cos back into tan? (A
1034 ;; call to trigreduce would do it, among other things.)
1037 ;;POLES -> ((mlist) ((mequal) ((%atan) foo) replacement) ......)
1038 ;;We can then use $SUBSTITUTE
1043 ul-ans)
1045 nil)))
1059 var ll ul)))
1076 llist)))))))))))))))
1097 nil)
1116 ;; e is of form poly(x)*exp(m*%i*x)
1117 ;; s is degree of denominator
1118 ;; adds e to bptu or bptd according to sign of m
1133 ;; check term is of form poly(x)*exp(m*%i*x)
1134 ;; n is degree of denominator
1147 term)
1155 term))
1187 ;; exponentialize sin and cos
1193 '$%i
1213 exp)
1229 ;; computes integral from minf to inf for expressions of the form
1230 ;; exp(%i*m*x)*r(x) by residues on either the upper half
1231 ;; plane or the lower half plane, depending on whether
1232 ;; m is positive or negative. [wang p. 77]
1233 ;;
1234 ;; exponentializes sin and cos before applying residue method.
1235 ;; can handle some expressions with poles on real line, such as
1236 ;; sin(x)*cos(x)/x.
1241 ratterms ratans
1242 plf bptu bptd s upans downans)
1247 var
1248 nil)
1251 ;;;Above tests for applicability of this method.
1260 ;; call to $expand can create rational terms
1261 ;; with no exp(m*%i*x)
1265 ;;;Function RIB sets up the values of BPTU and BPTD
1272 ;;;If there is BPTU then CSEMIUP must succeed.
1273 ;;;Likewise for BPTD.
1277 ;; The original integrand was already
1278 ;; determined to have no poles by initial-analysis.
1279 ;; If individual terms of the expansion have poles, the poles
1280 ;; must cancel each other out, so we can ignore them.
1283 ;; if integral of ratterms is divergent, ratans is nil,
1284 ;; and mtosc returns nil
1301 t)
1303 t)
1311 t)
1313 t)
1318 s nc dc
1319 ans r $savefactors checkfactors temp test-var)
1341 ;;;
1342 ;;;New section.
1370 findout
1374 on
1396 s)
1401 ;; Let's not signal an error here. Return nil so that we
1402 ;; eventually return a noun form if no other algorithm gives
1403 ;; a result.
1406 nil)))
1498 ;; We have an integral of the form p(x)*F(exp(x)), where
1499 ;; p(x) is a polynomial.
1501 ;; No polynomial
1505 ;; These limits must exist for the integral to converge.
1508 ;; This only handles the case when the F(z) is a
1509 ;; rational function.
1512 ;; If we get here, F(z) is not a rational function.
1513 ;; We transform it using the substitution x=log(y)
1514 ;; which gives us an integral of the form
1515 ;; p(log(y))*F(y)/y, which maxima should be able to
1516 ;; handle.
1519 ;; Give up. We don't know how to handle this.
1521 en
1535 exp)))
1537 ;;; given (b*x+a)^n+c returns (a b n c)
1545 nil)
1547 ;;;get high degree of poly
1549 ;;;diff down to linear.
1551 ;;;all the way to constant.
1554 ;;;get rid of factorial from differentiation.
1557 ;;;Sees if can be expressed as (a*x+b)^n + part freeof x.
1566 s)
1591 n
1592 d)))))
1595 on
1650 e nil)
1664 e))))))
1670 ;; Compute the integral(n/d,x,0,inf) by computing the negative of the
1671 ;; sum of residues of log(-x)*n/d over the poles of n/d inside the
1672 ;; keyhole contour. This contour is basically an disk with a slit
1673 ;; along the positive real axis. n/d must be a rational function.
1678 ;; Ok if not on the positive real axis.
1683 t)
1694 ;; Look at an expression e of the form sin(r*x)^k, where k is an
1695 ;; integer. Return the list (1 r k). (Not sure if the first element
1696 ;; of the list is always 1 because I'm not sure what partition is
1697 ;; trying to do here.)
1711 ;; Look at an expression e of the form sin(r*x) and return r.
1722 ;; integrate(a*sc(r*x)^k/x^n,x,0,inf).
1725 ;; Get the argument of the involved trig function.
1728 ;; I don't think this needs to be special.
1729 #+nil
1731 ;; Replace (1-cos(arg)^2) with sin(arg)^2.
1733 ;(m+t 1. (m- (m^t `((%cos) ,var) 2.)))
1734 ;; The code from above generates expressions with
1735 ;; a missing simp flag. Furthermore, the
1736 ;; substitution has to be done for the complete
1737 ;; argument of the trig function. (DK 02/2010)
1740 exp))
1745 ;; n is the power of the denominator.
1747 ;; The simple case.
1751 ;; Do this for a sum of such terms.
1753 c)))))))))
1755 ;; We have an integral of the form sin(r*x)^k/x^n. C is the list (1 r k).
1756 ;;
1757 ;; The substitution y=r*x converts this integral to
1758 ;;
1759 ;; r^(n-1)*integral(sin(y)^k/y^n,y,0,inf)
1760 ;;
1761 ;; (If r is negative, we need to negate the result.)
1762 ;;
1763 ;; The recursion Wang gives on p. 87 has a typo. The second term
1764 ;; should be subtracted from the first. This formula is given in G&R,
1765 ;; 3.82, formula 12.
1766 ;;
1767 ;; integrate(sin(x)^r/x^s,x) =
1768 ;; r*(r-1)/(s-1)/(s-2)*integrate(sin(x)^(r-2)/x^(s-2),x)
1769 ;; - r^2/(s-1)/(s-2)*integrate(sin(x)^r/x^(s-2),x)
1770 ;;
1771 ;; (Limits are assumed to be 0 to inf.)
1772 ;;
1773 ;; This recursion ends up with integrals with s = 1 or 2 and
1774 ;;
1775 ;; integrate(sin(x)^p/x,x,0,inf) = integrate(sin(x)^(p-1),x,0,%pi/2)
1776 ;;
1777 ;; with p > 0 and odd. This latter integral is known to maxima, and
1778 ;; it's value is beta(p/2,1/2)/2.
1779 ;;
1780 ;; integrate(sin(x)^2/x^2,x,0,inf) = %pi/2*binomial(q-3/2,q-1)
1781 ;;
1782 ;; where q >= 2.
1783 ;;
1785 ;; Compute sign(r)*r^(n-1)*integrate(sin(y)^k/y^n,y,0,inf)
1787 c
1793 recursion)
1794 ;; Recursion failed. Return the integrand
1795 ;; The following code generates expressions with a missing simp flag
1796 ;; for the sin function. Use better simplifying code. (DK 02/2010)
1797 ; (let ((integrand (div (pow `((%sin) ,(m* r var))
1798 ; k)
1799 ; (pow var n))))
1801 k)
1806 ;; integrate(sin(x)^n/x^2,x,0,inf) = pi/2*binomial(n-3/2,n-1).
1807 ;; Express in terms of Gamma functions, though.
1813 ;; integrate(sin(x)^n/x,x,0,inf) = beta((n+1)/2,1/2)/2, for n odd and
1814 ;; n > 0.
1817 half%pi)
1820 ;; This implements the recursion for computing
1821 ;; integrate(sin(y)^l/y^k,y,0,inf). (Note the change in notation from
1822 ;; the above!)
1828 ;; Integral diverges if l-(k-1) < 0.
1831 ;; If l + k is not even, return NIL. (Is this the right
1832 ;; thing to do?)
1833 nil)
1835 ;; We have integrate(sin(y)^l/y^2). Use sevn to evaluate.
1838 ;; We have integrate(sin(y)^l/y,y)
1844 ;; j = k-2, i = (k-1)*(k-2)
1845 ;;
1846 ;;
1847 ;; The main recursion:
1848 ;;
1849 ;; i(sin(y)^l/y^k)
1850 ;; = l*(l-1)/(k-1)/(k-2)*i(sin(y)^(l-2)/y^k)
1851 ;; - l^2/(k-1)/(k-1)*i(sin(y)^l/y^(k-2))
1859 ;; Returns the fractional part of a?
1863 ;; Why do we return 0 if a is a number? Perhaps we really
1864 ;; mean integer?
1867 ;; If we're here, this basically assumes a is a rational.
1868 ;; Compute the remainder and return the result.
1886 ;;; THE FOLLOWING FUNCTION IS FOR TRIG FUNCTIONS OF THE FOLLOWING TYPE:
1887 ;;; LOWER LIMIT=0 B A MULTIPLE OF %PI SCA FUNCTION OF SIN (X) COS (X)
1888 ;;; B<=%PI2
1892 ;; means there was no error
1895 ; returns cons of (integer_part . fractional_part) of a
1897 ;; I think we really want to compute how many full periods are in a
1898 ;; and the remainder.
1903 ; returns cons of (integer_part . fractional_part) of a
1905 ;; I think we really want to compute how many full periods are in a
1906 ;; and the remainder.
1913 ;; Return the integer part of r.
1915 ;; r - fpart(r)
1919 ;;;Try making exp(%i*var) --> yy, if result is rational then do integral
1920 ;;;around unit circle. Make corrections for limits of integration if possible.
1940 rat-form)))
1945 ;;; Do integrals of sin and cos. this routine makes sure lower limit
1946 ;;; is zero.
1948 ;; integrate(e,var,a,b)
1965 var e)
1966 cc))
1976 ;; Exit if b or a is not a constant or if the integrand
1977 ;; doesn't appear to have a period of 2 pi.
1980 ;; Multiples of 2*%pi in limits.
1987 ;; The integral is not over a full period (2*%pi) or multiple
1988 ;; of a full period.
1990 ;; Wang p. 111: The integral integrate(f(x),x,a,b) can be
1991 ;; written as:
1992 ;;
1993 ;; n * integrate(f,x,0,2*%pi) + integrate(f,x,0,c)
1994 ;; - integrate(f,x,0,d)
1995 ;;
1996 ;; for some integer n and d >= 0, c < 2*%pi because there exist
1997 ;; integers p and q such that a = 2 * p *%pi + d and b = 2 * q
1998 ;; * %pi + c. Then n = q - p.
2000 ;; Compute q and c for the upper limit b.
2007 ;; Compute p and d for the lower limit a.
2009 ;; avoid an extra trip around the circle - helps skip principal values
2014 ;; Compute -integrate(f,x,0,d)
2019 ;; Compute n = q - p (stored in nzp2)
2021 out
2022 ;; Compute n*integrate(f,x,0,2*%pi)
2024 ;; n = 0
2027 ;; n is not zero, so compute
2028 ;; integrate(f,x,0,2*%pi)
2030 ;; Unable to compute integrate(f,x,0,2*%pi)
2032 ;; Compute integrate(f,x,0,c)
2036 ;; All three pieces succeeded.
2039 ;; Less than 1 full period, so intsc can integrate it.
2040 ;; Apply the substitution to make the lower limit 0.
2041 ;; This is last resort because substitution often causes intsc to fail.
2043 limit-diff var))
2044 ;; nothing worked
2047 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)).
2048 ;; calls intsc with a wrapper to just return nil if integral is divergent,
2049 ;; rather than generating an error.
2054 nil
2055 ans)))
2057 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)). I (rtoy)
2058 ;; think this expects b to be less than 2*%pi.
2061 0
2068 ;; Partition the integrand SC into the factors that do not
2069 ;; contain VAR (the car part) and the parts that do (the cdr
2070 ;; part).
2075 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)).
2077 ;; Determine if sc is a product of sin's and cos's.
2081 ;; We have a product of sin's and cos's. We handle some
2082 ;; special cases here.
2084 ;; Wang p. 110, formula (1):
2085 ;; integrate(sin(x)^m*cos(x)^n, x, 0, %pi/2) =
2086 ;; gamma((m+1)/2)*gamma((n+1)/2)/2/gamma((n+m+2)/2)
2089 ;; Wang p. 110, near the bottom, says
2090 ;;
2091 ;; int(f(sin(x),cos(x)), x, 0, %pi) =
2092 ;; int(f(sin(x),cos(x)) + f(sin(x),-cos(x)),x,0,%pi/2)
2113 ;; Wang, p. 111 says
2114 ;;
2115 ;; int(f(sin(x),cos(x)),x,0,3*%pi/2) =
2116 ;; int(f(sin(x),cos(x)),x,0,%pi)
2117 ;; + int(f(-sin(x),-cos(x)),x,0,%pi/2)
2121 ;; We don't have a product of sin's and cos's.
2126 dn*)
2131 ;;;Is careful about substitution of limits where the denominator may be zero
2132 ;;;because of various assumptions made.
2151 nil
2152 ans)))
2158 nil
2159 ans)))
2161 ;; Determine whether E is of the form sin(x)^m*cos(x)^n and return the
2162 ;; list (m n).
2166 m n)
2185 ;; Says wether (m^t value exponent) has at least one real branch.
2186 ;; Only works for values of 1 and -1 now. Returns $yes $no
2187 ;; $unknown.
2198 ;; Compute beta((m+1)/2,(n+1)/2)/2.
2204 ;;Seems like Guys who call this don't agree on what it should return.
2217 ;; Check e for an expression of the form x^kk*(b*x^n+a)^l. If it
2218 ;; matches, Return the two values kk and (list l a n b).
2223 ;; We have f(x)^y. Look to see if f(x) has the desired
2224 ;; form. Then f(x)^y has the desired form too, with
2225 ;; suitably modified values.
2226 ;;
2227 ;; XXX: Should we ask for the sign of f(x) if y is not an
2228 ;; integer? This transformation we're going to do requires
2229 ;; that f(x)^y be real.
2231 e
2236 ;; Got a match. Adjust kk and cc appropriately.
2238 cc
2253 ;;(DEFUN BATAP (E)
2254 ;; (PROG (K C L)
2255 ;; (COND ((NOT (BATA0 E)) (RETURN NIL))
2256 ;; ((AND (EQUAL -1. (CADDDR C))
2257 ;; (EQ ($askSIGN (SETQ K (m+ 1. K)))
2258 ;; '$pos)
2259 ;; (EQ ($askSIGN (SETQ L (m+ 1. (CAR C))))
2260 ;; '$pos)
2261 ;; (ALIKE1 (CADR C)
2262 ;; (m^ UL (CADDR C)))
2263 ;; (SETQ E (CADR C))
2264 ;; (EQ ($askSIGN (SETQ C (CADDR C))) '$pos))
2265 ;; (RETURN (M// (m* (m^ UL (m+t K (m* C (m+t -1. L))))
2266 ;; `(($BETA) ,(SETQ NN* (M// K C))
2267 ;; ,(SETQ DN* L)))
2268 ;; C))))))
2271 ;; Integrals of the form i(log(x)^m*x^k*(a+b*x^n)^l,x,0,ul). There
2272 ;; are two cases to consider: One case has ul>0, b<0, a=-b*ul^n, k>-1,
2273 ;; l>-1, n>0, m a nonnegative integer. The second case has ul=inf, l < 0.
2274 ;;
2275 ;; These integrals are essentially partial derivatives of the Beta
2276 ;; function (i.e. the Eulerian integral of the first kind). Note
2277 ;; that, currently, with the default setting intanalysis:true, this
2278 ;; function might not even be called for some of these integrals.
2279 ;; However, this can be palliated by setting intanalysis:false.
2298 ;; We have i(x^kk/(d+cc*x^r)^s,x,0,inf) =
2299 ;; beta(aa,bb)/(cc^aa*d^bb*r). Compute this, and then
2300 ;; differentiate it m times to get the log term
2301 ;; incorporated.
2306 0
2307 m)))
2311 r))))
2320 ;; The result looks like B(k/n,l) ( ... ).
2321 ;; Perhaps, we should just $factor, instead of
2322 ;; pulling out beta like this.
2324 beta
2325 ($fullratsimp
2333 var m)
2335 beta))))))))))))
2338 ;;; If e is of the form given below, make the obvious change
2339 ;;; of variables (substituting ul*x^(1/n) for x) in order to reduce
2340 ;;; e to the usual form of the integrand in the Eulerian
2341 ;;; integral of the first kind.
2342 ;;; N. B: The old version of ZTO1 completely ignored this
2343 ;;; substitution; the log(x)s were just thrown in, which,
2344 ;;; of course would give wrong results.
2347 ;; Parse e
2351 ;; e=x^k*(a+b*x^n)^l
2353 c
2365 ;; Wang p. 71 gives the following formula for a beta function:
2366 ;;
2367 ;; integrate(x^(k-1)/(c*x^r+d)^s,x,0,inf)
2368 ;; = beta(a,b)/(c^a*d^b*r)
2369 ;;
2370 ;; where a = k/r > 0, b = s - a > 0, s > k > 0, r > 0, c*d > 0.
2371 ;;
2372 ;; This function matches this and returns k-1, d, r, c, a, b. And
2373 ;; also checks that all the conditions hold. If not, NIL is returned.
2374 ;;
2380 c
2397 ;; Handles beta integrals.
2405 nil)
2408 c
2409 ;; e = x^k*(d+c*x^al)^l.
2414 new-k)
2422 ;; Compute exp(d)*gamma((c+1)/b)/b/a^((c+1)/b). In essence, this is
2423 ;; the value of integrate(x^c*exp(d-a*x^b),x,0,inf).
2434 ;; Evaluates the contour integral of GRAND around the unit circle
2435 ;; using residues.
2441 ;; Is pt stricly inside the unit circle?
2447 ;; Is pt on the unit circle? (Use
2448 ;; the cached value computed
2449 ;; above.)
2470 ;; Wang 81-83. Unfortunately, the pdf version has page 82 as all
2471 ;; black, so here is, as best as I can tell, what Wang is doing.
2472 ;; Fortunately, p. 81 has the necessary hints.
2473 ;;
2474 ;; First consider integrate(exp(%i*k*x^n),x) around the closed contour
2475 ;; consisting of the real axis from 0 to R, the arc from the angle 0
2476 ;; to %pi/(2*n) and the ray from the arc back to the origin.
2477 ;;
2478 ;; There are no poles in this region, so the integral must be zero.
2479 ;; But consider the integral on the three parts. The real axis is the
2480 ;; integral we want. The return ray is
2481 ;;
2482 ;; exp(%i*%pi/2/n) * integrate(exp(%i*k*(t*exp(%i*%pi/2/n))^n),t,R,0)
2483 ;; = exp(%i*%pi/2/n) * integrate(exp(%i*k*t^n*exp(%i*%pi/2)),t,R,0)
2484 ;; = -exp(%i*%pi/2/n) * integrate(exp(-k*t^n),t,0,R)
2485 ;;
2486 ;; As R -> infinity, this last integral is gamma(1/n)/k^(1/n)/n.
2487 ;;
2488 ;; We assume the integral on the circular arc approaches 0 as R ->
2489 ;; infinity. (Need to prove this.)
2490 ;;
2491 ;; Thus, we have
2492 ;;
2493 ;; integrate(exp(%i*k*t^n),t,0,inf)
2494 ;; = exp(%i*%pi/2/n) * gamma(1/n)/k^(1/n)/n.
2495 ;;
2496 ;; Equating real and imaginary parts gives us the desired results:
2497 ;;
2498 ;; integrate(cos(k*t^n),t,0,inf) = G * cos(%pi/2/n)
2499 ;; integrate(sin(k*t^n),t,0,inf) = G * sin(%pi/2/n)
2500 ;;
2501 ;; where G = gamma(1/n)/k^(1/n)/n.
2502 ;;
2510 ;; Ok, we have cos(b*x^n) or sin(b*x^n), and we set e = (n
2511 ;; b)
2513 ;; n = 1. Give up. (Why not divergent?)
2514 nil)
2519 ;; s is the sign of b. Give up if it's zero.
2520 nil)
2522 ;; Give up if n-1 <= 0. (Why give up? Isn't the
2523 ;; integral divergent?)
2524 nil)
2526 ;; We can apply our formula now. g = gamma(1/n)/n/b^(1/n)
2533 e)))))))
2536 ;; this is the second part of the definite integral package
2543 ())
2545 var t)
2566 t)))))
2577 plm*))))
2596 ans))))
2604 d
2605 plm*)))))
2609 0
2612 ;; Handle integral(n(x)/d(x)*log(x)^m,x,0,inf). n and d are
2613 ;; polynomials.
2640 ans)))
2657 n
2661 ;;Makes a new poly such that np(x)-np(x+2*%i*%pi)=p(x).
2662 ;;Constructs general POLY of degree one higher than P with
2663 ;;arbitrary coeff. and then solves for coeffs by equating like powers
2664 ;;of the varibale of integration.
2665 ;;Can probably be made simpler now.
2673 ;;;Now, poly(x)-poly(x+2*%i*%pi)=p(x), P is the original poly.
2682 -1
2691 ;; Integrate(p(x)*f(exp(x))/g(exp(x)),x,minf,inf) by applying the
2692 ;; transformation y = exp(x) to get
2693 ;; integrate(p(log(y))*f(y)/g(y)/y,y,0,inf). This should be handled
2694 ;; by dintlog.
2701 ;; This implements Wang's algorithm in Chapter 5.2, pp. 98-100.
2702 ;;
2703 ;; This is a very brief description of the algorithm. Basically, we
2704 ;; have integrate(R(exp(x))*p(x),x,minf,inf), where R(x) is a rational
2705 ;; function and p(x) is a polynomial.
2706 ;;
2707 ;; We find a polynomial q(x) such that q(x) - q(x+2*%i*%pi) = p(x).
2708 ;; Then consider a contour integral of R(exp(z))*q(z) over a
2709 ;; rectangular contour. Opposite corners of the rectangle are (-R,
2710 ;; 2*%i*%pi) and (R, 0).
2711 ;;
2712 ;; Wang shows that this contour integral, in the limit, is the
2713 ;; integral of R(exp(x))*q(x)-R(exp(x))*q(x+2*%i*%pi), which is
2714 ;; exactly the integral we're looking for.
2715 ;;
2716 ;; Thus, to find the value of the contour integral, we just need the
2717 ;; residues of R(exp(z))*q(z). The only tricky part is that we want
2718 ;; the log function to have an imaginary part between 0 and 2*%pi
2719 ;; instead of -%pi to %pi.
2721 ;; We have R(exp(x))*p(x) represented as p(x)*pe(exp(x))/d(exp(x)).
2727 ;; At this point denom-exponential has converted d(exp(x)) to the
2728 ;; polynomial d(z), where z = exp(x).
2731 pe))
2734 ;; Find the poles of the denominator. denom-exponential is the
2735 ;; denominator of R(x).
2736 ;;
2737 ;; It seems as if polelist returns a list of several items.
2738 ;; The first element is a list consisting of the pole and (z -
2739 ;; pole). We don't care about this, so we take the rest of the
2740 ;; result.
2743 ;; The imaginary part is nonzero,
2744 ;; or the realpart is negative.
2748 ;; The realpart is not zero.
2750 ;; Not sure what this does.
2752 ;; No roots at all, so return
2756 ;; We have simple roots or roots in REGION1
2759 ;; The cadr of pl is the list of the simple poles of
2760 ;; denom-exponential. Take the log of them to find the
2761 ;; poles of the original expression. Then compute the
2762 ;; residues at each of these poles and sum them up and put
2763 ;; the result in B. (If no simple poles set B to 0.)
2769 ;; I think this handles the case of poles outside the
2770 ;; regions. The sum of these residues are placed in C.
2774 temp)))
2779 ;; We have the repeated poles of deonom-exponential, so we
2780 ;; need to convert them to the actual pole values for
2781 ;; R(exp(x)), by taking the log of the value of poles.
2786 ;; Compute the residues at all of these poles and sum
2787 ;; them up.
2790 poles))
2805 ;; Check to see if each term in exp that is of the form exp(k*x) has
2806 ;; an integer value for k.
2820 ;; Test to see if exp is of the form f(exp(x)), and if so, replace
2821 ;; exp(x) with 'z*. That is, basically return f(z*).
2825 exp)
2831 ;; Does e really need to be special here? Seems to be ok without
2832 ;; it; testsuite works.
2833 #+nil
2836 e)
2855 ;; Test to see if exp is of the form p(x)*f(exp(x)). If so, set p* to
2856 ;; be p(x) and set pe* to f(exp(x)).
2873 ;; I think this is looking at f(exp(x)) and tries to find some
2874 ;; rational function R and some number k such that f(exp(x)) =
2875 ;; R(exp(k*x)).
2892 ;; Integration by parts.
2893 ;;
2894 ;; integrate(u(x)*diff(v(x),x),x,a,b)
2895 ;; |b
2896 ;; = u(x)*v(x)| - integrate(v(x)*diff(u(x),x))
2897 ;; |a
2898 ;;
2900 ;;;SINCE ONLY CALLED FROM DINTLOG TO get RID OF LOGS - IF LOG REMAINS, QUIT
2904 nil)
2911 nil)
2919 ;; integrate(f(exp(k*x)),x,a,b), where f(z) is rational.
2920 ;;
2921 ;; See Wang p. 96-97.
2922 ;;
2923 ;; If the limits are minf to inf, we use the substitution y=exp(k*x)
2924 ;; to get integrate(f(y)/y,y,0,inf)/k. If the limits are 0 to inf,
2925 ;; use the substitution s+1=exp(k*x) to get
2926 ;; integrate(f(s+1)/(s+1),s,0,inf).
2933 ;; If we can integrate it directly, do so and take the
2934 ;; appropriate limits.
2935 )
2937 ;; ans is the list (f(x) exp(k*x)).
2940 ;; Use the substitution s + 1 = exp(k*x). The
2941 ;; integral becomes integrate(f(s+1)/(s+1),s,0,inf)
2944 ;; Use the substitution y=exp(k*x) because the
2945 ;; limits are minf to inf.
2947 ;; Apply the substitution and integrate it.
2950 ;; integrate(log(g(x))*f(x),x,0,inf)
2958 exp))
2960 ;; Make the substitution y=1/x. If the integrand has
2961 ;; exactly the same form, the answer has to be 0.
2967 ;; It's easy if we have the antiderivative.
2968 ;; but intsubs sometimes gives results containing %limit
2970 ;; Ok, the easy cases didn't work. We now try integration by
2971 ;; parts. Set ANS to f(x).
2974 ;; Bad. f(x) contains a log term, so we give up.
2980 var ll ul))))
2981 ;; The arg of the log function is the same as the
2982 ;; integration variable. We can do something a little
2983 ;; simpler than integration by parts. We have something
2984 ;; like f(x)*log(x). Consider f(x)*x^z. If we
2985 ;; differentiate this wrt to z, the integrand becomes
2986 ;; f(x)*log(x)*x^z. When we evaluate this at z = 0, we
2987 ;; get the desired integrand.
2988 ;;
2989 ;; So we need f(0) to be 0 at 0. If we can integrate
2990 ;; f(x)*x^z, then we differentiate the result and
2991 ;; evaluate it at z = 0.
2994 ;; Try integration by parts.
2997 ;; Compute diff(e,var,n) at the point pt.
3001 ;;; GGR and friends
3003 ;; MAYBPC returns (COEF EXPO CONST)
3004 ;;
3005 ;; This basically picks off b*x^n+a and returns the list
3006 ;; (b n a). It may also set the global *zd*.
3013 ;; At this point, e is of the form (a n b)
3018 ;; If we're here, b is complex, and n > 0. zn = imagpart(b).
3019 ;;
3020 ;; Set var to the same sign as zn.
3025 ;; zd = exp(var*%i*%pi*(1+nd)/(2*n). (ZD is special!)
3028 ;; Return zn, n, a.
3034 ;; We're here if realpart(b) <= 0, and n >= 0. Then return -b, n, a.
3037 ;; Integrate x^m*exp(b*x^n+a), with realpart(m) > -1.
3038 ;;
3039 ;; See Wang, pp. 84-85.
3040 ;;
3041 ;; I believe the formula Wang gives is incorrect. The derivation is
3042 ;; correct except for the last step.
3043 ;;
3044 ;; Let J = integrate(x^m*exp(%i*k*x^n),x,0,inf), with real k.
3045 ;;
3046 ;; Consider the case for k < 0. Take a sector of a circle bounded by
3047 ;; the real line and the angle -%pi/(2*n), and by the radii, r and R.
3048 ;; Since there are no poles inside this contour, the integral
3049 ;;
3050 ;; integrate(z^m*exp(%i*k*z^n),z) = 0
3051 ;;
3052 ;; Then J = exp(-%pi*%i*(m+1)/(2*n))*integrate(R^m*exp(k*R^n),R,0,inf)
3053 ;;
3054 ;; because the integral along the boundary is zero except for the part
3055 ;; on the real axis. (Proof?)
3056 ;;
3057 ;; Wang seems to say this last integral is gamma(s/n/(-k)^s) where s =
3058 ;; (m+1)/n. But that seems wrong. If we use the substitution R =
3059 ;; (y/(-k))^(1/n), we end up with the result:
3060 ;;
3061 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n).
3062 ;;
3063 ;; or gamma((m+1)/n)/k^((m+1)/n)/n.
3064 ;;
3065 ;; Note that this also handles the case of
3066 ;;
3067 ;; integrate(x^m*exp(-k*x^n),x,0,inf);
3068 ;;
3069 ;; where k is positive real number. A simple change of variables,
3070 ;; y=k*x^n, gives
3071 ;;
3072 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n))
3073 ;;
3074 ;; which is the same form above.
3093 ;; e = (m b n a). That is, the integral is of the form
3094 ;; x^m*exp(b*x^n+a). I think we want to compute
3095 ;; gamma((m+1)/n)/b^((m+1)/n)/n.
3096 ;;
3097 ;; FIXME: If n > m + 1, the integral converges. We need
3098 ;; to check for this.
3100 e
3103 ;; The derivation only holds if b is purely real or
3104 ;; purely imaginary. Give up if it's not.
3106 ;; Check for convergence. If b is complex, we need n -
3107 ;; m > 1. If b is real, we need b < 0.
3116 ;; If we're here, b must be negative.
3119 ;; Complex b. Take the imaginary part
3121 n a))
3122 ;; NOTE: *zd* (Ick!) is special and might be set by maybpc.
3124 ;; FIXME: Why do we set %emode here? Shouldn't we just
3125 ;; bind it? And why do we want it bound to T anyway?
3126 ;; Shouldn't the user control that? The same goes for
3127 ;; dosimp.
3128 ;;(setq $%emode t)
3134 ;; Match x^m*exp(b*x^n+a). If it does, return (list m b n a).
3139 ;; We're looking at something like exp(f(var)). See if it's
3140 ;; of the form b*x^n+a, and return (list 0 b n a). (The 0 is
3141 ;; so we can graft something onto it if needed.)
3145 ;; E should be the product of exactly 2 terms
3147 ;; Check to see if one of the terms is of the form
3148 ;; var^p. If so, make sure the realpart of p > -1. If
3149 ;; so, check the other term has the right form via
3150 ;; another call to ggr1.
3159 ;; Both terms have the right form and nn* contains the arg of
3160 ;; the exponential term. Put dn* as the car of nn*. The
3161 ;; result is something like (m b n a) when we have the
3162 ;; expression x^m*exp(b*x^n+a).
3166 ;; Match b*x^n+a. If a match is found, return the list (a n b).
3167 ;; Otherwise, return NIL
3180 ;; Match b*x^n. Return the list (n b) if found or NIL if not.
3200 ;;; given (b*x^n+a)^m returns (m a n b)
3213 ;;;Catches Unfactored forms.
3230 ;;;Is E = VAR raised to some power? If so return power or 0.
3236 ;;;Is E = VAR1 raised to some power? If so return power.
3257 ans)))))
3334 ;;;Looks at the IMAGINARY part of a log and puts it in the interval 0 2*%pi.
3337 ;; We take the $rectform above to make sure that the log is
3338 ;; expanded out for the situations where simplifying plog itself
3339 ;; doesn't do it. This should probably be considered a bug in the
3340 ;; plog simplifier and should be fixed there.
3353 ;;; Temporary fix for a lacking in taylor, which loses with %i in denom.
3354 ;;; Besides doesn't seem like a bad thing to do in general.
3358 ;; Multiply the denominator by it's conjugate to get rid of
3359 ;; %i.
3364 ;; If the new denominator still contains %i, just give up.
3366 exp
3367 new-exp)))
3370 ;;; LL and UL must be real otherwise this routine return $UNKNOWN.
3371 ;;; Returns $no $unknown or a list of poles in the interval (ll ul)
3372 ;;; for exp w.r.t. var.
3373 ;;; Form of list ((pole . multiplicity) (pole1 . multiplicity) ....)
3390 ;; this clause handles cases where we can't find the exact roots,
3391 ;; but we know that they occur outside the interval of integration.
3392 ;; example: integrate ((1+exp(t))/sqrt(t+exp(t)), t, 0, 1);
3408 ;; (multiplicity (cdar dummy)) (not used? -- cwh)
3418 pole-list))))))))))))
3421 ;;;Returns $YES if there is no pole and $NO if there is one.
3434 ;;;Takes care of forms that the ratio test fails on.
3450 (%einvolve
3485 ;; real values for ll and ul; place can be imaginary.
3492 '$no
3497 ;; returns true or nil
3499 ;; real values for ll and ul; place can be imaginary.
3515 ;; If *failures is set, we may have missed some roots.
3516 ;; We still return the roots that we have found.
3521 rootlist)
3528 rootlist)))))))))
3531 ;;; Is x > y. X or Y can be $MINF or $INF, zeroA or zeroB.
3571 nil)
3574 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3575 ;;;
3576 ;;; Integrate Definite Integrals involving log and exp functions. The algorithm
3577 ;;; are taken from the paper "Evaluation of CLasses of Definite Integrals ..."
3578 ;;; by K.O.Geddes et. al.
3579 ;;;
3580 ;;; 1. CASE: Integrals generated by the Gamma function.
3581 ;;;
3582 ;;; inf
3583 ;;; /
3584 ;;; [ w m s - m - 1
3585 ;;; I t log (t) expt(- t ) dt = s signum(s)
3586 ;;; ]
3587 ;;; /
3588 ;;; 0
3589 ;;; !
3590 ;;; m !
3591 ;;; d !
3592 ;;; (--- (gamma(z))! )
3593 ;;; m !
3594 ;;; dz ! w + 1
3595 ;;; !z = -----
3596 ;;; s
3597 ;;;
3598 ;;; The integral converges for:
3599 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3600 ;;;
3601 ;;; 2. CASE: Integrals generated by the Incomplete Gamma function.
3602 ;;;
3603 ;;; inf !
3604 ;;; / m !
3605 ;;; [ w m s d s !
3606 ;;; I t log (t) exp(- t ) dt = (--- (gamma_incomplete(a, x ))! )
3607 ;;; ] m !
3608 ;;; / da ! w + 1
3609 ;;; x !z = -----
3610 ;;; s
3611 ;;; - m - 1
3612 ;;; s signum(s)
3613 ;;;
3614 ;;; The integral converges for:
3615 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3616 ;;; The shown solution is valid for s>0. For s<0 gamma_incomplete has to be
3617 ;;; replaced by gamma(a) - gamma_incomplete(a,x^s).
3618 ;;;
3619 ;;; 3. CASE: Integrals generated by the beta function.
3620 ;;;
3621 ;;; 1
3622 ;;; /
3623 ;;; [ m s r n
3624 ;;; I log (1 - t) (1 - t) t log (t) dt =
3625 ;;; ]
3626 ;;; /
3627 ;;; 0
3628 ;;; !
3629 ;;; ! !
3630 ;;; n m ! !
3631 ;;; d d ! !
3632 ;;; --- (--- (beta(z, w))! )!
3633 ;;; n m ! !
3634 ;;; dz dw ! !
3635 ;;; !w = s + 1 !
3636 ;;; !z = r + 1
3637 ;;;
3638 ;;; The integral converges for:
3639 ;;; n, m = 0, 1, 2, ..., s > -1 and r > -1.
3640 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3644 ;;; Recognize c*z^w*log(z)^m*exp(-t^s)
3656 ;;; Recognize c*z^r*log(z)^n*(1-z)^s*log(1-z)^m
3669 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3676 ;; var1 is used as a parameter for differentiation. Add var1>0 to the
3677 ;; database, to get the desired simplification of the differentiation of
3678 ;; the gamma_incomplete function.
3686 ;; The integrand matches the cases 1 and 2.
3705 ;; Case 1: Generated by the Gamma function.
3712 var1
3717 ; derivate the involved hypergeometric
3718 ; functions.
3723 ;; Case 2: Generated by the Incomplete Gamma function.
3739 ;; Case 3: Generated by the Beta function.
3765 var2 m)
3767 var1 n)
3773 ;; Simplify result and set $gamma_expand to global value
3776 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;