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I got into trouble when I tried to calculate this integral while doing a quantum field theory ( QFT ) problem: $$ \int_{-\infty}^{+\infty} \sin\left(\,{a\cosh\left(\,{x}\,\right)}\right)\,{\rm d}x $$ The value is $\pi\operatorname{J}_{0}\left(\,{a}\,\right)$, where $\operatorname{J}_{n}\left(\,{x}\,\right)$ is the Bessel function of the first kind.

How can I prove it analytically?

Felix Marin
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PyroTechnics
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    There is a problem somewhere : either there is a typo in the integrand or the answer is wrong – Claude Leibovici Sep 19 '22 at 10:26
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    Agreed with @ClaudeLeibovici Checked also via W. Mathematica. For example, if $a = 1$, the numerical integration gives $1.821(...)$ where as $\pi J_0(1) = 2.4039(...)$ – Enrico M. Sep 19 '22 at 10:40
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    @Laplacian and @Claude Leibovici the answer is:$\pi J_0(a)$ is correct, try: NIntegrate[Sin[Cosh[x]], {x, -1000, 1000}, WorkingPrecision -> 30, MaxRecursion -> 20] for: $a=1$ – Mariusz Iwaniuk Sep 19 '22 at 13:23
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    Using only Mathematica: InverseLaplaceTransform[ Integrate[ LaplaceTransform[Sin[a Cosh[x]], a, s], {x, -Infinity, Infinity}, Assumptions -> s > 0], s, a] we have:\[Pi] BesselJ[0, a]. – Mariusz Iwaniuk Sep 19 '22 at 13:27
  • Yes, I found the same problem in Mathematica. Seems that Mathematica could run into unexpected problems while dealing with such oscillating but integrable functions. How can I prove this expression analytically? – PyroTechnics Sep 19 '22 at 13:29
  • Got it. Using Laplace transform according to a and the inverse. Thanks @MariuszIwaniuk – PyroTechnics Sep 19 '22 at 13:40
  • @MariuszIwaniuk Oh, that was unexpected! Thank you!! – Enrico M. Sep 19 '22 at 15:00
  • @MariuszIwaniuk. Thanks for your comment – Claude Leibovici Sep 20 '22 at 02:06

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