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MathGroup Archive 1999

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Re: Tough Integral

  • To: mathgroup at smc.vnet.net
  • Subject: [mg20285] Re: Tough Integral
  • From: "Bill Bertram" <wkb at ansto.gov.au>
  • Date: Mon, 11 Oct 1999 02:19:56 -0400
  • Organization: Australian Nuclear Science and Technology Organisation
  • References: <7tp7fj$95g@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Lawrence Walker wrote in message <7tp7fj$95g at smc.vnet.net>...
>Mathematica is able to calculate the following integral.
>
>In[1]:=
>Integrate[x^3/(E^x-1),{x,0,Infinity}]
>Out[1]=
>Pi^4/15
>
>For the life of me, I cannot solve this by hand.  Does any
>know or has any ideas.
>

Yes, I have one.

Rewrite the integral as


Integrate[ E^(-x) x^3/(1-E^(-x)),{x,0,Infinity}

Now  expand the 1/(1-E^(-x)) factor as a power series,  Sum[ E^(-k
x),{k,0,Infinity}} and integrate the result term by term. You will end up
with a sum of reciprical powers,( ie. a Zeta function) which gives you the
desired numerical result.

Cheers,
    Bill




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