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

$n$ 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!

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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
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    $\begingroup$ 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

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