Hi, can someone help me out with this question.(adsbygoogle = window.adsbygoogle || []).push({});

Using [tex]f(t) = t^2 = \frac{{\pi ^2 }}{3} + 4\sum\limits_{n = 1}^\infty {\frac{{\left( { - 1} \right)^n }}{{n^2 }}} \cos \left( {nt} \right), - \pi < t < \pi [/tex] and Parseval's Theorem deduce the identity [tex]\sum\limits_{n = 1}^\infty {\frac{1}{{n^4 }}} = \frac{{\pi ^4 }}{{90}}[/tex].

Hence calculate [tex]S = \int\limits_0^\infty {\frac{{x^3 }}{{e^x - 1}}} dx

[/tex]

Hint: Expand the denominator as a power series in exp(-x) and integrate term by term.

I obtained the sum but I can't evaluate the integral.

If I multiply the numerator and the denominator by exp(-x) the integrand becomes

[tex]

\frac{{x^3 e^{ - x} }}{{1 - e^{ - x} }}

[/tex]

[tex]

= \frac{{\sum\limits_{k = 0}^\infty {\frac{{\left( { - 1} \right)^k }}{{k!}}} x^{k + 3} }}{{\sum\limits_{k = 1}^\infty {\frac{{\left( { - 1} \right)^{k + 1} }}{{k!}}} x^k }}

[/tex]

When I write out the first few terms in the summations, no terms appear to cancel because the starting index for the summations are different. I'm not sure how to do this question so any help would be good thanks.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Calculate int_0^infinity x^3/(e^x -1) dx

**Physics Forums | Science Articles, Homework Help, Discussion**