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- Xref: sparky sci.math:14679 sci.math.symbolic:2929
- Newsgroups: sci.math,sci.math.symbolic
- Path: sparky!uunet!wri!news
- From: victor@tuamotu.wri.com (Victor Adamchik)
- Subject: Re: Help wanted in integration.
- Message-ID: <1992Nov10.031024.4001@wri.com>
- Sender: news@wri.com
- Nntp-Posting-Host: tuamotu.wri.com
- Organization: Wolfram Research, Inc.
- References: <Nov.6.00.08.18.1992.2647@gandalf.rutgers.edu>
- Date: Tue, 10 Nov 1992 03:10:24 GMT
- Lines: 52
-
- In article <Nov.6.00.08.18.1992.2647@gandalf.rutgers.edu>
- amarmahb@gandalf.rutgers.edu (Amar Mahboob Ali) writes:
- >
- > Hi
- >
- > Can anyone please help me in integrating the following.
- >
- > infinity
- >
- > / 4 2 2
- > | x sin (Pi a x) sin (Pi b x)
- > | ---------------------------- dx
- > | 2 2 2 2 2 2
- > / (x - d ) (x - c )
- >
- > -infinity
- >
- >
- > Where a,b,c and d are positive intergers.
- >
- > This function has double poles on the real axis. Hence theorems
- > related to computing such definite integrals dont seem to help me,
- > as they allow at the most a simple pole on the real axis. Is there
- > some other theorem that I can use? I would appretiate the least bit of
- > help on this.
- >
- > I am begining to think that there is no closed form solution.
- > Please help.
- >
- > Thanks
- >
- > Amar
-
- I evaluated your integral in the closed form and
- hope the following answer is a correct for integer a,b,c and d:
-
- if 0 < a <= b then (a*(c^2 + d^2)*Pi^2)/(16*(c^2 - d^2)^2)
- if 0 < b <= a then (b*(c^2 + d^2)*Pi^2)/(16*(c^2 - d^2)^2)
-
- I checked numerically (the precision was 6 digits) it for
- c = 2; d = 1; a = 3; b = 4
- c = 2; d = 1; a = 7; b = 4
- and
- c = 2; d = 4; a = 3; b = 4
-
- If you are interested to look at the proof send me email. I have
- used Mathematica to get that result.
-
- --
- Victor Adamchik
- victor@wri.com
-
-