# Can I flip the integral and sum here?

I have the convergent integral-sum:

$$\int_0^\infty \sum_{n\mathop=0}^\infty \frac {x^{4n+1}} {e^x - 1} \frac {(-1)^n} {(2n)!(4\pi)^{2n}}\mathrm d x$$

But is it the same as this?:

$$\sum_{n\mathop=0}^\infty \int_0^\infty \frac {x^{4n+1}} {e^x - 1} \frac {(-1)^n} {(2n)!(4\pi)^{2n}}\mathrm d x$$

We don't look at Fubini or Tonelli's theorems at high school so I'm not confident with it, but I don't believe Tonelli applies here as we have the $(-1)^n$ and I don't think Fubini works either because the absolute value wouldn't converge. Does this mean the integral and sum cannot be interchanged in the above case?

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We have (for $\alpha > 1$)

$$\int_0^\infty \frac{x^{\alpha-1}}{e^x-1}\,dx = \Gamma(\alpha)\zeta(\alpha),$$

and thus

$$\left\lvert \frac{(-1)^n}{(2n)!(4\pi)^{2n}}\int_0^\infty \frac{x^{4n+1}}{e^x-1}\,dx\right\rvert = \frac{(4n+1)!\zeta(4n+1)}{(2n)!(4\pi)^{2n}},$$

which does not converge to $0$, hence

$$\sum_{n=0}^\infty \int_0^\infty \frac{x^{4n+1}}{e^x-1}\frac{(-1)^n}{(2n)!(4\pi)^{2n}}\,dx$$

diverges.

Since

$$\int_0^\infty\sum_{n=0}^\infty \frac{x^{4n+1}}{e^x-1}\frac{(-1)^n}{(2n)!(4\pi)^{2n}}\,dx = \int_0^\infty \frac{x}{e^x-1} \cos \left(\frac{x^2}{4\pi}\right)\,dx$$

is finite, you cannot interchange summation and integration.

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Thanks for clarifying! – Superbus May 3 '14 at 16:17