| blob | 59547caeaa4ce34d262005efe7f7c99ad86ab46e |
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 $ind)
402 ans))))
404 ;; rewrites exp, the integrand in terms of var,
405 ;; into exp1, the integrand in terms of 'yx.
426 ;; D (not used? -- cwh)
439 ;; Look for integrals which involve log and exp functions.
440 ;; Maxima has a special algorithm to get general results.
451 %tan %sinh %cosh %tanh
452 %log %asin %acos %atan
453 %cot %acot %sec
454 %asec %csc %acsc
456 ;; Call ANTIDERIV with logabs disabled,
457 ;; because the Risch algorithm assumes
458 ;; the integral of 1/x is log(x), not log(abs(x)).
459 ;; Why not just assume logabs = false within RISCHINT itself?
460 ;; Well, there's at least one existing result which requires
461 ;; logabs = true in RISCHINT, so try to make a minimal change here instead.
516 ;;Should be a PROG inside of unwind-protect, but Multics has a compiler
517 ;;bug wrt. and I want to test this code now.
525 ;;;This seems((and (and (eq ul '$inf)
526 ;;;fairly losing (setq exp (subin (m+ ll var) exp))
527 ;;; (setq ll 0.))
528 ;;; (ztoinf exp var)))
533 ;; ((and (and (equal ul 1.)
534 ;; (equal ll 0.)) (zto1 exp)))
542 ;;;NOT COMPLETE for sin's and cos's.
564 ;; Call ANTIDERIV with logabs disabled,
565 ;; because the Risch algorithm assumes
566 ;; the integral of 1/x is log(x), not log(abs(x)).
567 ;; Why not just assume logabs = false within RISCHINT itself?
568 ;; Well, there's at least one existing result which requires
569 ;; logabs = true in RISCHINT, so try to make a minimal change here instead.
575 ;;;Recursion stopper
598 ;; adds up integrals of ranges between each pair of poles.
599 ;; checks if whole thing is divergent as limits of integration approach poles.
601 ;;; calling $logcontract causes antiderivative of 1/(1-x^5) to blow up
602 ;; (setq anti-deriv (cond ((involve anti-deriv '(%log))
603 ;; ($logcontract anti-deriv))
604 ;; (t anti-deriv)))
615 ;;Return section.
640 pole-list)
642 ;; Assumes EXP is a rational expression with no polynomial part and
643 ;; converts the finite integration to integration over a half-infinite
644 ;; interval. The substitution y = (x-a)/(b-x) is used. Equivalently,
645 ;; x = (b*y+a)/(y+1).
646 ;;
647 ;; (I'm guessing CV means Change Variable here.)
650 ;; FIXME! This is a hack. We apply the transformation with
651 ;; symbolic limits and then substitute the actual limits later.
652 ;; That way method-by-limits (usually?) sees a simpler
653 ;; integrand.
654 ;;
655 ;; See Bugs 938235 and 941457. These fail because $FACTOR is
656 ;; unable to factor the transformed result. This needs more
657 ;; work (in other places).
661 ;; If the limit is a number, use $substitute so we simplify
662 ;; the result. Do we really want to do this?
670 ()))
672 ;; Integrate rational functions over a finite interval by doing the
673 ;; polynomial part directly, and converting the rational part to an
674 ;; integral from 0 to inf. This is evaluated via residues.
677 ;; PQR divides the rational expression and returns the quotient
678 ;; and remainder
685 ;; No polynomial part
688 ans
691 ;; Only polynomial part
694 ;; A non-zero quotient and remainder. Combine the results
695 ;; together.
699 ans))
704 ;; I think this takes a rational expression E, and finds the
705 ;; polynomial part. A cons is returned. The car is the quotient and
706 ;; the cdr is the remainder.
731 ;;;If leadop isn't plus don't do anything.
741 t)
747 t)
802 ;; *Z* is a "zero parameter" for this package.
803 ;; Its also used to transform.
804 ;; limit(exp,var,val,dir) -- limit(exp,tvar,0,dir)
811 ;; EPSILON is used in principal value code to denote the familiar
812 ;; mathematical entity.
817 ;;; PRIN-INF Is a special symbol in the principal value code used to
818 ;;; denote an end-point which is proceeding to infinity.
841 ; We have minf <= ll = ul <= inf
842 )
844 ; We have minf <= ll < ul = inf
845 )
847 ; We have minf = ll < ul < inf
848 ;
849 ; Now substitute
850 ;
851 ; var -> -var
852 ; ll -> -ul
853 ; ul -> inf
854 ;
855 ; so that minf < ll < ul = inf
861 ; We have minf <= ul < ll
862 ;
863 ; Now substitute
864 ;
865 ; exp -> -exp
866 ; ll <-> ul
867 ;
868 ; so that minf <= ll < ul
871 t)))
885 ;; Substitute a and b into integral e
886 ;;
887 ;; Looks for discontinuties in integral, and works around them.
888 ;; For example, in
889 ;;
890 ;; integrate(x^(2*n)*exp(-(x)^2),x) ==>
891 ;; -gamma_incomplete((2*n+1)/2,x^2)*x^(2*n+1)*abs(x)^(-2*n-1)/2
892 ;;
893 ;; the integral has a discontinuity at x=0.
894 ;;
901 var a b)))))
915 total))))))
917 ;; look for terms with a negative exponent
918 ;;
919 ;; recursively traverses exp in order to find discontinuities such as
920 ;; erfc(1/x-x) at x=0
930 ;; returns list of places where exp might be discontinuous in var.
931 ;; list begins with ll and ends with ul, and include any values between
932 ;; ll and ul.
933 ;; return '$no or '$unknown if no discontinuities found.
950 ;; (multiplicity (cdar dummy)) ;; not used
955 ;; Carefully substitute the integration limits A and B into the
956 ;; expression E.
966 ;; Try easy substitutions. Return NIL if we can't.
969 ;; Infinite limits aren't easy
970 nil)
974 %gamma_incomplete %expintegral_ei)))
976 ;; It's easy if we have a polynomial. I (rtoy) think
977 ;; it's also easy if the denominator is free of the
978 ;; integration variable and also if the expression
979 ;; doesn't involve inverse functions.
980 ;;
981 ;; gamma_incomplete and expintegral_ie
982 ;; included because of discontinuity in
983 ;; gamma_incomplete(0, exp(%i*x)) and
984 ;; expintegral_ei(exp(%i*x))
985 ;;
986 ;; XXX: Are there other cases we've forgotten about?
987 ;;
988 ;; So just try to substitute the limits into the
989 ;; expression. If no errors are produced, we're done.
994 ;; no-err-sub has returned T. An error was catched.
995 nil)
1009 ;; check for divergent integral
1021 ;;;This function works only on things with ATAN's in them now.
1023 ;; POLES-IN-INTERVAL doesn't know about the poles of tan(x). Call
1024 ;; trigsimp to convert tan into sin/cos, which POLES-IN-INTERVAL
1025 ;; knows how to handle.
1026 ;;
1027 ;; XXX Should we fix POLES-IN-INTERVAL instead?
1028 ;;
1029 ;; XXX Is calling trigsimp too much? Should we just only try to
1030 ;; substitute sin/cos for tan?
1031 ;;
1032 ;; XXX Should the result try to convert sin/cos back into tan? (A
1033 ;; call to trigreduce would do it, among other things.)
1036 ;;POLES -> ((mlist) ((mequal) ((%atan) foo) replacement) ......)
1037 ;;We can then use $SUBSTITUTE
1042 ul-ans)
1044 nil)))
1058 var ll ul)))
1075 llist)))))))))))))))
1096 nil)
1115 ;; e is of form poly(x)*exp(m*%i*x)
1116 ;; s is degree of denominator
1117 ;; adds e to bptu or bptd according to sign of m
1132 ;; check term is of form poly(x)*exp(m*%i*x)
1133 ;; n is degree of denominator
1146 term)
1154 term))
1186 ;; exponentialize sin and cos
1192 '$%i
1212 exp)
1228 ;; computes integral from minf to inf for expressions of the form
1229 ;; exp(%i*m*x)*r(x) by residues on either the upper half
1230 ;; plane or the lower half plane, depending on whether
1231 ;; m is positive or negative. [wang p. 77]
1232 ;;
1233 ;; exponentializes sin and cos before applying residue method.
1234 ;; can handle some expressions with poles on real line, such as
1235 ;; sin(x)*cos(x)/x.
1240 ratterms ratans
1241 plf bptu bptd s upans downans)
1246 var
1247 nil)
1250 ;;;Above tests for applicability of this method.
1259 ;; call to $expand can create rational terms
1260 ;; with no exp(m*%i*x)
1264 ;;;Function RIB sets up the values of BPTU and BPTD
1271 ;;;If there is BPTU then CSEMIUP must succeed.
1272 ;;;Likewise for BPTD.
1276 ;; The original integrand was already
1277 ;; determined to have no poles by initial-analysis.
1278 ;; If individual terms of the expansion have poles, the poles
1279 ;; must cancel each other out, so we can ignore them.
1282 ;; if integral of ratterms is divergent, ratans is nil,
1283 ;; and mtosc returns nil
1300 t)
1302 t)
1310 t)
1312 t)
1317 s nc dc
1318 ans r $savefactors checkfactors temp test-var)
1340 ;;;
1341 ;;;New section.
1369 findout
1373 on
1395 s)
1400 ;; Let's not signal an error here. Return nil so that we
1401 ;; eventually return a noun form if no other algorithm gives
1402 ;; a result.
1405 nil)))
1497 ;; We have an integral of the form p(x)*F(exp(x)), where
1498 ;; p(x) is a polynomial.
1500 ;; No polynomial
1504 ;; These limits must exist for the integral to converge.
1507 ;; This only handles the case when the F(z) is a
1508 ;; rational function.
1511 ;; If we get here, F(z) is not a rational function.
1512 ;; We transform it using the substitution x=log(y)
1513 ;; which gives us an integral of the form
1514 ;; p(log(y))*F(y)/y, which maxima should be able to
1515 ;; handle.
1518 ;; Give up. We don't know how to handle this.
1520 en
1534 exp)))
1536 ;;; given (b*x+a)^n+c returns (a b n c)
1544 nil)
1546 ;;;get high degree of poly
1548 ;;;diff down to linear.
1550 ;;;all the way to constant.
1553 ;;;get rid of factorial from differentiation.
1556 ;;;Sees if can be expressed as (a*x+b)^n + part freeof x.
1565 s)
1590 n
1591 d)))))
1594 on
1649 e nil)
1663 e))))))
1669 ;; Compute the integral(n/d,x,0,inf) by computing the negative of the
1670 ;; sum of residues of log(-x)*n/d over the poles of n/d inside the
1671 ;; keyhole contour. This contour is basically an disk with a slit
1672 ;; along the positive real axis. n/d must be a rational function.
1677 ;; Ok if not on the positive real axis.
1682 t)
1693 ;; Look at an expression e of the form sin(r*x)^k, where k is an
1694 ;; integer. Return the list (1 r k). (Not sure if the first element
1695 ;; of the list is always 1 because I'm not sure what partition is
1696 ;; trying to do here.)
1710 ;; Look at an expression e of the form sin(r*x) and return r.
1721 ;; integrate(a*sc(r*x)^k/x^n,x,0,inf).
1724 ;; Get the argument of the involved trig function.
1727 ;; I don't think this needs to be special.
1728 #+nil
1730 ;; Replace (1-cos(arg)^2) with sin(arg)^2.
1732 ;(m+t 1. (m- (m^t `((%cos) ,var) 2.)))
1733 ;; The code from above generates expressions with
1734 ;; a missing simp flag. Furthermore, the
1735 ;; substitution has to be done for the complete
1736 ;; argument of the trig function. (DK 02/2010)
1739 exp))
1744 ;; n is the power of the denominator.
1746 ;; The simple case.
1750 ;; Do this for a sum of such terms.
1752 c)))))))))
1754 ;; We have an integral of the form sin(r*x)^k/x^n. C is the list (1 r k).
1755 ;;
1756 ;; The substitution y=r*x converts this integral to
1757 ;;
1758 ;; r^(n-1)*integral(sin(y)^k/y^n,y,0,inf)
1759 ;;
1760 ;; (If r is negative, we need to negate the result.)
1761 ;;
1762 ;; The recursion Wang gives on p. 87 has a typo. The second term
1763 ;; should be subtracted from the first. This formula is given in G&R,
1764 ;; 3.82, formula 12.
1765 ;;
1766 ;; integrate(sin(x)^r/x^s,x) =
1767 ;; r*(r-1)/(s-1)/(s-2)*integrate(sin(x)^(r-2)/x^(s-2),x)
1768 ;; - r^2/(s-1)/(s-2)*integrate(sin(x)^r/x^(s-2),x)
1769 ;;
1770 ;; (Limits are assumed to be 0 to inf.)
1771 ;;
1772 ;; This recursion ends up with integrals with s = 1 or 2 and
1773 ;;
1774 ;; integrate(sin(x)^p/x,x,0,inf) = integrate(sin(x)^(p-1),x,0,%pi/2)
1775 ;;
1776 ;; with p > 0 and odd. This latter integral is known to maxima, and
1777 ;; it's value is beta(p/2,1/2)/2.
1778 ;;
1779 ;; integrate(sin(x)^2/x^2,x,0,inf) = %pi/2*binomial(q-3/2,q-1)
1780 ;;
1781 ;; where q >= 2.
1782 ;;
1784 ;; Compute sign(r)*r^(n-1)*integrate(sin(y)^k/y^n,y,0,inf)
1786 c
1792 recursion)
1793 ;; Recursion failed. Return the integrand
1794 ;; The following code generates expressions with a missing simp flag
1795 ;; for the sin function. Use better simplifying code. (DK 02/2010)
1796 ; (let ((integrand (div (pow `((%sin) ,(m* r var))
1797 ; k)
1798 ; (pow var n))))
1800 k)
1805 ;; integrate(sin(x)^n/x^2,x,0,inf) = pi/2*binomial(n-3/2,n-1).
1806 ;; Express in terms of Gamma functions, though.
1812 ;; integrate(sin(x)^n/x,x,0,inf) = beta((n+1)/2,1/2)/2, for n odd and
1813 ;; n > 0.
1816 half%pi)
1819 ;; This implements the recursion for computing
1820 ;; integrate(sin(y)^l/y^k,y,0,inf). (Note the change in notation from
1821 ;; the above!)
1827 ;; Integral diverges if l-(k-1) < 0.
1830 ;; If l + k is not even, return NIL. (Is this the right
1831 ;; thing to do?)
1832 nil)
1834 ;; We have integrate(sin(y)^l/y^2). Use sevn to evaluate.
1837 ;; We have integrate(sin(y)^l/y,y)
1843 ;; j = k-2, i = (k-1)*(k-2)
1844 ;;
1845 ;;
1846 ;; The main recursion:
1847 ;;
1848 ;; i(sin(y)^l/y^k)
1849 ;; = l*(l-1)/(k-1)/(k-2)*i(sin(y)^(l-2)/y^k)
1850 ;; - l^2/(k-1)/(k-1)*i(sin(y)^l/y^(k-2))
1858 ;; Returns the fractional part of a?
1862 ;; Why do we return 0 if a is a number? Perhaps we really
1863 ;; mean integer?
1866 ;; If we're here, this basically assumes a is a rational.
1867 ;; Compute the remainder and return the result.
1885 ;;; THE FOLLOWING FUNCTION IS FOR TRIG FUNCTIONS OF THE FOLLOWING TYPE:
1886 ;;; LOWER LIMIT=0 B A MULTIPLE OF %PI SCA FUNCTION OF SIN (X) COS (X)
1887 ;;; B<=%PI2
1891 ;; means there was no error
1894 ; returns cons of (integer_part . fractional_part) of a
1896 ;; I think we really want to compute how many full periods are in a
1897 ;; and the remainder.
1902 ; returns cons of (integer_part . fractional_part) of a
1904 ;; I think we really want to compute how many full periods are in a
1905 ;; and the remainder.
1912 ;; Return the integer part of r.
1914 ;; r - fpart(r)
1918 ;;;Try making exp(%i*var) --> yy, if result is rational then do integral
1919 ;;;around unit circle. Make corrections for limits of integration if possible.
1939 rat-form)))
1944 ;;; Do integrals of sin and cos. this routine makes sure lower limit
1945 ;;; is zero.
1947 ;; integrate(e,var,a,b)
1963 var e)
1964 cc))
1974 ;; Exit if b or a is not a constant or if the integrand
1975 ;; doesn't appear to have a period of 2 pi.
1978 ;; Multiples of 2*%pi in limits.
1985 ;; The integral is not over a full period (2*%pi) or multiple
1986 ;; of a full period.
1988 ;; Wang p. 111: The integral integrate(f(x),x,a,b) can be
1989 ;; written as:
1990 ;;
1991 ;; n * integrate(f,x,0,2*%pi) + integrate(f,x,0,c)
1992 ;; - integrate(f,x,0,d)
1993 ;;
1994 ;; for some integer n and d >= 0, c < 2*%pi because there exist
1995 ;; integers p and q such that a = 2 * p *%pi + d and b = 2 * q
1996 ;; * %pi + c. Then n = q - p.
1998 ;; Compute q and c for the upper limit b.
2005 ;; Compute p and d for the lower limit a.
2007 ;; avoid an extra trip around the circle - helps skip principal values
2012 ;; Compute -integrate(f,x,0,d)
2017 ;; Compute n = q - p (stored in nzp2)
2019 out
2020 ;; Compute n*integrate(f,x,0,2*%pi)
2022 ;; n = 0
2025 ;; n is not zero, so compute
2026 ;; integrate(f,x,0,2*%pi)
2028 ;; Unable to compute integrate(f,x,0,2*%pi)
2030 ;; Compute integrate(f,x,0,c)
2034 ;; All three pieces succeeded.
2037 ;; Less than 1 full period, so intsc can integrate it.
2038 ;; Apply the substitution to make the lower limit 0.
2039 ;; This is last resort because substitution often causes intsc to fail.
2041 limit-diff var))
2042 ;; nothing worked
2045 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)).
2046 ;; calls intsc with a wrapper to just return nil if integral is divergent,
2047 ;; rather than generating an error.
2052 nil
2053 ans)))
2055 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)). I (rtoy)
2056 ;; think this expects b to be less than 2*%pi.
2059 0
2066 ;; Partition the integrand SC into the factors that do not
2067 ;; contain VAR (the car part) and the parts that do (the cdr
2068 ;; part).
2073 ;; integrate(sc, var, 0, b), where sc is f(sin(x), cos(x)).
2075 ;; Determine if sc is a product of sin's and cos's.
2079 ;; We have a product of sin's and cos's. We handle some
2080 ;; special cases here.
2082 ;; Wang p. 110, formula (1):
2083 ;; integrate(sin(x)^m*cos(x)^n, x, 0, %pi/2) =
2084 ;; gamma((m+1)/2)*gamma((n+1)/2)/2/gamma((n+m+2)/2)
2087 ;; Wang p. 110, near the bottom, says
2088 ;;
2089 ;; int(f(sin(x),cos(x)), x, 0, %pi) =
2090 ;; int(f(sin(x),cos(x)) + f(sin(x),-cos(x)),x,0,%pi/2)
2111 ;; Wang, p. 111 says
2112 ;;
2113 ;; int(f(sin(x),cos(x)),x,0,3*%pi/2) =
2114 ;; int(f(sin(x),cos(x)),x,0,%pi)
2115 ;; + int(f(-sin(x),-cos(x)),x,0,%pi/2)
2119 ;; We don't have a product of sin's and cos's.
2124 dn*)
2129 ;;;Is careful about substitution of limits where the denominator may be zero
2130 ;;;because of various assumptions made.
2149 nil
2150 ans)))
2156 nil
2157 ans)))
2159 ;; Determine whether E is of the form sin(x)^m*cos(x)^n and return the
2160 ;; list (m n).
2164 m n)
2183 ;; Says wether (m^t value exponent) has at least one real branch.
2184 ;; Only works for values of 1 and -1 now. Returns $yes $no
2185 ;; $unknown.
2196 ;; Compute beta((m+1)/2,(n+1)/2)/2.
2202 ;;Seems like Guys who call this don't agree on what it should return.
2215 ;; Check e for an expression of the form x^kk*(b*x^n+a)^l. If it
2216 ;; matches, Return the two values kk and (list l a n b).
2221 ;; We have f(x)^y. Look to see if f(x) has the desired
2222 ;; form. Then f(x)^y has the desired form too, with
2223 ;; suitably modified values.
2224 ;;
2225 ;; XXX: Should we ask for the sign of f(x) if y is not an
2226 ;; integer? This transformation we're going to do requires
2227 ;; that f(x)^y be real.
2229 e
2234 ;; Got a match. Adjust kk and cc appropriately.
2236 cc
2251 ;;(DEFUN BATAP (E)
2252 ;; (PROG (K C L)
2253 ;; (COND ((NOT (BATA0 E)) (RETURN NIL))
2254 ;; ((AND (EQUAL -1. (CADDDR C))
2255 ;; (EQ ($askSIGN (SETQ K (m+ 1. K)))
2256 ;; '$pos)
2257 ;; (EQ ($askSIGN (SETQ L (m+ 1. (CAR C))))
2258 ;; '$pos)
2259 ;; (ALIKE1 (CADR C)
2260 ;; (m^ UL (CADDR C)))
2261 ;; (SETQ E (CADR C))
2262 ;; (EQ ($askSIGN (SETQ C (CADDR C))) '$pos))
2263 ;; (RETURN (M// (m* (m^ UL (m+t K (m* C (m+t -1. L))))
2264 ;; `(($BETA) ,(SETQ NN* (M// K C))
2265 ;; ,(SETQ DN* L)))
2266 ;; C))))))
2269 ;; Integrals of the form i(log(x)^m*x^k*(a+b*x^n)^l,x,0,ul). There
2270 ;; are two cases to consider: One case has ul>0, b<0, a=-b*ul^n, k>-1,
2271 ;; l>-1, n>0, m a nonnegative integer. The second case has ul=inf, l < 0.
2272 ;;
2273 ;; These integrals are essentially partial derivatives of the Beta
2274 ;; function (i.e. the Eulerian integral of the first kind). Note
2275 ;; that, currently, with the default setting intanalysis:true, this
2276 ;; function might not even be called for some of these integrals.
2277 ;; However, this can be palliated by setting intanalysis:false.
2296 ;; We have i(x^kk/(d+cc*x^r)^s,x,0,inf) =
2297 ;; beta(aa,bb)/(cc^aa*d^bb*r). Compute this, and then
2298 ;; differentiate it m times to get the log term
2299 ;; incorporated.
2304 0
2305 m)))
2309 r))))
2318 ;; The result looks like B(k/n,l) ( ... ).
2319 ;; Perhaps, we should just $factor, instead of
2320 ;; pulling out beta like this.
2322 beta
2323 ($fullratsimp
2331 var m)
2333 beta))))))))))))
2336 ;;; If e is of the form given below, make the obvious change
2337 ;;; of variables (substituting ul*x^(1/n) for x) in order to reduce
2338 ;;; e to the usual form of the integrand in the Eulerian
2339 ;;; integral of the first kind.
2340 ;;; N. B: The old version of ZTO1 completely ignored this
2341 ;;; substitution; the log(x)s were just thrown in, which,
2342 ;;; of course would give wrong results.
2345 ;; Parse e
2349 ;; e=x^k*(a+b*x^n)^l
2351 c
2363 ;; Wang p. 71 gives the following formula for a beta function:
2364 ;;
2365 ;; integrate(x^(k-1)/(c*x^r+d)^s,x,0,inf)
2366 ;; = beta(a,b)/(c^a*d^b*r)
2367 ;;
2368 ;; where a = k/r > 0, b = s - a > 0, s > k > 0, r > 0, c*d > 0.
2369 ;;
2370 ;; This function matches this and returns k-1, d, r, c, a, b. And
2371 ;; also checks that all the conditions hold. If not, NIL is returned.
2372 ;;
2378 c
2395 ;; Handles beta integrals.
2403 nil)
2406 c
2407 ;; e = x^k*(d+c*x^al)^l.
2412 new-k)
2420 ;; Compute exp(d)*gamma((c+1)/b)/b/a^((c+1)/b). In essence, this is
2421 ;; the value of integrate(x^c*exp(d-a*x^b),x,0,inf).
2432 ;; Evaluates the contour integral of GRAND around the unit circle
2433 ;; using residues.
2439 ;; Is pt stricly inside the unit circle?
2445 ;; Is pt on the unit circle? (Use
2446 ;; the cached value computed
2447 ;; above.)
2468 ;; Wang 81-83. Unfortunately, the pdf version has page 82 as all
2469 ;; black, so here is, as best as I can tell, what Wang is doing.
2470 ;; Fortunately, p. 81 has the necessary hints.
2471 ;;
2472 ;; First consider integrate(exp(%i*k*x^n),x) around the closed contour
2473 ;; consisting of the real axis from 0 to R, the arc from the angle 0
2474 ;; to %pi/(2*n) and the ray from the arc back to the origin.
2475 ;;
2476 ;; There are no poles in this region, so the integral must be zero.
2477 ;; But consider the integral on the three parts. The real axis is the
2478 ;; integral we want. The return ray is
2479 ;;
2480 ;; exp(%i*%pi/2/n) * integrate(exp(%i*k*(t*exp(%i*%pi/2/n))^n),t,R,0)
2481 ;; = exp(%i*%pi/2/n) * integrate(exp(%i*k*t^n*exp(%i*%pi/2)),t,R,0)
2482 ;; = -exp(%i*%pi/2/n) * integrate(exp(-k*t^n),t,0,R)
2483 ;;
2484 ;; As R -> infinity, this last integral is gamma(1/n)/k^(1/n)/n.
2485 ;;
2486 ;; We assume the integral on the circular arc approaches 0 as R ->
2487 ;; infinity. (Need to prove this.)
2488 ;;
2489 ;; Thus, we have
2490 ;;
2491 ;; integrate(exp(%i*k*t^n),t,0,inf)
2492 ;; = exp(%i*%pi/2/n) * gamma(1/n)/k^(1/n)/n.
2493 ;;
2494 ;; Equating real and imaginary parts gives us the desired results:
2495 ;;
2496 ;; integrate(cos(k*t^n),t,0,inf) = G * cos(%pi/2/n)
2497 ;; integrate(sin(k*t^n),t,0,inf) = G * sin(%pi/2/n)
2498 ;;
2499 ;; where G = gamma(1/n)/k^(1/n)/n.
2500 ;;
2508 ;; Ok, we have cos(b*x^n) or sin(b*x^n), and we set e = (n
2509 ;; b)
2511 ;; n = 1. Give up. (Why not divergent?)
2512 nil)
2517 ;; s is the sign of b. Give up if it's zero.
2518 nil)
2520 ;; Give up if n-1 <= 0. (Why give up? Isn't the
2521 ;; integral divergent?)
2522 nil)
2524 ;; We can apply our formula now. g = gamma(1/n)/n/b^(1/n)
2531 e)))))))
2534 ;; this is the second part of the definite integral package
2541 ())
2543 var t)
2564 t)))))
2575 plm*))))
2594 ans))))
2602 d
2603 plm*)))))
2607 0
2610 ;; Handle integral(n(x)/d(x)*log(x)^m,x,0,inf). n and d are
2611 ;; polynomials.
2638 ans)))
2655 n
2659 ;;Makes a new poly such that np(x)-np(x+2*%i*%pi)=p(x).
2660 ;;Constructs general POLY of degree one higher than P with
2661 ;;arbitrary coeff. and then solves for coeffs by equating like powers
2662 ;;of the varibale of integration.
2663 ;;Can probably be made simpler now.
2671 ;;;Now, poly(x)-poly(x+2*%i*%pi)=p(x), P is the original poly.
2680 -1
2689 ;; Integrate(p(x)*f(exp(x))/g(exp(x)),x,minf,inf) by applying the
2690 ;; transformation y = exp(x) to get
2691 ;; integrate(p(log(y))*f(y)/g(y)/y,y,0,inf). This should be handled
2692 ;; by dintlog.
2699 ;; This implements Wang's algorithm in Chapter 5.2, pp. 98-100.
2700 ;;
2701 ;; This is a very brief description of the algorithm. Basically, we
2702 ;; have integrate(R(exp(x))*p(x),x,minf,inf), where R(x) is a rational
2703 ;; function and p(x) is a polynomial.
2704 ;;
2705 ;; We find a polynomial q(x) such that q(x) - q(x+2*%i*%pi) = p(x).
2706 ;; Then consider a contour integral of R(exp(z))*q(z) over a
2707 ;; rectangular contour. Opposite corners of the rectangle are (-R,
2708 ;; 2*%i*%pi) and (R, 0).
2709 ;;
2710 ;; Wang shows that this contour integral, in the limit, is the
2711 ;; integral of R(exp(x))*q(x)-R(exp(x))*q(x+2*%i*%pi), which is
2712 ;; exactly the integral we're looking for.
2713 ;;
2714 ;; Thus, to find the value of the contour integral, we just need the
2715 ;; residues of R(exp(z))*q(z). The only tricky part is that we want
2716 ;; the log function to have an imaginary part between 0 and 2*%pi
2717 ;; instead of -%pi to %pi.
2719 ;; We have R(exp(x))*p(x) represented as p(x)*pe(exp(x))/d(exp(x)).
2725 ;; At this point denom-exponential has converted d(exp(x)) to the
2726 ;; polynomial d(z), where z = exp(x).
2729 pe))
2732 ;; Find the poles of the denominator. denom-exponential is the
2733 ;; denominator of R(x).
2734 ;;
2735 ;; It seems as if polelist returns a list of several items.
2736 ;; The first element is a list consisting of the pole and (z -
2737 ;; pole). We don't care about this, so we take the rest of the
2738 ;; result.
2741 ;; The imaginary part is nonzero,
2742 ;; or the realpart is negative.
2746 ;; The realpart is not zero.
2748 ;; Not sure what this does.
2750 ;; No roots at all, so return
2754 ;; We have simple roots or roots in REGION1
2757 ;; The cadr of pl is the list of the simple poles of
2758 ;; denom-exponential. Take the log of them to find the
2759 ;; poles of the original expression. Then compute the
2760 ;; residues at each of these poles and sum them up and put
2761 ;; the result in B. (If no simple poles set B to 0.)
2767 ;; I think this handles the case of poles outside the
2768 ;; regions. The sum of these residues are placed in C.
2772 temp)))
2777 ;; We have the repeated poles of deonom-exponential, so we
2778 ;; need to convert them to the actual pole values for
2779 ;; R(exp(x)), by taking the log of the value of poles.
2784 ;; Compute the residues at all of these poles and sum
2785 ;; them up.
2788 poles))
2803 ;; Check to see if each term in exp that is of the form exp(k*x) has
2804 ;; an integer value for k.
2818 ;; Test to see if exp is of the form f(exp(x)), and if so, replace
2819 ;; exp(x) with 'z*. That is, basically return f(z*).
2823 exp)
2829 ;; Does e really need to be special here? Seems to be ok without
2830 ;; it; testsuite works.
2831 #+nil
2834 e)
2853 ;; Test to see if exp is of the form p(x)*f(exp(x)). If so, set p* to
2854 ;; be p(x) and set pe* to f(exp(x)).
2871 ;; I think this is looking at f(exp(x)) and tries to find some
2872 ;; rational function R and some number k such that f(exp(x)) =
2873 ;; R(exp(k*x)).
2890 ;; Integration by parts.
2891 ;;
2892 ;; integrate(u(x)*diff(v(x),x),x,a,b)
2893 ;; |b
2894 ;; = u(x)*v(x)| - integrate(v(x)*diff(u(x),x))
2895 ;; |a
2896 ;;
2898 ;;;SINCE ONLY CALLED FROM DINTLOG TO get RID OF LOGS - IF LOG REMAINS, QUIT
2902 nil)
2909 nil)
2917 ;; integrate(f(exp(k*x)),x,a,b), where f(z) is rational.
2918 ;;
2919 ;; See Wang p. 96-97.
2920 ;;
2921 ;; If the limits are minf to inf, we use the substitution y=exp(k*x)
2922 ;; to get integrate(f(y)/y,y,0,inf)/k. If the limits are 0 to inf,
2923 ;; use the substitution s+1=exp(k*x) to get
2924 ;; integrate(f(s+1)/(s+1),s,0,inf).
2931 ;; If we can integrate it directly, do so and take the
2932 ;; appropriate limits.
2933 )
2935 ;; ans is the list (f(x) exp(k*x)).
2938 ;; Use the substitution s + 1 = exp(k*x). The
2939 ;; integral becomes integrate(f(s+1)/(s+1),s,0,inf)
2942 ;; Use the substitution y=exp(k*x) because the
2943 ;; limits are minf to inf.
2945 ;; Apply the substitution and integrate it.
2948 ;; integrate(log(g(x))*f(x),x,0,inf)
2956 exp))
2958 ;; Make the substitution y=1/x. If the integrand has
2959 ;; exactly the same form, the answer has to be 0.
2965 ;; It's easy if we have the antiderivative.
2966 ;; but intsubs sometimes gives results containing %limit
2968 ;; Ok, the easy cases didn't work. We now try integration by
2969 ;; parts. Set ANS to f(x).
2972 ;; Bad. f(x) contains a log term, so we give up.
2978 var ll ul))))
2979 ;; The arg of the log function is the same as the
2980 ;; integration variable. We can do something a little
2981 ;; simpler than integration by parts. We have something
2982 ;; like f(x)*log(x). Consider f(x)*x^z. If we
2983 ;; differentiate this wrt to z, the integrand becomes
2984 ;; f(x)*log(x)*x^z. When we evaluate this at z = 0, we
2985 ;; get the desired integrand.
2986 ;;
2987 ;; So we need f(0) to be 0 at 0. If we can integrate
2988 ;; f(x)*x^z, then we differentiate the result and
2989 ;; evaluate it at z = 0.
2992 ;; Try integration by parts.
2995 ;; Compute diff(e,var,n) at the point pt.
2999 ;;; GGR and friends
3001 ;; MAYBPC returns (COEF EXPO CONST)
3002 ;;
3003 ;; This basically picks off b*x^n+a and returns the list
3004 ;; (b n a). It may also set the global *zd*.
3011 ;; At this point, e is of the form (a n b)
3016 ;; If we're here, b is complex, and n > 0. zn = imagpart(b).
3017 ;;
3018 ;; Set var to the same sign as zn.
3023 ;; zd = exp(var*%i*%pi*(1+nd)/(2*n). (ZD is special!)
3026 ;; Return zn, n, a.
3032 ;; We're here if realpart(b) <= 0, and n >= 0. Then return -b, n, a.
3035 ;; Integrate x^m*exp(b*x^n+a), with realpart(m) > -1.
3036 ;;
3037 ;; See Wang, pp. 84-85.
3038 ;;
3039 ;; I believe the formula Wang gives is incorrect. The derivation is
3040 ;; correct except for the last step.
3041 ;;
3042 ;; Let J = integrate(x^m*exp(%i*k*x^n),x,0,inf), with real k.
3043 ;;
3044 ;; Consider the case for k < 0. Take a sector of a circle bounded by
3045 ;; the real line and the angle -%pi/(2*n), and by the radii, r and R.
3046 ;; Since there are no poles inside this contour, the integral
3047 ;;
3048 ;; integrate(z^m*exp(%i*k*z^n),z) = 0
3049 ;;
3050 ;; Then J = exp(-%pi*%i*(m+1)/(2*n))*integrate(R^m*exp(k*R^n),R,0,inf)
3051 ;;
3052 ;; because the integral along the boundary is zero except for the part
3053 ;; on the real axis. (Proof?)
3054 ;;
3055 ;; Wang seems to say this last integral is gamma(s/n/(-k)^s) where s =
3056 ;; (m+1)/n. But that seems wrong. If we use the substitution R =
3057 ;; (y/(-k))^(1/n), we end up with the result:
3058 ;;
3059 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n).
3060 ;;
3061 ;; or gamma((m+1)/n)/k^((m+1)/n)/n.
3062 ;;
3063 ;; Note that this also handles the case of
3064 ;;
3065 ;; integrate(x^m*exp(-k*x^n),x,0,inf);
3066 ;;
3067 ;; where k is positive real number. A simple change of variables,
3068 ;; y=k*x^n, gives
3069 ;;
3070 ;; integrate(y^((m+1)/n-1)*exp(-y),y,0,inf)/(n*k^((m+1)/n))
3071 ;;
3072 ;; which is the same form above.
3091 ;; e = (m b n a). That is, the integral is of the form
3092 ;; x^m*exp(b*x^n+a). I think we want to compute
3093 ;; gamma((m+1)/n)/b^((m+1)/n)/n.
3094 ;;
3095 ;; FIXME: If n > m + 1, the integral converges. We need
3096 ;; to check for this.
3098 e
3101 ;; The derivation only holds if b is purely real or
3102 ;; purely imaginary. Give up if it's not.
3104 ;; Check for convergence. If b is complex, we need n -
3105 ;; m > 1. If b is real, we need b < 0.
3114 ;; If we're here, b must be negative.
3117 ;; Complex b. Take the imaginary part
3119 n a))
3120 ;; NOTE: *zd* (Ick!) is special and might be set by maybpc.
3122 ;; FIXME: Why do we set %emode here? Shouldn't we just
3123 ;; bind it? And why do we want it bound to T anyway?
3124 ;; Shouldn't the user control that? The same goes for
3125 ;; dosimp.
3126 ;;(setq $%emode t)
3132 ;; Match x^m*exp(b*x^n+a). If it does, return (list m b n a).
3137 ;; We're looking at something like exp(f(var)). See if it's
3138 ;; of the form b*x^n+a, and return (list 0 b n a). (The 0 is
3139 ;; so we can graft something onto it if needed.)
3143 ;; E should be the product of exactly 2 terms
3145 ;; Check to see if one of the terms is of the form
3146 ;; var^p. If so, make sure the realpart of p > -1. If
3147 ;; so, check the other term has the right form via
3148 ;; another call to ggr1.
3157 ;; Both terms have the right form and nn* contains the arg of
3158 ;; the exponential term. Put dn* as the car of nn*. The
3159 ;; result is something like (m b n a) when we have the
3160 ;; expression x^m*exp(b*x^n+a).
3164 ;; Match b*x^n+a. If a match is found, return the list (a n b).
3165 ;; Otherwise, return NIL
3178 ;; Match b*x^n. Return the list (n b) if found or NIL if not.
3198 ;;; given (b*x^n+a)^m returns (m a n b)
3211 ;;;Catches Unfactored forms.
3228 ;;;Is E = VAR raised to some power? If so return power or 0.
3234 ;;;Is E = VAR1 raised to some power? If so return power.
3255 ans)))))
3332 ;;;Looks at the IMAGINARY part of a log and puts it in the interval 0 2*%pi.
3335 ;; We take the $rectform above to make sure that the log is
3336 ;; expanded out for the situations where simplifying plog itself
3337 ;; doesn't do it. This should probably be considered a bug in the
3338 ;; plog simplifier and should be fixed there.
3351 ;;; Temporary fix for a lacking in taylor, which loses with %i in denom.
3352 ;;; Besides doesn't seem like a bad thing to do in general.
3356 ;; Multiply the denominator by it's conjugate to get rid of
3357 ;; %i.
3361 ;; If the new denominator still contains %i, just give
3362 ;; up. Otherwise, multiply the numerator by the
3363 ;; conjugate and divide by the new denominator.
3365 exp
3369 ;;; LL and UL must be real otherwise this routine return $UNKNOWN.
3370 ;;; Returns $no $unknown or a list of poles in the interval (ll ul)
3371 ;;; for exp w.r.t. var.
3372 ;;; Form of list ((pole . multiplicity) (pole1 . multiplicity) ....)
3389 ;; this clause handles cases where we can't find the exact roots,
3390 ;; but we know that they occur outside the interval of integration.
3391 ;; example: integrate ((1+exp(t))/sqrt(t+exp(t)), t, 0, 1);
3407 ;; (multiplicity (cdar dummy)) (not used? -- cwh)
3417 pole-list))))))))))))
3420 ;;;Returns $YES if there is no pole and $NO if there is one.
3433 ;;;Takes care of forms that the ratio test fails on.
3449 (%einvolve
3484 ;; real values for ll and ul; place can be imaginary.
3491 '$no
3496 ;; returns true or nil
3498 ;; real values for ll and ul; place can be imaginary.
3514 ;; If *failures is set, we may have missed some roots.
3515 ;; We still return the roots that we have found.
3520 rootlist)
3527 rootlist)))))))))
3530 ;;; Is x > y. X or Y can be $MINF or $INF, zeroA or zeroB.
3570 nil)
3573 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3574 ;;;
3575 ;;; Integrate Definite Integrals involving log and exp functions. The algorithm
3576 ;;; are taken from the paper "Evaluation of CLasses of Definite Integrals ..."
3577 ;;; by K.O.Geddes et. al.
3578 ;;;
3579 ;;; 1. CASE: Integrals generated by the Gamma function.
3580 ;;;
3581 ;;; inf
3582 ;;; /
3583 ;;; [ w m s - m - 1
3584 ;;; I t log (t) expt(- t ) dt = s signum(s)
3585 ;;; ]
3586 ;;; /
3587 ;;; 0
3588 ;;; !
3589 ;;; m !
3590 ;;; d !
3591 ;;; (--- (gamma(z))! )
3592 ;;; m !
3593 ;;; dz ! w + 1
3594 ;;; !z = -----
3595 ;;; s
3596 ;;;
3597 ;;; The integral converges for:
3598 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3599 ;;;
3600 ;;; 2. CASE: Integrals generated by the Incomplete Gamma function.
3601 ;;;
3602 ;;; inf !
3603 ;;; / m !
3604 ;;; [ w m s d s !
3605 ;;; I t log (t) exp(- t ) dt = (--- (gamma_incomplete(a, x ))! )
3606 ;;; ] m !
3607 ;;; / da ! w + 1
3608 ;;; x !z = -----
3609 ;;; s
3610 ;;; - m - 1
3611 ;;; s signum(s)
3612 ;;;
3613 ;;; The integral converges for:
3614 ;;; s # 0, m = 0, 1, 2, ... and realpart((w+1)/s) > 0.
3615 ;;; The shown solution is valid for s>0. For s<0 gamma_incomplete has to be
3616 ;;; replaced by gamma(a) - gamma_incomplete(a,x^s).
3617 ;;;
3618 ;;; 3. CASE: Integrals generated by the beta function.
3619 ;;;
3620 ;;; 1
3621 ;;; /
3622 ;;; [ m s r n
3623 ;;; I log (1 - t) (1 - t) t log (t) dt =
3624 ;;; ]
3625 ;;; /
3626 ;;; 0
3627 ;;; !
3628 ;;; ! !
3629 ;;; n m ! !
3630 ;;; d d ! !
3631 ;;; --- (--- (beta(z, w))! )!
3632 ;;; n m ! !
3633 ;;; dz dw ! !
3634 ;;; !w = s + 1 !
3635 ;;; !z = r + 1
3636 ;;;
3637 ;;; The integral converges for:
3638 ;;; n, m = 0, 1, 2, ..., s > -1 and r > -1.
3639 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3643 ;;; Recognize c*z^w*log(z)^m*exp(-t^s)
3655 ;;; Recognize c*z^r*log(z)^n*(1-z)^s*log(1-z)^m
3668 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
3675 ;; var1 is used as a parameter for differentiation. Add var1>0 to the
3676 ;; database, to get the desired simplification of the differentiation of
3677 ;; the gamma_incomplete function.
3685 ;; The integrand matches the cases 1 and 2.
3704 ;; Case 1: Generated by the Gamma function.
3711 var1
3716 ; derivate the involved hypergeometric
3717 ; functions.
3722 ;; Case 2: Generated by the Incomplete Gamma function.
3738 ;; Case 3: Generated by the Beta function.
3764 var2 m)
3766 var1 n)
3772 ;; Simplify result and set $gamma_expand to global value
3775 ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;