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From: makdisi@yale.ARPA (Makdisi)
Newsgroups: net.math,net.physics
Subject: Re: value of an integral
Message-ID: <896@yale.ARPA>
Date: Sun, 23-Feb-86 01:38:45 EST
Article-I.D.: yale.896
Posted: Sun Feb 23 01:38:45 1986
Date-Received: Wed, 26-Feb-86 07:58:12 EST
References: <823@drux2.UUCP>
Reply-To: makdisi@yale-cheops.UUCP (Kamal Khuri-Makdisi)
Organization: Yale University CS Dept., New Haven CT
Lines: 37
Xref: linus net.math:2515 net.physics:3612
Summary: integral of s^3/[exp(s)-1] by elementary means -- expand into series

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The integral of s^3/[exp(s)-1] from 0 to infinity can be calculated by
elementary means (and a little trickery), except that one needs the result that
the sum of 1/n^4, n = 1 to infinity, is pi^2/90.  Here we go:

Write the integrand as s^3*exp(-s)/[1-exp(-s)], and then expand 1/[1-exp(-s)]
into 1 + exp(-s) + exp(-2s) + exp(-ns) + ... .  This is justified since s > 0,
so exp(-s) < 1 .  The integrand will then be the sum of s^3*exp(-n*s), n = 1 to
infinity (remember the exp(-s) term in the numerator).  What remains to be done
is to show that the integral of each term in the integrand is 6/n^4, so that the
sum of the integrals of the terms is 6*pi^2/90, or pi^2/15.  In fact, the
indefinite integral of s^3*exp(-n*s) is
	   3     2
       /  s    3s    6s     6   \   -ns
    - (  --- + --- + --- + ---   ) e    + C          [note the minus sign!]
       \         2     3     4  /
	  n     n     n     n
from integrating by parts three times.  The definite integral of each term from
0 to infinity is therefore 6/n^4, since as s tends to plus infinity, exp(-ns)
"outdoes" the polynomial, and the product tends to 0.

The only proof I know that the sum of 1/n^4, n = 1 to infinity, is pi^2/90
involves Fourier series -- anyone know a more elementary way of proving this?

P.S.  1/[1-exp(-s)] = 1 + exp(-s) + [exp(-s)]^2 + [exp(-s)]^3 + ...
      =  1 + exp(-s) + exp(-2s) + exp(-3s) + ... , just in case that step wasn't
      clear.
--
						Kamal Khuri-Makdisi
						makdisi@yale-cheops.UUCP

      LAMAKDISIDKAMAL         LAMAKDISIDKAMAL         LAMAKDISIDKAMAL
-- 
						Kamal Khuri-Makdisi
						makdisi@yale-cheops.ARPA

      LAMAKDISIDKAMAL         LAMAKDISIDKAMAL         LAMAKDISIDKAMAL