$$\int_0^{\infty} \frac{\sin(nx)e^{-anx}}{x^2-\pi^2}\,dx$$

$n$ is an integer and $a>0$.

I came across this integral while solving an another problem but I have no idea about evaluating it. I tried to use $\sin(nx)=\Im(e^{inx})$ but that doesn't help. Wolfram Alpha returns nothing.

Any help is appreciated. Many thanks!

  • $\begingroup$ Set $k=n(-a+i)$, then: $$\Im(\int_{0}^\infty\frac{e^{kx}}{x^2-\pi^2}dx)=\Im(\int_{0}^\infty\frac{e^{-anx}(\cos(nx)+i\sin(nx))}{x^2-\pi^2})$$ $$=\int_{0}^\infty\frac{\sin(nx)e^{-anx}}{x^2-\pi^2} dx$$ $\endgroup$ – Ethan Jun 8 '14 at 8:29
  • $\begingroup$ @Ethan: Yes, I know that but I don't think it is easy to solve $$\int_0^{\infty} \frac{e^{kx}}{x^2-\pi^2}\,dx$$ or maybe I am missing something? $\endgroup$ – Pranav Arora Jun 8 '14 at 8:32
  • $\begingroup$ Not sure, but you could always send $a\rightarrow \frac{a}{n}$ and then your problem is really just evaluating $$\int_{0}^\infty \frac{\sin(nx)e^{-ax}}{x^2-\pi^2} dx$$ Where $n$ is an integer and $a>0$, though I don't think that's necessarily easier to work with either. $\endgroup$ – Ethan Jun 8 '14 at 8:34

Based on results from Mathematica, I conjecture that the integral equals

$$ \frac{e^{-a n \pi }}{8\pi} \left[ 4 e^{2 a n \pi } \pi +2 i e^{2 a n \pi }\, \text{Ei}[-(-i+a) n \pi ]-2 i\, \text{Ei}[(-i+a) n \pi ]-2 i e^{2 a n \pi } \text{Ei}[-(i+a) n \pi ]+2 i\, \text{Ei}[(i+a) n \pi ] \right] $$ It seems that there is no simpler expression.

  • $\begingroup$ Darn! I expected a nicer result to use it for this problem: math.stackexchange.com/questions/822477/… . Can you please check if the derivative of the expression you wrote wrt $a$ results in a nice expression? Many thanks! $\endgroup$ – Pranav Arora Jun 8 '14 at 11:48
  • $\begingroup$ @Pranav The derivative is $\frac{1}{4} e^{-n a \pi } \left(n e^{2n a \pi } (2 \pi +i \text{Ei}[-n (-i+a) \pi ]-i \text{Ei}[-n (i+a) \pi ])+n i (\text{Ei}[n (-i+a) \pi ]-\text{Ei}[n (i+a) \pi ])\right) $ after simplification by Mathematica. $\endgroup$ – user111187 Jun 8 '14 at 11:51
  • $\begingroup$ Oops, that's no better. :P Anyways, I will check your post as the answer. BTW, can you please look at the problem I linked to? I saw on integrals and series board that someone solved it by residue calculus, any ideas about to solve the problem without residue calculus? $\endgroup$ – Pranav Arora Jun 8 '14 at 11:57
  • 1
    $\begingroup$ I looked at it but it looks very hard without residues. I might try it later when I have more time. $\endgroup$ – user111187 Jun 8 '14 at 12:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.