| blob | 02edb075571810c3f6e1c4947ea2eb6124c05938 |
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 ;;
26 ;; nointegrate is a macsyma level flag which inhibits indefinite
27 ;;integration.
28 ;; abconv is a macsyma level flag which inhibits the absolute
29 ;;convergence test.
30 ;;
31 ;; $defint is the top level function that takes the user input
32 ;;and does minor changes to make the integrand ready for the package.
33 ;;
34 ;; next comes defint, which is the function that does the
35 ;;integration. it is often called recursively from the bowels of the
36 ;;package. defint does some of the easy cases and dispatches to:
37 ;;
38 ;; dintegrate. this program first sees if the limits of
39 ;;integration are 0,inf or minf,inf. if so it sends the problem to
40 ;;ztoinf or mtoinf, respectivly.
41 ;; else, dintegrate tries:
42 ;;
43 ;; intsc1 - does integrals of sin's or cos's or exp(%i var)'s
44 ;; when the interval is 0,2 %pi or 0,%pi.
45 ;; method is conversion to rational function and find
46 ;; residues in the unit circle. [wang, pp 107-109]
47 ;;
48 ;; ratfnt - does rational functions over finite interval by
49 ;; doing polynomial part directly, and converting
50 ;; the rational part to an integral on 0,inf and finding
51 ;; the answer by residues.
52 ;;
53 ;; zto1 - i(x^(k-1)*(1-x)^(l-1),x,0,1) = beta(k,l) or
54 ;; i(log(x)*x^(x-1)*(1-x)^(l-1),x,0,1) = psi...
55 ;; [wang, pp 116,117]
56 ;;
57 ;; dintrad- i(x^m/(a*x^2+b*x+c)^(n+3/2),x,0,inf) [wang, p 74]
58 ;;
59 ;; dintlog- i(log(g(x))*f(x),x,0,inf) = 0 (by symmetry) or
60 ;; tries an integration by parts. (only routine to
61 ;; try integration by parts) [wang, pp 93-95]
62 ;;
63 ;; dintexp- i(f(exp(k*x)),x,a,inf) = i(f(x+1)/(x+1),x,0,inf)
64 ;; or i(f(x)/x,x,0,inf)/k. First case hold for a=0;
65 ;; the second for a=minf. [wang 96-97]
66 ;;
67 ;;dintegrate also tries indefinite integration based on certain
68 ;;predicates (such as abconv) and tries breaking up the integrand
69 ;;over a sum or tries a change of variable.
70 ;;
71 ;; ztoinf is the routine for doing integrals over the range 0,inf.
72 ;; it goes over a series of routines and sees if any will work:
73 ;;
74 ;; scaxn - sc(b*x^n) (sc stands for sin or cos) [wang, pp 81-83]
75 ;;
76 ;; ssp - a*sc^n(r*x)/x^m [wang, pp 86,87]
77 ;;
78 ;; zmtorat- rational function. done by multiplication by plog(-x)
79 ;; and finding the residues over the keyhole contour
80 ;; [wang, pp 59-61]
81 ;;
82 ;; log*rat- r(x)*log^n(x) [wang, pp 89-92]
83 ;;
84 ;; logquad0 log(x)/(a*x^2+b*x+c) uses formula
85 ;; i(log(x)/(x^2+2*x*a*cos(t)+a^2),x,0,inf) =
86 ;; t*log(a)/sin(t). a better formula might be
87 ;; i(log(x)/(x+b)/(x+c),x,0,inf) =
88 ;; (log^2(b)-log^2(c))/(2*(b-c))
89 ;;
90 ;; batapp - x^(p-1)/(b*x^n+a)^m uses formula related to the beta
91 ;; function [wang, p 71]
92 ;; there is also a special case when m=1 and a*b<0
93 ;; see [wang, p 65]
94 ;;
95 ;; sinnu - x^-a*n(x)/d(x) [wang, pp 69-70]
96 ;;
97 ;; ggr - x^r*exp(a*x^n+b)
98 ;;
99 ;; dintexp- see dintegrate
100 ;;
101 ;; ztoinf also tries 1/2*mtoinf if the integrand is an even function
102 ;;
103 ;; mtoinf is the routine for doing integrals on minf,inf.
104 ;; it too tries a series of routines and sees if any succeed.
105 ;;
106 ;; scaxn - when the integrand is an even function, see ztoinf
107 ;;
108 ;; mtosc - exp(%i*m*x)*r(x) by residues on either the upper half
109 ;; plane or the lower half plane, depending on whether
110 ;; m is positive or negative.
111 ;;
112 ;; zmtorat- does rational function by finding residues in upper
113 ;; half plane
114 ;;
115 ;; dintexp- see dintegrate
116 ;;
117 ;; rectzto%pi2 - poly(x)*rat(exp(x)) by finding residues in
118 ;; rectangle [wang, pp98-100]
119 ;;
120 ;; ggrm - x^r*exp((x+a)^n+b)
121 ;;
122 ;; mtoinf also tries 2*ztoinf if the integrand is an even function.
128 *nodiverg exp1
133 $trigexpandplus $trigexpandtimes
139 ;;;rsn* is in comdenom. does a ratsimp of numerator.
140 ;expvar
142 ;impvar
144 $logabs $tlimswitch $maxposex $maxnegex
145 $trigsign $savefactors $radexpand $breakup $%emode
146 $float $exptsubst dosimp context rp-polylogp
147 %p%i half%pi %pi2 half%pi3 varlist genvar
148 $domain $m1pbranch errorsw
149 limitp $algebraic
150 ;;LIMITP T Causes $ASKSIGN to do special things
151 ;;For DEFINT like eliminate epsilon look for prin-inf
152 ;;take realpart and imagpart.
153 integer-info
154 ;;If LIMITP is non-null ask-integer conses
155 ;;its assumptions onto this list.
156 generate-atan2))
157 ;If this switch is () then RPART returns ATAN's
158 ;instead of ATAN2's
162 ;;These are really defined in LIMIT but DEFINT uses them also.
169 "When @code{true}, definite integration tries to find poles in the integrand
180 ;; Not really sure what this is meant to do, but it's used by MTORAT,
181 ;; KEYHOLE, and POLELIST.
186 ;; Distribute $defint over equations, lists, and matrices.
218 (merror (intl:gettext "defint: variable of integration must be a simple or subscripted variable.~%defint: found ~M") var))
247 ;; If antideriv can't do it, returns nil
248 ;; use limit to evaluate every answer returned by antideriv.
251 ;;;Hack the expression up for exponentials.
254 ;; Is this expr a candidate for SININT ?
263 t)))
264 ;;ERF type things go here.
269 t))))))
289 ;; We have some subscripted variable that's not our variable
290 ;; (I think), so it's deg-lessp.
291 ;;
292 ;; FIXME: Is this the best way to handle this? Are there
293 ;; other cases we're mising here?
294 t)))
304 ;; This routine tries to take a limit a couple of ways.
310 ans
313 var
314 0
324 ;; test whether fun2 is inverse of fun1 at val
332 ;; integration change of variable
340 ())
342 ;; This is a hack! If nv is of the form b*x^n+a, we can
343 ;; solve the equation manually instead of using solve.
344 ;; Why? Because solve asks us for the sign of yx and
345 ;; that's bogus.
347 ;; Solve yx = b*x^n+a, for x. Any root will do. So we
348 ;; have x = ((yx-a)/b)^(1/n).
350 d
356 ))))
373 ;; d: original variable (var) as a function of 'yx
374 ;; ind: boolean flag
375 ;; nv: new variable ('yx) as a function of original variable (var)
384 ;; converts limits of integration to values for new variable 'yx
394 ;; wrapper around limit, returns nil if
395 ;; limit not found (nounform returned), or undefined ($und or $ind)
402 ans))))
404 ;; rewrites exp, the integrand in terms of var,
405 ;; into exp1, the integrand in terms of 'yx.
426 ;; D (not used? -- cwh)
439 ;; Look for integrals which involve log and exp functions.
440 ;; Maxima has a special algorithm to get general results.
451 %tan %sinh %cosh %tanh
452 %log %asin %acos %atan
453 %cot %acot %sec
454 %asec %csc %acsc
510 ;;Should be a PROG inside of unwind-protect, but Multics has a compiler
511 ;;bug wrt. and I want to test this code now.
519 ;;;This seems((and (and (eq ul '$inf)
520 ;;;fairly losing (setq exp (subin (m+ ll var) exp))
521 ;;; (setq ll 0.))
522 ;;; (ztoinf exp var)))
527 ;; ((and (and (equal ul 1.)
528 ;; (equal ll 0.)) (zto1 exp)))
536 ;;;NOT COMPLETE for sin's and cos's.
563 ;;;Recursion stopper
586 ;; adds up integrals of ranges between each pair of poles.
587 ;; checks if whole thing is divergent as limits of integration approach poles.
589 ;;; calling $logcontract causes antiderivative of 1/(1-x^5) to blow up
590 ;; (setq anti-deriv (cond ((involve anti-deriv '(%log))
591 ;; ($logcontract anti-deriv))
592 ;; (t anti-deriv)))
603 ;;Return section.
628 pole-list)
630 ;; Assumes EXP is a rational expression with no polynomial part and
631 ;; converts the finite integration to integration over a half-infinite
632 ;; interval. The substitution y = (x-a)/(b-x) is used. Equivalently,
633 ;; x = (b*y+a)/(y+1).
634 ;;
635 ;; (I'm guessing CV means Change Variable here.)
638 ;; FIXME! This is a hack. We apply the transformation with
639 ;; symbolic limits and then substitute the actual limits later.
640 ;; That way method-by-limits (usually?) sees a simpler
641 ;; integrand.
642 ;;
643 ;; See Bugs 938235 and 941457. These fail because $FACTOR is
644 ;; unable to factor the transformed result. This needs more
645 ;; work (in other places).
649 ;; If the limit is a number, use $substitute so we simplify
650 ;; the result. Do we really want to do this?
658 ()))
660 ;; Integrate rational functions over a finite interval by doing the
661 ;; polynomial part directly, and converting the rational part to an
662 ;; integral from 0 to inf. This is evaluated via residues.
665 ;; PQR divides the rational expression and returns the quotient
666 ;; and remainder
673 ;; No polynomial part
676 ans
679 ;; Only polynomial part
682 ;; A non-zero quotient and remainder. Combine the results
683 ;; together.
687 ans))
692 ;; I think this takes a rational expression E, and finds the
693 ;; polynomial part. A cons is returned. The car is the quotient and
694 ;; the cdr is the remainder.
719 ;;;If leadop isn't plus don't do anything.
729 t)
735 t)
790 ;; *Z* is a "zero parameter" for this package.
791 ;; Its also used to transform.
792 ;; limit(exp,var,val,dir) -- limit(exp,tvar,0,dir)
799 ;; EPSILON is used in principal value code to denote the familiar
800 ;; mathematical entity.
805 ;;; PRIN-INF Is a special symbol in the principal value code used to
806 ;;; denote an end-point which is proceeding to infinity.
837 ;;Fix limits so that ll < ul.
858 ;; Substitute a and b into integral e
859 ;;
860 ;; Looks for discontinuties in integral, and works around them.
861 ;; For example, in
862 ;;
863 ;; integrate(x^(2*n)*exp(-(x)^2),x) ==>
864 ;; -gamma_incomplete((2*n+1)/2,x^2)*x^(2*n+1)*abs(x)^(-2*n-1)/2
865 ;;
866 ;; the integral has a discontinuity at x=0.
867 ;;
874 var a b)))))
888 total))))))
890 ;; look for terms with a negative exponent
891 ;;
892 ;; recursively traverses exp in order to find discontinuities such as
893 ;; erfc(1/x-x) at x=0
903 ;; returns list of places where exp might be discontinuous in var.
904 ;; list begins with ll and ends with ul, and include any values between
905 ;; ll and ul.
906 ;; return '$no or '$unknown if no discontinuities found.
923 ;; (multiplicity (cdar dummy)) ;; not used
928 ;; Carefully substitute the integration limits A and B into the
929 ;; expression E.
939 ;; Try easy substitutions. Return NIL if we can't.
942 ;; Infinite limits aren't easy
943 nil)
947 %gamma_incomplete %expintegral_ei)))
949 ;; It's easy if we have a polynomial. I (rtoy) think
950 ;; it's also easy if the denominator is free of the
951 ;; integration variable and also if the expression
952 ;; doesn't involve inverse functions.
953 ;;
954 ;; gamma_incomplete and expintegral_ie
955 ;; included because of discontinuity in
956 ;; gamma_incomplete(0, exp(%i*x)) and
957 ;; expintegral_ei(exp(%i*x))
958 ;;
959 ;; XXX: Are there other cases we've forgotten about?
960 ;;
961 ;; So just try to substitute the limits into the
962 ;; expression. If no errors are produced, we're done.
967 ;; no-err-sub has returned T. An error was catched.
968 nil)
990 ;;;This function works only on things with ATAN's in them now.
992 ;; POLES-IN-INTERVAL doesn't know about the poles of tan(x). Call
993 ;; trigsimp to convert tan into sin/cos, which POLES-IN-INTERVAL
994 ;; knows how to handle.
995 ;;
996 ;; XXX Should we fix POLES-IN-INTERVAL instead?
997 ;;
998 ;; XXX Is calling trigsimp too much? Should we just only try to
999 ;; substitute sin/cos for tan?
1000 ;;
1001 ;; XXX Should the result try to convert sin/cos back into tan? (A
1002 ;; call to trigreduce would do it, among other things.)
1005 ;;POLES -> ((mlist) ((mequal) ((%atan) foo) replacement) ......)
1006 ;;We can then use $SUBSTITUTE
1011 ul-ans
1015 nil)))
1029 var ll ul)))
1046 llist)))))))))))))))
1067 nil)
1086 ;; e is of form poly(x)*exp(m*%i*x)
1087 ;; s is degree of denominator
1088 ;; adds e to bptu or bptd according to sign of m
1103 ;; check term is of form poly(x)*exp(m*%i*x)
1104 ;; n is degree of denominator
1117 term)
1124 term))
1156 ;; exponentialize sin and cos
1162 '$%i
1182 exp)
1198 ;; computes integral from minf to inf for expressions of the form
1199 ;; exp(%i*m*x)*r(x) by residues on either the upper half
1200 ;; plane or the lower half plane, depending on whether
1201 ;; m is positive or negative. [wang p. 77]
1202 ;;
1203 ;; exponentializes sin and cos before applying residue method.
1204 ;; can handle some expressions with poles on real line, such as
1205 ;; sin(x)*cos(x)/x.
1210 ratterms ratans
1211 plf bptu bptd s upans downans)
1216 var
1217 nil)
1220 ;;;Above tests for applicability of this method.
1229 ;; call to $expand can create rational terms
1230 ;; with no exp(m*%i*x)
1234 ;;;Function RIB sets up the values of BPTU and BPTD
1241 ;;;If there is BPTU then CSEMIUP must succeed.
1242 ;;;Likewise for BPTD.
1246 ;; The original integrand was already
1247 ;; determined to have no poles by initial-analysis.
1248 ;; If individual terms of the expansion have poles, the poles
1249 ;; must cancel each other out, so we can ignore them.
1252 ;; if integral of ratterms is divergent, ratans is nil,
1253 ;; and mtosc returns nil
1270 t)
1272 t)
1280 t)
1282 t)
1287 s nc dc
1288 ans r $savefactors checkfactors temp test-var)
1310 ;;;
1311 ;;;New section.
1339 findout
1343 on
1365 s)
1370 ;; Let's not signal an error here. Return nil so that we
1371 ;; eventually return a noun form if no other algorithm gives
1372 ;; a result.
1375 nil)))
1461 s
1468 ;; We have an integral of the form p(x)*F(exp(x)), where
1469 ;; p(x) is a polynomial.
1471 ;; No polynomial
1475 ;; These limits must exist for the integral to converge.
1478 ;; This only handles the case when the F(z) is a
1479 ;; rational function.
1482 ;; If we get here, F(z) is not a rational function.
1483 ;; We transform it using the substitution x=log(y)
1484 ;; which gives us an integral of the form
1485 ;; p(log(y))*F(y)/y, which maxima should be able to
1486 ;; handle.
1489 ;; Give up. We don't know how to handle this.
1491 en
1505 exp)))
1507 ;;; given (b*x+a)^n+c returns (a b n c)
1515 nil)
1517 ;;;get high degree of poly
1519 ;;;diff down to linear.
1521 ;;;all the way to constant.
1524 ;;;get rid of factorial from differentiation.
1527 ;;;Sees if can be expressed as (a*x+b)^n + part freeof x.
1536 s)
1560 n
1561 d)))))
1564 on
1619 e nil)
1633 e))))))
1639 ;; Compute the integral(n/d,x,0,inf) by computing the negative of the
1640 ;; sum of residues of log(-x)*n/d over the poles of n/d inside the
1641 ;; keyhole contour. This contour is basically an disk with a slit
1642 ;; along the positive real axis. n/d must be a rational function.
1647 ;; Ok if not on the positive real axis.
1652 t)
1663 ;; Look at an expression e of the form sin(r*x)^k, where k is an
1664 ;; integer. Return the list (1 r k). (Not sure if the first element
1665 ;; of the list is always 1 because I'm not sure what partition is
1666 ;; trying to do here.)
1680 ;; Look at an expression e of the form sin(r*x) and return r.
1691 ;; integrate(a*sc(r*x)^k/x^n,x,0,inf).
1694 ;; Get the argument of the involved trig function.
1697 ;; I don't think this needs to be special.
1698 #+nil
1700 ;; Replace (1-cos(arg)^2) with sin(arg)^2.
1702 ;(m+t 1. (m- (m^t `((%cos) ,var) 2.)))
1703 ;; The code from above generates expressions with
1704 ;; a missing simp flag. Furthermore, the
1705 ;; substitution has to be done for the complete
1706 ;; argument of the trig function. (DK 02/2010)
1709 exp))
1714 ;; n is the power of the denominator.
1716 ;; The simple case.
1720 ;; Do this for a sum of such terms.
1722 c)))))))))
1724 ;; We have an integral of the form sin(r*x)^k/x^n. C is the list (1 r k).
1725 ;;
1726 ;; The substitution y=r*x converts this integral to
1727 ;;
1728 ;; r^(n-1)*integral(sin(y)^k/y^n,y,0,inf)
1729 ;;
1730 ;; (If r is negative, we need to negate the result.)
1731 ;;
1732 ;; The recursion Wang gives on p. 87 has a typo. The second term
1733 ;; should be subtracted from the first. This formula is given in G&R,
1734 ;; 3.82, formula 12.
1735 ;;
1736 ;; integrate(sin(x)^r/x^s,x) =
1737 ;; r*(r-1)/(s-1)/(s-2)*integrate(sin(x)^(r-2)/x^(s-2),x)
1738 ;; - r^2/(s-1)/(s-2)*integrate(sin(x)^r/x^(s-2),x)
1739 ;;
1740 ;; (Limits are assumed to be 0 to inf.)
1741 ;;
1742 ;; This recursion ends up with integrals with s = 1 or 2 and
1743 ;;
1744 ;; integrate(sin(x)^p/x,x,0,inf) = integrate(sin(x)^(p-1),x,0,%pi/2)
1745 ;;
1746 ;; with p > 0 and odd. This latter integral is known to maxima, and
1747 ;; it's value is beta(p/2,1/2)/2.
1748 ;;
1749 ;; integrate(sin(x)^2/x^2,x,0,inf) = %pi/2*binomial(q-3/2,q-1)
1750 ;;
1751 ;; where q >= 2.
1752 ;;
1754 ;; Compute sign(r)*r^(n-1)*integrate(sin(y)^k/y^n,y,0,inf)
1756 c
1762 recursion)
1763 ;; Recursion failed. Return the integrand
1764 ;; The following code generates expressions with a missing simp flag
1765 ;; for the sin function. Use better simplifying code. (DK 02/2010)
1766 ; (let ((integrand (div (pow `((%sin) ,(m* r var))
1767 ; k)
1768 ; (pow var n))))
1770 k)
1775 ;; integrate(sin(x)^n/x^2,x,0,inf) = pi/2*binomial(n-3/2,n-1).
1776 ;; Express in terms of Gamma functions, though.
1782 ;; integrate(sin(x)^n/x,x,0,inf) = beta((n+1)/2,1/2)/2, for n odd and
1783 ;; n > 0.
1786 half%pi)
1789 ;; This implements the recursion for computing
1790 ;; integrate(sin(y)^l/y^k,y,0,inf). (Note the change in notation from
1791 ;; the above!)
1797 ;; Integral diverges if l-(k-1) < 0.
1800 ;; If l + k is not even, return NIL. (Is this the right
1801 ;; thing to do?)
1802 nil)
1804 ;; We have integrate(sin(y)^l/y^2). Use sevn to evaluate.
1807 ;; We have integrate(sin(y)^l/y,y)
1813 ;; j = k-2, i = (k-1)*(k-2)
1814 ;;
1815 ;;
1816 ;; The main recursion:
1817 ;;
1818 ;; i(sin(y)^l/y^k)
1819 ;; = l*(l-1)/(k-1)/(k-2)*i(sin(y)^(l-2)/y^k)
1820 ;; - l^2/(k-1)/(k-1)*i(sin(y)^l/y^(k-2))
1828 ;; Returns the fractional part of a?
1832 ;; Why do we return 0 if a is a number? Perhaps we really
1833 ;; mean integer?
1836 ;; If we're here, this basically assumes a is a rational.
1837 ;; Compute the remainder and return the result.
1855 ;;; THE FOLLOWING FUNCTION IS FOR TRIG FUNCTIONS OF THE FOLLOWING TYPE:
1856 ;;; LOWER LIMIT=0 B A MULTIPLE OF %PI SCA FUNCTION OF SIN (X) COS (X)
1857 ;;; B<=%PI2
1861 ;; means there was no error
1864 ; returns cons of (integer_part . fractional_part) of a
1866 ;; I think we really want to compute how many full periods are in a
1867 ;; and the remainder.
1873 ;; Return the integer part of r.
1875 ;; r - fpart(r)
1879 ;;;Try making exp(%i*var) --> yy, if result is rational then do integral
1880 ;;;around unit circle. Make corrections for limits of integration if possible.
1900 rat-form)))
1905 ;;; Do integrals of sin and cos. this routine makes sure lower limit
1906 ;;; is zero.
1908 ;; integrate(e,var,a,b)
1924 var e)
1925 cc))
1933 ;; Exit if b or a is not a constant or if the integrand
1934 ;; doesn't appear to have a period of 2 pi.
1937 ;; Multiples of 2*%pi in limits.
1944 ;; The integral is not over a full period (2*%pi) or multiple
1945 ;; of a full period.
1947 ;; Wang p. 111: The integral integrate(f(x),x,a,b) can be
1948 ;; written as:
1949 ;;
1950 ;; n * integrate(f,x,0,2*%pi) + integrate(f,x,0,c)
1951 ;; - integrate(f,x,0,d)
1952 ;;
1953 ;; for some integer n and d >= 0, c < 2*%pi because there exist
1954 ;; integers p and q such that a = 2 * p *%pi + d and b = 2 * q
1955 ;; * %pi + c. Then n = q - p.
1957 ;; Compute q and c for the upper limit b.
1964 ;; Compute p and d for the lower limit a.
1966 ;; Compute -integrate(f,x,0,d)
1971 ;; Compute n = q - p (stored in nzp2)
1973 out
1974 ;; Compute n*integrate(f,x,0,2*%pi)
1976 ;; n = 0
1979 ;; n is not zero, so compute
1980 ;; integrate(f,x,0,2*%pi)
1982 ;; Unable to compute integrate(f,x,0,2*%pi)
1984 ;; Compute integrate(f,x,0,c)
1988 ;; All three pieces succeeded.
1991 ;; Less than 1 full period, so intsc can integrate it.
1992 ;; Apply the substitution to make the lower limit 0.
1993 ;; This is last resort because substitution often causes intsc to fail.
1995 limit-diff var))
1996 ;; nothing worked
1999 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)).
2000 ;; calls intsc with a wrapper to just return nil if integral is divergent,
2001 ;; rather than generating an error.
2006 nil
2007 ans)))
2009 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)). I (rtoy)
2010 ;; think this expects b to be less than 2*%pi.
2013 0
2020 ;; Partition the integrand SC into the factors that do not
2021 ;; contain VAR (the car part) and the parts that do (the cdr
2022 ;; part).
2027 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)).
2029 ;; Determine if sc is a product of sin's and cos's.
2033 ;; We have a product of sin's and cos's. We handle some
2034 ;; special cases here.
2036 ;; Wang p. 110, formula (1):
2037 ;; integrate(sin(x)^m*cos(x)^n, x, 0, %pi/2) =
2038 ;; gamma((m+1)/2)*gamma((n+1)/2)/2/gamma((n+m+2)/2)
2041 ;; Wang p. 110, near the bottom, says
2042 ;;
2043 ;; int(f(sin(x),cos(x)), x, 0, %pi) =
2044 ;; int(f(sin(x),cos(x)) + f(sin(x),-cos(x)),x,0,%pi/2)
2065 ;; Wang, p. 111 says
2066 ;;
2067 ;; int(f(sin(x),cos(x)),x,0,3*%pi/2) =
2068 ;; int(f(sin(x),cos(x)),x,0,%pi)
2069 ;; + int(f(-sin(x),-cos(x)),x,0,%pi/2)
2073 ;; We don't have a product of sin's and cos's.
2078 dn*)
2083 ;;;Is careful about substitution of limits where the denominator may be zero
2084 ;;;because of various assumptions made.
2103 nil
2104 ans)))
2110 nil
2111 ans)))
2113 ;; Determine whether E is of the form sin(x)^m*cos(x)^n and return the
2114 ;; list (m n).
2118 m n)
2137 ;; Says wether (m^t value exponent) has at least one real branch.
2138 ;; Only works for values of 1 and -1 now. Returns $yes $no
2139 ;; $unknown.
2150 ;; Compute beta((m+1)/2,(n+1)/2)/2.
2156 ;;Seems like Guys who call this don't agree on what it should return.
2169 ;; Check e for an expression of the form x^kk*(b*x^n+a)^l. If it
2170 ;; matches, Return the two values kk and (list l a n b).
2175 ;; We have f(x)^y. Look to see if f(x) has the desired
2176 ;; form. Then f(x)^y has the desired form too, with
2177 ;; suitably modified values.
2178 ;;
2179 ;; XXX: Should we ask for the sign of f(x) if y is not an
2180 ;; integer? This transformation we're going to do requires
2181 ;; that f(x)^y be real.
2183 e
2188 ;; Got a match. Adjust kk and cc appropriately.
2190 cc
2205 ;;(DEFUN BATAP (E)
2206 ;; (PROG (K C L)
2207 ;; (COND ((NOT (BATA0 E)) (RETURN NIL))
2208 ;; ((AND (EQUAL -1. (CADDDR C))
2209 ;; (EQ ($askSIGN (SETQ K (m+ 1. K)))
2210 ;; '$pos)
2211 ;; (EQ ($askSIGN (SETQ L (m+ 1. (CAR C))))
2212 ;; '$pos)
2213 ;; (ALIKE1 (CADR C)
2214 ;; (m^ UL (CADDR C)))
2215 ;; (SETQ E (CADR C))
2216 ;; (EQ ($askSIGN (SETQ C (CADDR C))) '$pos))
2217 ;; (RETURN (M// (m* (m^ UL (m+t K (m* C (m+t -1. L))))
2218 ;; `(($BETA) ,(SETQ NN* (M// K C))
2219 ;; ,(SETQ DN* L)))
2220 ;; C))))))
2223 ;; Integrals of the form i(log(x)^m*x^k*(a+b*x^n)^l,x,0,ul). There
2224 ;; are two cases to consider: One case has ul>0, b<0, a=-b*ul^n, k>-1,
2225 ;; l>-1, n>0, m a nonnegative integer. The second case has ul=inf, l < 0.
2226 ;;
2227 ;; These integrals are essentially partial derivatives of the Beta
2228 ;; function (i.e. the Eulerian integral of the first kind). Note
2229 ;; that, currently, with the default setting intanalysis:true, this
2230 ;; function might not even be called for some of these integrals.
2231 ;; However, this can be palliated by setting intanalysis:false.
2250 ;; We have i(x^kk/(d+cc*x^r)^s,x,0,inf) =
2251 ;; beta(aa,bb)/(cc^aa*d^bb*r). Compute this, and then
2252 ;; differentiate it m times to get the log term
2253 ;; incorporated.
2258 0
2259 m)))
2263 r))))
2272 ;; The result looks like B(k/n,l) ( ... ).
2273 ;; Perhaps, we should just $factor, instead of
2274 ;; pulling out beta like this.
2276 beta
2277 ($fullratsimp
2285 var m)
2287 beta))))))))))))
2290 ;;; If e is of the form given below, make the obvious change
2291 ;;; of variables (substituting ul*x^(1/n) for x) in order to reduce
2292 ;;; e to the usual form of the integrand in the Eulerian
2293 ;;; integral of the first kind.
2294 ;;; N. B: The old version of ZTO1 completely ignored this
2295 ;;; substitution; the log(x)s were just thrown in, which,
2296 ;;; of course would give wrong results.
2299 ;; Parse e
2303 ;; e=x^k*(a+b*x^n)^l
2305 c
2317 ;; Wang p. 71 gives the following formula for a beta function:
2318 ;;
2319 ;; integrate(x^(k-1)/(c*x^r+d)^s,x,0,inf)
2320 ;; = beta(a,b)/(c^a*d^b*r)
2321 ;;
2322 ;; where a = k/r > 0, b = s - a > 0, s > k > 0, r > 0, c*d > 0.
2323 ;;
2324 ;; This function matches this and returns k-1, d, r, c, a, b. And
2325 ;; also checks that all the conditions hold. If not, NIL is returned.
2326 ;;
2332 c
2349 ;; Handles beta integrals.
2357 nil)
2360 c
2361 ;; e = x^k*(d+c*x^al)^l.
2366 new-k)
2374 ;; Compute exp(d)*gamma((c+1)/b)/b/a^((c+1)/b). In essence, this is
2375 ;; the value of integrate(x^c*exp(d-a*x^b),x,0,inf).
2386 ;; Evaluates the contour integral of GRAND around the unit circle
2387 ;; using residues.
2393 ;; Is pt stricly inside the unit circle?
2399 ;; Is pt on the unit circle? (Use
2400 ;; the cached value computed
2401 ;; above.)
2422 ;; Wang 81-83. Unfortunately, the pdf version has page 82 as all
2423 ;; black, so here is, as best as I can tell, what Wang is doing.
2424 ;; Fortunately, p. 81 has the necessary hints.
2425 ;;
2426 ;; First consider integrate(exp(%i*k*x^n),x) around the closed contour
2427 ;; consisting of the real axis from 0 to R, the arc from the angle 0
2428 ;; to %pi/(2*n) and the ray from the arc back to the origin.
2429 ;;
2430 ;; There are no poles in this region, so the integral must be zero.
2431 ;; But consider the integral on the three parts. The real axis is the
2432 ;; integral we want. The return ray is
2433 ;;
2434 ;; exp(%i*%pi/2/n) * integrate(exp(%i*k*(t*exp(%i*%pi/2/n))^n),t,R,0)
2435 ;; = exp(%i*%pi/2/n) * integrate(exp(%i*k*t^n*exp(%i*%pi/2)),t,R,0)
2436 ;; = -exp(%i*%pi/2/n) * integrate(exp(-k*t^n),t,0,R)
2437 ;;
2438 ;; As R -> infinity, this last integral is gamma(1/n)/k^(1/n)/n.
2439 ;;
2440 ;; We assume the integral on the circular arc approaches 0 as R ->
2441 ;; infinity. (Need to prove this.)
2442 ;;
2443 ;; Thus, we have
2444 ;;
2445 ;; integrate(exp(%i*k*t^n),t,0,inf)
2446 ;; = exp(%i*%pi/2/n) * gamma(1/n)/k^(1/n)/n.
2447 ;;
2448 ;; Equating real and imaginary parts gives us the desired results:
2449 ;;
2450 ;; integrate(cos(k*t^n),t,0,inf) = G * cos(%pi/2/n)
2451 ;; integrate(sin(k*t^n),t,0,inf) = G * sin(%pi/2/n)
2452 ;;
2453 ;; where G = gamma(1/n)/k^(1/n)/n.
2454 ;;
2462 ;; Ok, we have cos(b*x^n) or sin(b*x^n), and we set e = (n
2463 ;; b)
2465 ;; n = 1. Give up. (Why not divergent?)
2466 nil)
2471 ;; s is the sign of b. Give up if it's zero.
2472 nil)
2474 ;; Give up if n-1 <= 0. (Why give up? Isn't the
2475 ;; integral divergent?)
2476 nil)
2478 ;; We can apply our formula now. g = gamma(1/n)/n/b^(1/n)
2485 e)))))))
2488 ;; this is the second part of the definite integral package
2495 ())
2497 var t)
2518 t)))))
2529 plm*))))
2548 ans))))
2556 d
2557 plm*)))))
2561 0
2564 ;; Handle integral(n(x)/d(x)*log(x)^m,x,0,inf). n and d are
2565 ;; polynomials.
2592 ans)))
2609 n
2613 ;;Makes a new poly such that np(x)-np(x+2*%i*%pi)=p(x).
2614 ;;Constructs general POLY of degree one higher than P with
2615 ;;arbitrary coeff. and then solves for coeffs by equating like powers
2616 ;;of the varibale of integration.
2617 ;;Can probably be made simpler now.
2625 ;;;Now, poly(x)-poly(x+2*%i*%pi)=p(x), P is the original poly.
2634 -1
2643 ;; Integrate(p(x)*f(exp(x))/g(exp(x)),x,minf,inf) by applying the
2644 ;; transformation y = exp(x) to get
2645 ;; integrate(p(log(y))*f(y)/g(y)/y,y,0,inf). This should be handled
2646 ;; by dintlog.
2653 ;; This implements Wang's algorithm in Chapter 5.2, pp. 98-100.
2654 ;;
2655 ;; This is a very brief description of the algorithm. Basically, we
2656 ;; have integrate(R(exp(x))*p(x),x,minf,inf), where R(x) is a rational
2657 ;; function and p(x) is a polynomial.
2658 ;;
2659 ;; We find a polynomial q(x) such that q(x) - q(x+2*%i*%pi) = p(x).
2660 ;; Then consider a contour integral of R(exp(z))*q(z) over a
2661 ;; rectangular contour. Opposite corners of the rectangle are (-R,
2662 ;; 2*%i*%pi) and (R, 0).
2663 ;;
2664 ;; Wang shows that this contour integral, in the limit, is the
2665 ;; integral of R(exp(x))*q(x)-R(exp(x))*q(x+2*%i*%pi), which is
2666 ;; exactly the integral we're looking for.
2667 ;;
2668 ;; Thus, to find the value of the contour integral, we just need the
2669 ;; residues of R(exp(z))*q(z). The only tricky part is that we want
2670 ;; the log function to have an imaginary part between 0 and 2*%pi
2671 ;; instead of -%pi to %pi.
2673 ;; We have R(exp(x))*p(x) represented as p(x)*pe(exp(x))/d(exp(x)).
2679 ;; At this point denom-exponential has converted d(exp(x)) to the
2680 ;; polynomial d(z), where z = exp(x).
2683 pe))
2686 ;; Find the poles of the denominator. denom-exponential is the
2687 ;; denominator of R(x).
2688 ;;
2689 ;; It seems as if polelist returns a list of several items.
2690 ;; The first element is a list consisting of the pole and (z -
2691 ;; pole). We don't care about this, so we take the rest of the
2692 ;; result.
2695 ;; The imaginary part is nonzero,
2696 ;; or the realpart is negative.
2700 ;; The realpart is not zero.
2702 ;; Not sure what this does.
2704 ;; No roots at all, so return
2708 ;; We have simple roots or roots in REGION1
2711 ;; The cadr of pl is the list of the simple poles of
2712 ;; denom-exponential. Take the log of them to find the
2713 ;; poles of the original expression. Then compute the
2714 ;; residues at each of these poles and sum them up and put
2715 ;; the result in B. (If no simple poles set B to 0.)
2721 ;; I think this handles the case of poles outside the
2722 ;; regions. The sum of these residues are placed in C.
2726 temp)))
2731 ;; We have the repeated poles of deonom-exponential, so we
2732 ;; need to convert them to the actual pole values for
2733 ;; R(exp(x)), by taking the log of the value of poles.
2738 ;; Compute the residues at all of these poles and sum
2739 ;; them up.
2742 poles))
2757 ;; Check to see if each term in exp that is of the form exp(k*x) has
2758 ;; an integer value for k.
2772 ;; Test to see if exp is of the form f(exp(x)), and if so, replace
2773 ;; exp(x) with 'z*. That is, basically return f(z*).
2777 exp)
2783 ;; Does e really need to be special here? Seems to be ok without
2784 ;; it; testsuite works.
2785 #+nil
2788 e)
2807 ;; Test to see if exp is of the form p(x)*f(exp(x)). If so, set p* to
2808 ;; be p(x) and set pe* to f(exp(x)).
2825 ;; I think this is looking at f(exp(x)) and tries to find some
2826 ;; rational function R and some number k such that f(exp(x)) =
2827 ;; R(exp(k*x)).
2844 ;; Integration by parts.
2845 ;;
2846 ;; integrate(u(x)*diff(v(x),x),x,a,b)
2847 ;; |b
2848 ;; = u(x)*v(x)| - integrate(v(x)*diff(u(x),x))
2849 ;; |a
2850 ;;
2852 ;;;SINCE ONLY CALLED FROM DINTLOG TO get RID OF LOGS - IF LOG REMAINS, QUIT
2856 nil)
2863 nil)
2871 ;; integrate(f(exp(k*x)),x,a,b), where f(z) is rational.
2872 ;;
2873 ;; See Wang p. 96-97.
2874 ;;
2875 ;; If the limits are minf to inf, we use the substitution y=exp(k*x)
2876 ;; to get integrate(f(y)/y,y,0,inf)/k. If the limits are 0 to inf,
2877 ;; use the substitution s+1=exp(k*x) to get
2878 ;; integrate(f(s+1)/(s+1),s,0,inf).
2885 ;; If we can integrate it directly, do so and take the
2886 ;; appropriate limits.
2887 )
2889 ;; ans is the list (f(x) exp(k*x)).
2892 ;; Use the substitution s + 1 = exp(k*x). The
2893 ;; integral becomes integrate(f(s+1)/(s+1),s,0,inf)
2896 ;; Use the substitution y=exp(k*x) because the
2897 ;; limits are minf to inf.
2899 ;; Apply the substitution and integrate it.
2902 ;; integrate(log(g(x))*f(x),x,0,inf)
2910 exp))
2912 ;; Make the substitution y=1/x. If the integrand has
2913 ;; exactly the same form, the answer has to be 0.
2919 ;; It's easy if we have the antiderivative.
2920 ;; but intsubs sometimes gives results containing %limit
2922 ;; Ok, the easy cases didn't work. We now try integration by
2923 ;; parts. Set ANS to f(x).
2926 ;; Bad. f(x) contains a log term, so we give up.
2932 var ll ul))))
2933 ;; The arg of the log function is the same as the
2934 ;; integration variable. We can do something a little
2935 ;; simpler than integration by parts. We have something
2936 ;; like f(x)*log(x). Consider f(x)*x^z. If we
2937 ;; differentiate this wrt to z, the integrand becomes
2938 ;; f(x)*log(x)*x^z. When we evaluate this at z = 0, we
2939 ;; get the desired integrand.
2940 ;;
2941 ;; So we need f(0) to be 0 at 0. If we can integrate
2942 ;; f(x)*x^z, then we differentiate the result and
2943 ;; evaluate it at z = 0.
2946 ;; Try integration by parts.
2949 ;; Compute diff(e,var,n) at the point pt.
2953 ;;; GGR and friends
2955 ;; MAYBPC returns (COEF EXPO CONST)
2956 ;;
2957 ;; This basically picks off b*x^n+a and returns the list
2958 ;; (b n a). It may also set the global *zd*.
2965 ;; At this point, e is of the form (a n b)
2970 ;; If we're here, b is complex, and n > 0. zn = imagpart(b).
2971 ;;
2972 ;; Set var to the same sign as zn.
2977 ;; zd = exp(var*%i*%pi*(1+nd)/(2*n). (ZD is special!)
2980 ;; Return zn, n, a.
2986 ;; We're here if realpart(b) <= 0, and n >= 0. Then return -b, n, a.
2989 ;; Integrate x^m*exp(b*x^n+a), with realpart(m) > -1.
2990 ;;
2991 ;; See Wang, pp. 84-85.
2992 ;;
2993 ;; I believe the formula Wang gives is incorrect. The derivation is
2994 ;; correct except for the last step.
2995 ;;
2996 ;; Let J = integrate(x^m*exp(%i*k*x^n),x,0,inf), with real k.
2997 ;;
2998 ;; Consider the case for k < 0. Take a sector of a circle bounded by
2999 ;; the real line and the angle -%pi/(2*n), and by the radii, r and R.
3000 ;; Since there are no poles inside this contour, the integral
3001 ;;
3002 ;; integrate(z^m*exp(%i*k*z^n),z) = 0
3003 ;;
3004 ;; Then J = exp(-%pi*%i*(m+1)/(2*n))*integrate(R^m*exp(k*R^n),R,0,inf)
3005 ;;
3006 ;; because the integral along the boundary is zero except for the part
3007 ;; on the real axis. (Proof?)
3008 ;;
3009 ;; Wang seems to say this last integral is gamma(s/n/(-k)^s) where s =
3010 ;; (m+1)/n. But that seems wrong. If we use the substitution R =
3011 ;; (y/(-k))^(1/n), we end up with the result:
3012 ;;
3013 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n).
3014 ;;
3015 ;; or gamma((m+1)/n)/k^((m+1)/n)/n.
3016 ;;
3017 ;; Note that this also handles the case of
3018 ;;
3019 ;; integrate(x^m*exp(-k*x^n),x,0,inf);
3020 ;;
3021 ;; where k is positive real number. A simple change of variables,
3022 ;; y=k*x^n, gives
3023 ;;
3024 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n))
3025 ;;
3026 ;; which is the same form above.
3045 ;; e = (m b n a). That is, the integral is of the form
3046 ;; x^m*exp(b*x^n+a). I think we want to compute
3047 ;; gamma((m+1)/n)/b^((m+1)/n)/n.
3048 ;;
3049 ;; FIXME: If n > m + 1, the integral converges. We need
3050 ;; to check for this.
3052 e
3055 ;; The derivation only holds if b is purely real or
3056 ;; purely imaginary. Give up if it's not.
3058 ;; Check for convergence. If b is complex, we need n -
3059 ;; m > 1. If b is real, we need b < 0.
3068 ;; If we're here, b must be negative.
3071 ;; Complex b. Take the imaginary part
3073 n a))
3074 ;; NOTE: *zd* (Ick!) is special and might be set by maybpc.
3076 ;; FIXME: Why do we set %emode here? Shouldn't we just
3077 ;; bind it? And why do we want it bound to T anyway?
3078 ;; Shouldn't the user control that? The same goes for
3079 ;; dosimp.
3080 ;;(setq $%emode t)
3086 ;; Match x^m*exp(b*x^n+a). If it does, return (list m b n a).
3091 ;; We're looking at something like exp(f(var)). See if it's
3092 ;; of the form b*x^n+a, and return (list 0 b n a). (The 0 is
3093 ;; so we can graft something onto it if needed.)
3097 ;; E should be the product of exactly 2 terms
3099 ;; Check to see if one of the terms is of the form
3100 ;; var^p. If so, make sure the realpart of p > -1. If
3101 ;; so, check the other term has the right form via
3102 ;; another call to ggr1.
3111 ;; Both terms have the right form and nn* contains the arg of
3112 ;; the exponential term. Put dn* as the car of nn*. The
3113 ;; result is something like (m b n a) when we have the
3114 ;; expression x^m*exp(b*x^n+a).
3118 ;; Match b*x^n+a. If a match is found, return the list (a n b).
3119 ;; Otherwise, return NIL
3132 ;; Match b*x^n. Return the list (n b) if found or NIL if not.
3152 ;;; given (b*x^n+a)^m returns (m a n b)
3165 ;;;Catches Unfactored forms.
3182 ;;;Is E = VAR raised to some power? If so return power or 0.
3188 ;;;Is E = VAR1 raised to some power? If so return power.
3209 ans)))))
3286 ;;;Looks at the IMAGINARY part of a log and puts it in the interval 0 2*%pi.
3289 ;; We take the $rectform above to make sure that the log is
3290 ;; expanded out for the situations where simplifying plog itself
3291 ;; doesn't do it. This should probably be considered a bug in the
3292 ;; plog simplifier and should be fixed there.
3305 ;;; Temporary fix for a lacking in taylor, which loses with %i in denom.
3306 ;;; Besides doesn't seem like a bad thing to do in general.
3310 ;; Multiply the denominator by it's conjugate to get rid of
3311 ;; %i.
3315 ;; If the new denominator still contains %i, just give
3316 ;; up. Otherwise, multiply the numerator by the
3317 ;; conjugate and divide by the new denominator.
3319 exp
3323 ;;; LL and UL must be real otherwise this routine return $UNKNOWN.
3324 ;;; Returns $no $unknown or a list of poles in the interval (ll ul)
3325 ;;; for exp w.r.t. var.
3326 ;;; Form of list ((pole . multiplicity) (pole1 . multiplicity) ....)
3343 ;; this clause handles cases where we can't find the exact roots,
3344 ;; but we know that they occur outside the interval of integration.
3345 ;; example: integrate ((1+exp(t))/sqrt(t+exp(t)), t, 0, 1);
3361 ;; (multiplicity (cdar dummy)) (not used? -- cwh)
3371 pole-list))))))))))))
3374 ;;;Returns $YES if there is no pole and $NO if there is one.
3387 ;;;Takes care of forms that the ratio test fails on.
3403 (%einvolve
3438 ;; real values for ll and ul; place can be imaginary.
3445 '$no
3450 ;; returns true or nil
3452 ;; real values for ll and ul; place can be imaginary.
3468 ;; If *failures is set, we may have missed some roots.
3469 ;; We still return the roots that we have found.
3474 rootlist)
3481 rootlist)))))))))
3484 ;;; Is x > y. X or Y can be $MINF or $INF, zeroA or zeroB.
3524 nil)
3527 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3528 ;;;
3529 ;;; Integrate Definite Integrals involving log and exp functions. The algorithm
3530 ;;; are taken from the paper "Evaluation of CLasses of Definite Integrals ..."
3531 ;;; by K.O.Geddes et. al.
3532 ;;;
3533 ;;; 1. CASE: Integrals generated by the Gamma funtion.
3534 ;;;
3535 ;;; inf
3536 ;;; /
3537 ;;; [ w m s - m - 1
3538 ;;; I t log (t) expt(- t ) dt = s signum(s)
3539 ;;; ]
3540 ;;; /
3541 ;;; 0
3542 ;;; !
3543 ;;; m !
3544 ;;; d !
3545 ;;; (--- (gamma(z))! )
3546 ;;; m !
3547 ;;; dz ! w + 1
3548 ;;; !z = -----
3549 ;;; s
3550 ;;;
3551 ;;; The integral converges for:
3552 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3553 ;;;
3554 ;;; 2. CASE: Integrals generated by the Incomplete Gamma function.
3555 ;;;
3556 ;;; inf !
3557 ;;; / m !
3558 ;;; [ w m s d s !
3559 ;;; I t log (t) exp(- t ) dt = (--- (gamma_incomplete(a, x ))! )
3560 ;;; ] m !
3561 ;;; / da ! w + 1
3562 ;;; x !z = -----
3563 ;;; s
3564 ;;; - m - 1
3565 ;;; s signum(s)
3566 ;;;
3567 ;;; The integral converges for:
3568 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3569 ;;; The shown solution is valid for s>0. For s<0 gamma_incomplete has to be
3570 ;;; replaced by gamma(a) - gamma_incomplete(a,x^s).
3571 ;;;
3572 ;;; 3. CASE: Integrals generated by the beta function.
3573 ;;;
3574 ;;; 1
3575 ;;; /
3576 ;;; [ m s r n
3577 ;;; I log (1 - t) (1 - t) t log (t) dt =
3578 ;;; ]
3579 ;;; /
3580 ;;; 0
3581 ;;; !
3582 ;;; ! !
3583 ;;; n m ! !
3584 ;;; d d ! !
3585 ;;; --- (--- (beta(z, w))! )!
3586 ;;; n m ! !
3587 ;;; dz dw ! !
3588 ;;; !w = s + 1 !
3589 ;;; !z = r + 1
3590 ;;;
3591 ;;; The integral converges for:
3592 ;;; n, m = 0, 1, 2, ..., s > -1 and r > -1.
3593 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3597 ;;; Recognize c*z^w*log(z)^m*exp(-t^s)
3609 ;;; Recognize c*z^r*log(z)^n*(1-z)^s*log(1-z)^m
3622 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3629 ;; var1 is used as a parameter for differentiation. Add var1>0 to the
3630 ;; database, to get the desired simplification of the differentiation of
3631 ;; the gamma_incomplete function.
3639 ;; The integrand matches the cases 1 and 2.
3658 ;; Case 1: Generated by the Gamma function.
3665 var1
3670 ; derivate the involved hypergeometric
3671 ; functions.
3676 ;; Case 2: Generated by the Incomplete Gamma function.
3692 ;; Case 3: Generated by the Beta function.
3718 var2 m)
3720 var1 n)
3726 ;; Simplify result and set $gamma_expand to global value
3729 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;