| blob | c8360657b86a9cf542be35f248506919d0339b5c |
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, respectively.
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
132 factors rlm*
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.
829 ; We have minf <= ll = ul <= inf
830 )
832 ; We have minf <= ll < ul = inf
833 )
835 ; We have minf = ll < ul < inf
836 ;
837 ; Now substitute
838 ;
839 ; var -> -var
840 ; ll -> -ul
841 ; ul -> inf
842 ;
843 ; so that minf < ll < ul = inf
849 ; We have minf <= ul < ll
850 ;
851 ; Now substitute
852 ;
853 ; exp -> -exp
854 ; ll <-> ul
855 ;
856 ; so that minf <= ll < ul
859 t)))
873 ;; Substitute a and b into integral e
874 ;;
875 ;; Looks for discontinuties in integral, and works around them.
876 ;; For example, in
877 ;;
878 ;; integrate(x^(2*n)*exp(-(x)^2),x) ==>
879 ;; -gamma_incomplete((2*n+1)/2,x^2)*x^(2*n+1)*abs(x)^(-2*n-1)/2
880 ;;
881 ;; the integral has a discontinuity at x=0.
882 ;;
889 var a b)))))
903 total))))))
905 ;; look for terms with a negative exponent
906 ;;
907 ;; recursively traverses exp in order to find discontinuities such as
908 ;; erfc(1/x-x) at x=0
918 ;; returns list of places where exp might be discontinuous in var.
919 ;; list begins with ll and ends with ul, and include any values between
920 ;; ll and ul.
921 ;; return '$no or '$unknown if no discontinuities found.
938 ;; (multiplicity (cdar dummy)) ;; not used
943 ;; Carefully substitute the integration limits A and B into the
944 ;; expression E.
954 ;; Try easy substitutions. Return NIL if we can't.
957 ;; Infinite limits aren't easy
958 nil)
962 %gamma_incomplete %expintegral_ei)))
964 ;; It's easy if we have a polynomial. I (rtoy) think
965 ;; it's also easy if the denominator is free of the
966 ;; integration variable and also if the expression
967 ;; doesn't involve inverse functions.
968 ;;
969 ;; gamma_incomplete and expintegral_ie
970 ;; included because of discontinuity in
971 ;; gamma_incomplete(0, exp(%i*x)) and
972 ;; expintegral_ei(exp(%i*x))
973 ;;
974 ;; XXX: Are there other cases we've forgotten about?
975 ;;
976 ;; So just try to substitute the limits into the
977 ;; expression. If no errors are produced, we're done.
982 ;; no-err-sub has returned T. An error was catched.
983 nil)
997 ;; check for divergent integral
1009 ;;;This function works only on things with ATAN's in them now.
1011 ;; POLES-IN-INTERVAL doesn't know about the poles of tan(x). Call
1012 ;; trigsimp to convert tan into sin/cos, which POLES-IN-INTERVAL
1013 ;; knows how to handle.
1014 ;;
1015 ;; XXX Should we fix POLES-IN-INTERVAL instead?
1016 ;;
1017 ;; XXX Is calling trigsimp too much? Should we just only try to
1018 ;; substitute sin/cos for tan?
1019 ;;
1020 ;; XXX Should the result try to convert sin/cos back into tan? (A
1021 ;; call to trigreduce would do it, among other things.)
1024 ;;POLES -> ((mlist) ((mequal) ((%atan) foo) replacement) ......)
1025 ;;We can then use $SUBSTITUTE
1030 ul-ans)
1032 nil)))
1046 var ll ul)))
1063 llist)))))))))))))))
1084 nil)
1103 ;; e is of form poly(x)*exp(m*%i*x)
1104 ;; s is degree of denominator
1105 ;; adds e to bptu or bptd according to sign of m
1120 ;; check term is of form poly(x)*exp(m*%i*x)
1121 ;; n is degree of denominator
1134 term)
1142 term))
1174 ;; exponentialize sin and cos
1180 '$%i
1200 exp)
1216 ;; computes integral from minf to inf for expressions of the form
1217 ;; exp(%i*m*x)*r(x) by residues on either the upper half
1218 ;; plane or the lower half plane, depending on whether
1219 ;; m is positive or negative. [wang p. 77]
1220 ;;
1221 ;; exponentializes sin and cos before applying residue method.
1222 ;; can handle some expressions with poles on real line, such as
1223 ;; sin(x)*cos(x)/x.
1228 ratterms ratans
1229 plf bptu bptd s upans downans)
1234 var
1235 nil)
1238 ;;;Above tests for applicability of this method.
1247 ;; call to $expand can create rational terms
1248 ;; with no exp(m*%i*x)
1252 ;;;Function RIB sets up the values of BPTU and BPTD
1259 ;;;If there is BPTU then CSEMIUP must succeed.
1260 ;;;Likewise for BPTD.
1264 ;; The original integrand was already
1265 ;; determined to have no poles by initial-analysis.
1266 ;; If individual terms of the expansion have poles, the poles
1267 ;; must cancel each other out, so we can ignore them.
1270 ;; if integral of ratterms is divergent, ratans is nil,
1271 ;; and mtosc returns nil
1288 t)
1290 t)
1298 t)
1300 t)
1305 s nc dc
1306 ans r $savefactors checkfactors temp test-var)
1328 ;;;
1329 ;;;New section.
1357 findout
1361 on
1383 s)
1388 ;; Let's not signal an error here. Return nil so that we
1389 ;; eventually return a noun form if no other algorithm gives
1390 ;; a result.
1393 nil)))
1485 ;; We have an integral of the form p(x)*F(exp(x)), where
1486 ;; p(x) is a polynomial.
1488 ;; No polynomial
1492 ;; These limits must exist for the integral to converge.
1495 ;; This only handles the case when the F(z) is a
1496 ;; rational function.
1499 ;; If we get here, F(z) is not a rational function.
1500 ;; We transform it using the substitution x=log(y)
1501 ;; which gives us an integral of the form
1502 ;; p(log(y))*F(y)/y, which maxima should be able to
1503 ;; handle.
1506 ;; Give up. We don't know how to handle this.
1508 en
1522 exp)))
1524 ;;; given (b*x+a)^n+c returns (a b n c)
1532 nil)
1534 ;;;get high degree of poly
1536 ;;;diff down to linear.
1538 ;;;all the way to constant.
1541 ;;;get rid of factorial from differentiation.
1544 ;;;Sees if can be expressed as (a*x+b)^n + part freeof x.
1553 s)
1578 n
1579 d)))))
1582 on
1637 e nil)
1651 e))))))
1657 ;; Compute the integral(n/d,x,0,inf) by computing the negative of the
1658 ;; sum of residues of log(-x)*n/d over the poles of n/d inside the
1659 ;; keyhole contour. This contour is basically an disk with a slit
1660 ;; along the positive real axis. n/d must be a rational function.
1665 ;; Ok if not on the positive real axis.
1670 t)
1681 ;; Look at an expression e of the form sin(r*x)^k, where k is an
1682 ;; integer. Return the list (1 r k). (Not sure if the first element
1683 ;; of the list is always 1 because I'm not sure what partition is
1684 ;; trying to do here.)
1698 ;; Look at an expression e of the form sin(r*x) and return r.
1709 ;; integrate(a*sc(r*x)^k/x^n,x,0,inf).
1712 ;; Get the argument of the involved trig function.
1715 ;; I don't think this needs to be special.
1716 #+nil
1718 ;; Replace (1-cos(arg)^2) with sin(arg)^2.
1720 ;(m+t 1. (m- (m^t `((%cos) ,var) 2.)))
1721 ;; The code from above generates expressions with
1722 ;; a missing simp flag. Furthermore, the
1723 ;; substitution has to be done for the complete
1724 ;; argument of the trig function. (DK 02/2010)
1727 exp))
1732 ;; n is the power of the denominator.
1734 ;; The simple case.
1738 ;; Do this for a sum of such terms.
1740 c)))))))))
1742 ;; We have an integral of the form sin(r*x)^k/x^n. C is the list (1 r k).
1743 ;;
1744 ;; The substitution y=r*x converts this integral to
1745 ;;
1746 ;; r^(n-1)*integral(sin(y)^k/y^n,y,0,inf)
1747 ;;
1748 ;; (If r is negative, we need to negate the result.)
1749 ;;
1750 ;; The recursion Wang gives on p. 87 has a typo. The second term
1751 ;; should be subtracted from the first. This formula is given in G&R,
1752 ;; 3.82, formula 12.
1753 ;;
1754 ;; integrate(sin(x)^r/x^s,x) =
1755 ;; r*(r-1)/(s-1)/(s-2)*integrate(sin(x)^(r-2)/x^(s-2),x)
1756 ;; - r^2/(s-1)/(s-2)*integrate(sin(x)^r/x^(s-2),x)
1757 ;;
1758 ;; (Limits are assumed to be 0 to inf.)
1759 ;;
1760 ;; This recursion ends up with integrals with s = 1 or 2 and
1761 ;;
1762 ;; integrate(sin(x)^p/x,x,0,inf) = integrate(sin(x)^(p-1),x,0,%pi/2)
1763 ;;
1764 ;; with p > 0 and odd. This latter integral is known to maxima, and
1765 ;; it's value is beta(p/2,1/2)/2.
1766 ;;
1767 ;; integrate(sin(x)^2/x^2,x,0,inf) = %pi/2*binomial(q-3/2,q-1)
1768 ;;
1769 ;; where q >= 2.
1770 ;;
1772 ;; Compute sign(r)*r^(n-1)*integrate(sin(y)^k/y^n,y,0,inf)
1774 c
1780 recursion)
1781 ;; Recursion failed. Return the integrand
1782 ;; The following code generates expressions with a missing simp flag
1783 ;; for the sin function. Use better simplifying code. (DK 02/2010)
1784 ; (let ((integrand (div (pow `((%sin) ,(m* r var))
1785 ; k)
1786 ; (pow var n))))
1788 k)
1793 ;; integrate(sin(x)^n/x^2,x,0,inf) = pi/2*binomial(n-3/2,n-1).
1794 ;; Express in terms of Gamma functions, though.
1800 ;; integrate(sin(x)^n/x,x,0,inf) = beta((n+1)/2,1/2)/2, for n odd and
1801 ;; n > 0.
1804 half%pi)
1807 ;; This implements the recursion for computing
1808 ;; integrate(sin(y)^l/y^k,y,0,inf). (Note the change in notation from
1809 ;; the above!)
1815 ;; Integral diverges if l-(k-1) < 0.
1818 ;; If l + k is not even, return NIL. (Is this the right
1819 ;; thing to do?)
1820 nil)
1822 ;; We have integrate(sin(y)^l/y^2). Use sevn to evaluate.
1825 ;; We have integrate(sin(y)^l/y,y)
1831 ;; j = k-2, i = (k-1)*(k-2)
1832 ;;
1833 ;;
1834 ;; The main recursion:
1835 ;;
1836 ;; i(sin(y)^l/y^k)
1837 ;; = l*(l-1)/(k-1)/(k-2)*i(sin(y)^(l-2)/y^k)
1838 ;; - l^2/(k-1)/(k-1)*i(sin(y)^l/y^(k-2))
1846 ;; Returns the fractional part of a?
1850 ;; Why do we return 0 if a is a number? Perhaps we really
1851 ;; mean integer?
1854 ;; If we're here, this basically assumes a is a rational.
1855 ;; Compute the remainder and return the result.
1873 ;;; THE FOLLOWING FUNCTION IS FOR TRIG FUNCTIONS OF THE FOLLOWING TYPE:
1874 ;;; LOWER LIMIT=0 B A MULTIPLE OF %PI SCA FUNCTION OF SIN (X) COS (X)
1875 ;;; B<=%PI2
1879 ;; means there was no error
1882 ; returns cons of (integer_part . fractional_part) of a
1884 ;; I think we really want to compute how many full periods are in a
1885 ;; and the remainder.
1890 ; returns cons of (integer_part . fractional_part) of a
1892 ;; I think we really want to compute how many full periods are in a
1893 ;; and the remainder.
1900 ;; Return the integer part of r.
1902 ;; r - fpart(r)
1906 ;;;Try making exp(%i*var) --> yy, if result is rational then do integral
1907 ;;;around unit circle. Make corrections for limits of integration if possible.
1927 rat-form)))
1932 ;;; Do integrals of sin and cos. this routine makes sure lower limit
1933 ;;; is zero.
1935 ;; integrate(e,var,a,b)
1951 var e)
1952 cc))
1962 ;; Exit if b or a is not a constant or if the integrand
1963 ;; doesn't appear to have a period of 2 pi.
1966 ;; Multiples of 2*%pi in limits.
1973 ;; The integral is not over a full period (2*%pi) or multiple
1974 ;; of a full period.
1976 ;; Wang p. 111: The integral integrate(f(x),x,a,b) can be
1977 ;; written as:
1978 ;;
1979 ;; n * integrate(f,x,0,2*%pi) + integrate(f,x,0,c)
1980 ;; - integrate(f,x,0,d)
1981 ;;
1982 ;; for some integer n and d >= 0, c < 2*%pi because there exist
1983 ;; integers p and q such that a = 2 * p *%pi + d and b = 2 * q
1984 ;; * %pi + c. Then n = q - p.
1986 ;; Compute q and c for the upper limit b.
1993 ;; Compute p and d for the lower limit a.
1995 ;; avoid an extra trip around the circle - helps skip principal values
2000 ;; Compute -integrate(f,x,0,d)
2005 ;; Compute n = q - p (stored in nzp2)
2007 out
2008 ;; Compute n*integrate(f,x,0,2*%pi)
2010 ;; n = 0
2013 ;; n is not zero, so compute
2014 ;; integrate(f,x,0,2*%pi)
2016 ;; Unable to compute integrate(f,x,0,2*%pi)
2018 ;; Compute integrate(f,x,0,c)
2022 ;; All three pieces succeeded.
2025 ;; Less than 1 full period, so intsc can integrate it.
2026 ;; Apply the substitution to make the lower limit 0.
2027 ;; This is last resort because substitution often causes intsc to fail.
2029 limit-diff var))
2030 ;; nothing worked
2033 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)).
2034 ;; calls intsc with a wrapper to just return nil if integral is divergent,
2035 ;; rather than generating an error.
2040 nil
2041 ans)))
2043 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)). I (rtoy)
2044 ;; think this expects b to be less than 2*%pi.
2047 0
2054 ;; Partition the integrand SC into the factors that do not
2055 ;; contain VAR (the car part) and the parts that do (the cdr
2056 ;; part).
2061 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)).
2063 ;; Determine if sc is a product of sin's and cos's.
2067 ;; We have a product of sin's and cos's. We handle some
2068 ;; special cases here.
2070 ;; Wang p. 110, formula (1):
2071 ;; integrate(sin(x)^m*cos(x)^n, x, 0, %pi/2) =
2072 ;; gamma((m+1)/2)*gamma((n+1)/2)/2/gamma((n+m+2)/2)
2075 ;; Wang p. 110, near the bottom, says
2076 ;;
2077 ;; int(f(sin(x),cos(x)), x, 0, %pi) =
2078 ;; int(f(sin(x),cos(x)) + f(sin(x),-cos(x)),x,0,%pi/2)
2099 ;; Wang, p. 111 says
2100 ;;
2101 ;; int(f(sin(x),cos(x)),x,0,3*%pi/2) =
2102 ;; int(f(sin(x),cos(x)),x,0,%pi)
2103 ;; + int(f(-sin(x),-cos(x)),x,0,%pi/2)
2107 ;; We don't have a product of sin's and cos's.
2112 dn*)
2117 ;;;Is careful about substitution of limits where the denominator may be zero
2118 ;;;because of various assumptions made.
2137 nil
2138 ans)))
2144 nil
2145 ans)))
2147 ;; Determine whether E is of the form sin(x)^m*cos(x)^n and return the
2148 ;; list (m n).
2152 m n)
2171 ;; Says wether (m^t value exponent) has at least one real branch.
2172 ;; Only works for values of 1 and -1 now. Returns $yes $no
2173 ;; $unknown.
2184 ;; Compute beta((m+1)/2,(n+1)/2)/2.
2190 ;;Seems like Guys who call this don't agree on what it should return.
2203 ;; Check e for an expression of the form x^kk*(b*x^n+a)^l. If it
2204 ;; matches, Return the two values kk and (list l a n b).
2209 ;; We have f(x)^y. Look to see if f(x) has the desired
2210 ;; form. Then f(x)^y has the desired form too, with
2211 ;; suitably modified values.
2212 ;;
2213 ;; XXX: Should we ask for the sign of f(x) if y is not an
2214 ;; integer? This transformation we're going to do requires
2215 ;; that f(x)^y be real.
2217 e
2222 ;; Got a match. Adjust kk and cc appropriately.
2224 cc
2239 ;;(DEFUN BATAP (E)
2240 ;; (PROG (K C L)
2241 ;; (COND ((NOT (BATA0 E)) (RETURN NIL))
2242 ;; ((AND (EQUAL -1. (CADDDR C))
2243 ;; (EQ ($askSIGN (SETQ K (m+ 1. K)))
2244 ;; '$pos)
2245 ;; (EQ ($askSIGN (SETQ L (m+ 1. (CAR C))))
2246 ;; '$pos)
2247 ;; (ALIKE1 (CADR C)
2248 ;; (m^ UL (CADDR C)))
2249 ;; (SETQ E (CADR C))
2250 ;; (EQ ($askSIGN (SETQ C (CADDR C))) '$pos))
2251 ;; (RETURN (M// (m* (m^ UL (m+t K (m* C (m+t -1. L))))
2252 ;; `(($BETA) ,(SETQ NN* (M// K C))
2253 ;; ,(SETQ DN* L)))
2254 ;; C))))))
2257 ;; Integrals of the form i(log(x)^m*x^k*(a+b*x^n)^l,x,0,ul). There
2258 ;; are two cases to consider: One case has ul>0, b<0, a=-b*ul^n, k>-1,
2259 ;; l>-1, n>0, m a nonnegative integer. The second case has ul=inf, l < 0.
2260 ;;
2261 ;; These integrals are essentially partial derivatives of the Beta
2262 ;; function (i.e. the Eulerian integral of the first kind). Note
2263 ;; that, currently, with the default setting intanalysis:true, this
2264 ;; function might not even be called for some of these integrals.
2265 ;; However, this can be palliated by setting intanalysis:false.
2284 ;; We have i(x^kk/(d+cc*x^r)^s,x,0,inf) =
2285 ;; beta(aa,bb)/(cc^aa*d^bb*r). Compute this, and then
2286 ;; differentiate it m times to get the log term
2287 ;; incorporated.
2292 0
2293 m)))
2297 r))))
2306 ;; The result looks like B(k/n,l) ( ... ).
2307 ;; Perhaps, we should just $factor, instead of
2308 ;; pulling out beta like this.
2310 beta
2311 ($fullratsimp
2319 var m)
2321 beta))))))))))))
2324 ;;; If e is of the form given below, make the obvious change
2325 ;;; of variables (substituting ul*x^(1/n) for x) in order to reduce
2326 ;;; e to the usual form of the integrand in the Eulerian
2327 ;;; integral of the first kind.
2328 ;;; N. B: The old version of ZTO1 completely ignored this
2329 ;;; substitution; the log(x)s were just thrown in, which,
2330 ;;; of course would give wrong results.
2333 ;; Parse e
2337 ;; e=x^k*(a+b*x^n)^l
2339 c
2351 ;; Wang p. 71 gives the following formula for a beta function:
2352 ;;
2353 ;; integrate(x^(k-1)/(c*x^r+d)^s,x,0,inf)
2354 ;; = beta(a,b)/(c^a*d^b*r)
2355 ;;
2356 ;; where a = k/r > 0, b = s - a > 0, s > k > 0, r > 0, c*d > 0.
2357 ;;
2358 ;; This function matches this and returns k-1, d, r, c, a, b. And
2359 ;; also checks that all the conditions hold. If not, NIL is returned.
2360 ;;
2366 c
2383 ;; Handles beta integrals.
2391 nil)
2394 c
2395 ;; e = x^k*(d+c*x^al)^l.
2400 new-k)
2408 ;; Compute exp(d)*gamma((c+1)/b)/b/a^((c+1)/b). In essence, this is
2409 ;; the value of integrate(x^c*exp(d-a*x^b),x,0,inf).
2420 ;; Evaluates the contour integral of GRAND around the unit circle
2421 ;; using residues.
2427 ;; Is pt stricly inside the unit circle?
2433 ;; Is pt on the unit circle? (Use
2434 ;; the cached value computed
2435 ;; above.)
2456 ;; Wang 81-83. Unfortunately, the pdf version has page 82 as all
2457 ;; black, so here is, as best as I can tell, what Wang is doing.
2458 ;; Fortunately, p. 81 has the necessary hints.
2459 ;;
2460 ;; First consider integrate(exp(%i*k*x^n),x) around the closed contour
2461 ;; consisting of the real axis from 0 to R, the arc from the angle 0
2462 ;; to %pi/(2*n) and the ray from the arc back to the origin.
2463 ;;
2464 ;; There are no poles in this region, so the integral must be zero.
2465 ;; But consider the integral on the three parts. The real axis is the
2466 ;; integral we want. The return ray is
2467 ;;
2468 ;; exp(%i*%pi/2/n) * integrate(exp(%i*k*(t*exp(%i*%pi/2/n))^n),t,R,0)
2469 ;; = exp(%i*%pi/2/n) * integrate(exp(%i*k*t^n*exp(%i*%pi/2)),t,R,0)
2470 ;; = -exp(%i*%pi/2/n) * integrate(exp(-k*t^n),t,0,R)
2471 ;;
2472 ;; As R -> infinity, this last integral is gamma(1/n)/k^(1/n)/n.
2473 ;;
2474 ;; We assume the integral on the circular arc approaches 0 as R ->
2475 ;; infinity. (Need to prove this.)
2476 ;;
2477 ;; Thus, we have
2478 ;;
2479 ;; integrate(exp(%i*k*t^n),t,0,inf)
2480 ;; = exp(%i*%pi/2/n) * gamma(1/n)/k^(1/n)/n.
2481 ;;
2482 ;; Equating real and imaginary parts gives us the desired results:
2483 ;;
2484 ;; integrate(cos(k*t^n),t,0,inf) = G * cos(%pi/2/n)
2485 ;; integrate(sin(k*t^n),t,0,inf) = G * sin(%pi/2/n)
2486 ;;
2487 ;; where G = gamma(1/n)/k^(1/n)/n.
2488 ;;
2496 ;; Ok, we have cos(b*x^n) or sin(b*x^n), and we set e = (n
2497 ;; b)
2499 ;; n = 1. Give up. (Why not divergent?)
2500 nil)
2505 ;; s is the sign of b. Give up if it's zero.
2506 nil)
2508 ;; Give up if n-1 <= 0. (Why give up? Isn't the
2509 ;; integral divergent?)
2510 nil)
2512 ;; We can apply our formula now. g = gamma(1/n)/n/b^(1/n)
2519 e)))))))
2522 ;; this is the second part of the definite integral package
2529 ())
2531 var t)
2552 t)))))
2563 plm*))))
2582 ans))))
2590 d
2591 plm*)))))
2595 0
2598 ;; Handle integral(n(x)/d(x)*log(x)^m,x,0,inf). n and d are
2599 ;; polynomials.
2626 ans)))
2643 n
2647 ;;Makes a new poly such that np(x)-np(x+2*%i*%pi)=p(x).
2648 ;;Constructs general POLY of degree one higher than P with
2649 ;;arbitrary coeff. and then solves for coeffs by equating like powers
2650 ;;of the varibale of integration.
2651 ;;Can probably be made simpler now.
2659 ;;;Now, poly(x)-poly(x+2*%i*%pi)=p(x), P is the original poly.
2668 -1
2677 ;; Integrate(p(x)*f(exp(x))/g(exp(x)),x,minf,inf) by applying the
2678 ;; transformation y = exp(x) to get
2679 ;; integrate(p(log(y))*f(y)/g(y)/y,y,0,inf). This should be handled
2680 ;; by dintlog.
2687 ;; This implements Wang's algorithm in Chapter 5.2, pp. 98-100.
2688 ;;
2689 ;; This is a very brief description of the algorithm. Basically, we
2690 ;; have integrate(R(exp(x))*p(x),x,minf,inf), where R(x) is a rational
2691 ;; function and p(x) is a polynomial.
2692 ;;
2693 ;; We find a polynomial q(x) such that q(x) - q(x+2*%i*%pi) = p(x).
2694 ;; Then consider a contour integral of R(exp(z))*q(z) over a
2695 ;; rectangular contour. Opposite corners of the rectangle are (-R,
2696 ;; 2*%i*%pi) and (R, 0).
2697 ;;
2698 ;; Wang shows that this contour integral, in the limit, is the
2699 ;; integral of R(exp(x))*q(x)-R(exp(x))*q(x+2*%i*%pi), which is
2700 ;; exactly the integral we're looking for.
2701 ;;
2702 ;; Thus, to find the value of the contour integral, we just need the
2703 ;; residues of R(exp(z))*q(z). The only tricky part is that we want
2704 ;; the log function to have an imaginary part between 0 and 2*%pi
2705 ;; instead of -%pi to %pi.
2707 ;; We have R(exp(x))*p(x) represented as p(x)*pe(exp(x))/d(exp(x)).
2713 ;; At this point denom-exponential has converted d(exp(x)) to the
2714 ;; polynomial d(z), where z = exp(x).
2717 pe))
2720 ;; Find the poles of the denominator. denom-exponential is the
2721 ;; denominator of R(x).
2722 ;;
2723 ;; It seems as if polelist returns a list of several items.
2724 ;; The first element is a list consisting of the pole and (z -
2725 ;; pole). We don't care about this, so we take the rest of the
2726 ;; result.
2729 ;; The imaginary part is nonzero,
2730 ;; or the realpart is negative.
2734 ;; The realpart is not zero.
2736 ;; Not sure what this does.
2738 ;; No roots at all, so return
2742 ;; We have simple roots or roots in REGION1
2745 ;; The cadr of pl is the list of the simple poles of
2746 ;; denom-exponential. Take the log of them to find the
2747 ;; poles of the original expression. Then compute the
2748 ;; residues at each of these poles and sum them up and put
2749 ;; the result in B. (If no simple poles set B to 0.)
2755 ;; I think this handles the case of poles outside the
2756 ;; regions. The sum of these residues are placed in C.
2760 temp)))
2765 ;; We have the repeated poles of deonom-exponential, so we
2766 ;; need to convert them to the actual pole values for
2767 ;; R(exp(x)), by taking the log of the value of poles.
2772 ;; Compute the residues at all of these poles and sum
2773 ;; them up.
2776 poles))
2791 ;; Check to see if each term in exp that is of the form exp(k*x) has
2792 ;; an integer value for k.
2806 ;; Test to see if exp is of the form f(exp(x)), and if so, replace
2807 ;; exp(x) with 'z*. That is, basically return f(z*).
2811 exp)
2817 ;; Does e really need to be special here? Seems to be ok without
2818 ;; it; testsuite works.
2819 #+nil
2822 e)
2841 ;; Test to see if exp is of the form p(x)*f(exp(x)). If so, set p* to
2842 ;; be p(x) and set pe* to f(exp(x)).
2859 ;; I think this is looking at f(exp(x)) and tries to find some
2860 ;; rational function R and some number k such that f(exp(x)) =
2861 ;; R(exp(k*x)).
2878 ;; Integration by parts.
2879 ;;
2880 ;; integrate(u(x)*diff(v(x),x),x,a,b)
2881 ;; |b
2882 ;; = u(x)*v(x)| - integrate(v(x)*diff(u(x),x))
2883 ;; |a
2884 ;;
2886 ;;;SINCE ONLY CALLED FROM DINTLOG TO get RID OF LOGS - IF LOG REMAINS, QUIT
2890 nil)
2897 nil)
2905 ;; integrate(f(exp(k*x)),x,a,b), where f(z) is rational.
2906 ;;
2907 ;; See Wang p. 96-97.
2908 ;;
2909 ;; If the limits are minf to inf, we use the substitution y=exp(k*x)
2910 ;; to get integrate(f(y)/y,y,0,inf)/k. If the limits are 0 to inf,
2911 ;; use the substitution s+1=exp(k*x) to get
2912 ;; integrate(f(s+1)/(s+1),s,0,inf).
2919 ;; If we can integrate it directly, do so and take the
2920 ;; appropriate limits.
2921 )
2923 ;; ans is the list (f(x) exp(k*x)).
2926 ;; Use the substitution s + 1 = exp(k*x). The
2927 ;; integral becomes integrate(f(s+1)/(s+1),s,0,inf)
2930 ;; Use the substitution y=exp(k*x) because the
2931 ;; limits are minf to inf.
2933 ;; Apply the substitution and integrate it.
2936 ;; integrate(log(g(x))*f(x),x,0,inf)
2944 exp))
2946 ;; Make the substitution y=1/x. If the integrand has
2947 ;; exactly the same form, the answer has to be 0.
2953 ;; It's easy if we have the antiderivative.
2954 ;; but intsubs sometimes gives results containing %limit
2956 ;; Ok, the easy cases didn't work. We now try integration by
2957 ;; parts. Set ANS to f(x).
2960 ;; Bad. f(x) contains a log term, so we give up.
2966 var ll ul))))
2967 ;; The arg of the log function is the same as the
2968 ;; integration variable. We can do something a little
2969 ;; simpler than integration by parts. We have something
2970 ;; like f(x)*log(x). Consider f(x)*x^z. If we
2971 ;; differentiate this wrt to z, the integrand becomes
2972 ;; f(x)*log(x)*x^z. When we evaluate this at z = 0, we
2973 ;; get the desired integrand.
2974 ;;
2975 ;; So we need f(0) to be 0 at 0. If we can integrate
2976 ;; f(x)*x^z, then we differentiate the result and
2977 ;; evaluate it at z = 0.
2980 ;; Try integration by parts.
2983 ;; Compute diff(e,var,n) at the point pt.
2987 ;;; GGR and friends
2989 ;; MAYBPC returns (COEF EXPO CONST)
2990 ;;
2991 ;; This basically picks off b*x^n+a and returns the list
2992 ;; (b n a). It may also set the global *zd*.
2999 ;; At this point, e is of the form (a n b)
3004 ;; If we're here, b is complex, and n > 0. zn = imagpart(b).
3005 ;;
3006 ;; Set var to the same sign as zn.
3011 ;; zd = exp(var*%i*%pi*(1+nd)/(2*n). (ZD is special!)
3014 ;; Return zn, n, a.
3020 ;; We're here if realpart(b) <= 0, and n >= 0. Then return -b, n, a.
3023 ;; Integrate x^m*exp(b*x^n+a), with realpart(m) > -1.
3024 ;;
3025 ;; See Wang, pp. 84-85.
3026 ;;
3027 ;; I believe the formula Wang gives is incorrect. The derivation is
3028 ;; correct except for the last step.
3029 ;;
3030 ;; Let J = integrate(x^m*exp(%i*k*x^n),x,0,inf), with real k.
3031 ;;
3032 ;; Consider the case for k < 0. Take a sector of a circle bounded by
3033 ;; the real line and the angle -%pi/(2*n), and by the radii, r and R.
3034 ;; Since there are no poles inside this contour, the integral
3035 ;;
3036 ;; integrate(z^m*exp(%i*k*z^n),z) = 0
3037 ;;
3038 ;; Then J = exp(-%pi*%i*(m+1)/(2*n))*integrate(R^m*exp(k*R^n),R,0,inf)
3039 ;;
3040 ;; because the integral along the boundary is zero except for the part
3041 ;; on the real axis. (Proof?)
3042 ;;
3043 ;; Wang seems to say this last integral is gamma(s/n/(-k)^s) where s =
3044 ;; (m+1)/n. But that seems wrong. If we use the substitution R =
3045 ;; (y/(-k))^(1/n), we end up with the result:
3046 ;;
3047 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n).
3048 ;;
3049 ;; or gamma((m+1)/n)/k^((m+1)/n)/n.
3050 ;;
3051 ;; Note that this also handles the case of
3052 ;;
3053 ;; integrate(x^m*exp(-k*x^n),x,0,inf);
3054 ;;
3055 ;; where k is positive real number. A simple change of variables,
3056 ;; y=k*x^n, gives
3057 ;;
3058 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n))
3059 ;;
3060 ;; which is the same form above.
3079 ;; e = (m b n a). That is, the integral is of the form
3080 ;; x^m*exp(b*x^n+a). I think we want to compute
3081 ;; gamma((m+1)/n)/b^((m+1)/n)/n.
3082 ;;
3083 ;; FIXME: If n > m + 1, the integral converges. We need
3084 ;; to check for this.
3086 e
3089 ;; The derivation only holds if b is purely real or
3090 ;; purely imaginary. Give up if it's not.
3092 ;; Check for convergence. If b is complex, we need n -
3093 ;; m > 1. If b is real, we need b < 0.
3102 ;; If we're here, b must be negative.
3105 ;; Complex b. Take the imaginary part
3107 n a))
3108 ;; NOTE: *zd* (Ick!) is special and might be set by maybpc.
3110 ;; FIXME: Why do we set %emode here? Shouldn't we just
3111 ;; bind it? And why do we want it bound to T anyway?
3112 ;; Shouldn't the user control that? The same goes for
3113 ;; dosimp.
3114 ;;(setq $%emode t)
3120 ;; Match x^m*exp(b*x^n+a). If it does, return (list m b n a).
3125 ;; We're looking at something like exp(f(var)). See if it's
3126 ;; of the form b*x^n+a, and return (list 0 b n a). (The 0 is
3127 ;; so we can graft something onto it if needed.)
3131 ;; E should be the product of exactly 2 terms
3133 ;; Check to see if one of the terms is of the form
3134 ;; var^p. If so, make sure the realpart of p > -1. If
3135 ;; so, check the other term has the right form via
3136 ;; another call to ggr1.
3145 ;; Both terms have the right form and nn* contains the arg of
3146 ;; the exponential term. Put dn* as the car of nn*. The
3147 ;; result is something like (m b n a) when we have the
3148 ;; expression x^m*exp(b*x^n+a).
3152 ;; Match b*x^n+a. If a match is found, return the list (a n b).
3153 ;; Otherwise, return NIL
3166 ;; Match b*x^n. Return the list (n b) if found or NIL if not.
3186 ;;; given (b*x^n+a)^m returns (m a n b)
3199 ;;;Catches Unfactored forms.
3216 ;;;Is E = VAR raised to some power? If so return power or 0.
3222 ;;;Is E = VAR1 raised to some power? If so return power.
3243 ans)))))
3320 ;;;Looks at the IMAGINARY part of a log and puts it in the interval 0 2*%pi.
3323 ;; We take the $rectform above to make sure that the log is
3324 ;; expanded out for the situations where simplifying plog itself
3325 ;; doesn't do it. This should probably be considered a bug in the
3326 ;; plog simplifier and should be fixed there.
3339 ;;; Temporary fix for a lacking in taylor, which loses with %i in denom.
3340 ;;; Besides doesn't seem like a bad thing to do in general.
3344 ;; Multiply the denominator by it's conjugate to get rid of
3345 ;; %i.
3349 ;; If the new denominator still contains %i, just give
3350 ;; up. Otherwise, multiply the numerator by the
3351 ;; conjugate and divide by the new denominator.
3353 exp
3357 ;;; LL and UL must be real otherwise this routine return $UNKNOWN.
3358 ;;; Returns $no $unknown or a list of poles in the interval (ll ul)
3359 ;;; for exp w.r.t. var.
3360 ;;; Form of list ((pole . multiplicity) (pole1 . multiplicity) ....)
3377 ;; this clause handles cases where we can't find the exact roots,
3378 ;; but we know that they occur outside the interval of integration.
3379 ;; example: integrate ((1+exp(t))/sqrt(t+exp(t)), t, 0, 1);
3395 ;; (multiplicity (cdar dummy)) (not used? -- cwh)
3405 pole-list))))))))))))
3408 ;;;Returns $YES if there is no pole and $NO if there is one.
3421 ;;;Takes care of forms that the ratio test fails on.
3437 (%einvolve
3472 ;; real values for ll and ul; place can be imaginary.
3479 '$no
3484 ;; returns true or nil
3486 ;; real values for ll and ul; place can be imaginary.
3502 ;; If *failures is set, we may have missed some roots.
3503 ;; We still return the roots that we have found.
3508 rootlist)
3515 rootlist)))))))))
3518 ;;; Is x > y. X or Y can be $MINF or $INF, zeroA or zeroB.
3558 nil)
3561 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3562 ;;;
3563 ;;; Integrate Definite Integrals involving log and exp functions. The algorithm
3564 ;;; are taken from the paper "Evaluation of CLasses of Definite Integrals ..."
3565 ;;; by K.O.Geddes et. al.
3566 ;;;
3567 ;;; 1. CASE: Integrals generated by the Gamma function.
3568 ;;;
3569 ;;; inf
3570 ;;; /
3571 ;;; [ w m s - m - 1
3572 ;;; I t log (t) expt(- t ) dt = s signum(s)
3573 ;;; ]
3574 ;;; /
3575 ;;; 0
3576 ;;; !
3577 ;;; m !
3578 ;;; d !
3579 ;;; (--- (gamma(z))! )
3580 ;;; m !
3581 ;;; dz ! w + 1
3582 ;;; !z = -----
3583 ;;; s
3584 ;;;
3585 ;;; The integral converges for:
3586 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3587 ;;;
3588 ;;; 2. CASE: Integrals generated by the Incomplete Gamma function.
3589 ;;;
3590 ;;; inf !
3591 ;;; / m !
3592 ;;; [ w m s d s !
3593 ;;; I t log (t) exp(- t ) dt = (--- (gamma_incomplete(a, x ))! )
3594 ;;; ] m !
3595 ;;; / da ! w + 1
3596 ;;; x !z = -----
3597 ;;; s
3598 ;;; - m - 1
3599 ;;; s signum(s)
3600 ;;;
3601 ;;; The integral converges for:
3602 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3603 ;;; The shown solution is valid for s>0. For s<0 gamma_incomplete has to be
3604 ;;; replaced by gamma(a) - gamma_incomplete(a,x^s).
3605 ;;;
3606 ;;; 3. CASE: Integrals generated by the beta function.
3607 ;;;
3608 ;;; 1
3609 ;;; /
3610 ;;; [ m s r n
3611 ;;; I log (1 - t) (1 - t) t log (t) dt =
3612 ;;; ]
3613 ;;; /
3614 ;;; 0
3615 ;;; !
3616 ;;; ! !
3617 ;;; n m ! !
3618 ;;; d d ! !
3619 ;;; --- (--- (beta(z, w))! )!
3620 ;;; n m ! !
3621 ;;; dz dw ! !
3622 ;;; !w = s + 1 !
3623 ;;; !z = r + 1
3624 ;;;
3625 ;;; The integral converges for:
3626 ;;; n, m = 0, 1, 2, ..., s > -1 and r > -1.
3627 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3631 ;;; Recognize c*z^w*log(z)^m*exp(-t^s)
3643 ;;; Recognize c*z^r*log(z)^n*(1-z)^s*log(1-z)^m
3656 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3663 ;; var1 is used as a parameter for differentiation. Add var1>0 to the
3664 ;; database, to get the desired simplification of the differentiation of
3665 ;; the gamma_incomplete function.
3673 ;; The integrand matches the cases 1 and 2.
3692 ;; Case 1: Generated by the Gamma function.
3699 var1
3704 ; derivate the involved hypergeometric
3705 ; functions.
3710 ;; Case 2: Generated by the Incomplete Gamma function.
3726 ;; Case 3: Generated by the Beta function.
3752 var2 m)
3754 var1 n)
3760 ;; Simplify result and set $gamma_expand to global value
3763 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;