Calculate int_0^infinity x^3/(e^x -1) dx

[Benny]

Hi, can someone help me out with this question.

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

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

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

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

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

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.

 

[siddharth]

The integral is
$$S = \int\limits_0^\infty {\frac{{x^3 }}{{e^x - 1}}} dx$$
which you wrote as
$$\int\limits_0^\infty {\frac{x^3 e^{-x}}{1-e^{-x}} dx$$
$$\int\limits_0^\infty \left(x^3 e^{-x}\right)\left(1-e^{-x} \right)^{-1} dx$$
Now, expand
$$\left(1-e^{-x}\right)^{-1}$$
with the binomial theorem (don't use the taylor expansion for [itex]e^{-x}[/tex] in the numerator like you did), and integrate term by term (ie, integrate the nth term and sum it. You'll need to repeatedly use integration by parts). Then you'll be able to evaluate the resulting sum using the identity you proved.

 

[Benny]

Thanks for the response. I found out how to do this question from someone a day or two ago. Since 0 < exp(-x) < 1 for positive x I can use the geometric series to get a nice expression for the integrand. Then all I need to do is integrate by parts 3 times and switch the order of summation and integration. There's a slight error in my working at the moment but I should be able to fix it up.