4

Show that $$\int_0^\infty \frac{\sin x}{e^x-1}\,dx=\sum_{n=1}^\infty \frac{1}{n^2+1}.$$

Thoughts: I think I have to use the dominated convergence theorem, but I don't see how.. I tried expanding $\frac{1}{1-e^x}=1+e^x+e^{2x}+\ldots$ but then realised this works only $|e^x|<1$.

Math1000
  • 36,983
Dr. Heinz Doofenshmirtz
  • 2,360
  • $| e^{x} | = e^{x} < 1$ means $x < 0$, so its not applicable to the integral in the form written. Maybe you can write it into an integral of $(-\infty,0)$? Where does this problem come from? – ViktorStein Nov 26 '19 at 22:41
  • Also notice that at integer values, the functions $\frac{\sin(x)}{e^x + 1}$ and $\frac{1}{x^2 + 1}$ take very similar values. Maybe you can use that integration is the limit of Riemann sums. – ViktorStein Nov 26 '19 at 22:48
  • 2
    Hint: $\frac{1}{e^x-1}=\frac{e^{-x}}{1-e^{-x}}$ – Dispersion Nov 26 '19 at 23:02

1 Answers1

3

We have $1/(e^x-1)=e^{-x}/(1-e^{-x})$, so we can write $$\frac{\sin x}{e^x-1}=\sum_{n\ge1}e^{-nx}\sin(x).$$ Thus $$I=\int_0^{\infty}\frac{\sin x}{e^x-1}dx=\sum_{n\ge1}\int_0^\infty e^{-nx}\sin(x)dx.$$ For this final integral, use $\sin(x)=\Im (e^{ix})$, so that $$I=\sum_{n\ge1}\int_0^\infty e^{-nx}\Im( e^{ix})dx=\sum_{n\ge1}\Im\int_0^\infty e^{-(n-i)x}dx=\sum_{n\ge1}\Im\left(\frac{1}{n-i}\right).$$ It is then easy to show that $$\Im\left(\frac{1}{n-i}\right)=\frac{1}{n^2+1}.$$

clathratus
  • 17,161