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This is what I tried:

$$\sum_{k=0}^{\infty}\int_{-\infty}^{\infty}\dfrac{(x-n\sinh^2 x)^k}{k!}\ dx$$

$$\sum_{k=0}^{\infty}\frac1{k!}\int_{-\infty}^{\infty}\sum_{r=0}^{k}\binom{k}{r}x^r(-n\sinh^2 x)^{k-r}\ dx$$

After this I have no idea how to proceed

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    $\begingroup$ FWIW, I tried $n=1,2,3$ in Wolfram Alpha and in each case I got $\sqrt{\frac\pi n}$ $\endgroup$
    – saulspatz
    Aug 13, 2021 at 15:51
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    $\begingroup$ You left out a factor of $(-1)^{k-r}$, but the individual integrals diverge, so this doesn't look like a viable approach. $\endgroup$
    – saulspatz
    Aug 13, 2021 at 16:34
  • $\begingroup$ ok ,i edited it if it's not viable approach then how should i proceed $\endgroup$
    – Alex Rubin
    Aug 13, 2021 at 18:10

1 Answer 1

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$$\ldots=\int_0^\infty(e^x+e^{-x})e^{-n\sinh^2 x}\,dx\ \underset{\sinh x=y}{\phantom{\big[}=\phantom{\big]}}\ 2\int_0^\infty e^{-ny^2}\,dy=\sqrt{\pi/n}.$$

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    $\begingroup$ Very nice trick. $\endgroup$
    – K.defaoite
    Aug 13, 2021 at 21:11

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