Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP
Posting-Version: version B 2.10.2 9/18/84; site faron.UUCP
Path: utzoo!linus!faron!bs
From: bs@faron.UUCP (Robert D. Silverman)
Newsgroups: net.math,net.physics
Subject: Re: value of an integral
Message-ID: <486@faron.UUCP>
Date: Mon, 24-Feb-86 12:28:08 EST
Article-I.D.: faron.486
Posted: Mon Feb 24 12:28:08 1986
Date-Received: Wed, 26-Feb-86 06:10:11 EST
References: <823@drux2.UUCP> <896@yale.ARPA>
Organization: The MITRE Coporation, Bedford, MA
Lines: 47
Xref: linus net.math:2516 net.physics:3613

> Expires:
> Followup-To:
> Distribution:
> Keywords:
> 
> 
> 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

One minor detail: One needs to establish that the series converges uniformly
in order to justify the term by term integration.

Bob Silverman