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This problem and its solution were found in a book by Cornel Ioan Valean (p. 206 ch. 3.41). \begin{align} \int_0^\infty \frac{\sin(\sin(x))e^{\cos(x)}}{x}dx=\frac{\pi}{2}(e-1) \end{align}

Its solution is quite clever, but seems to be a little bit quick. The problem is swiftly reduced to the following, after which the integral and the sum are switched.

\begin{align} \int_0^\infty \frac{1}{x}\sum_{n=1}^{\infty}\frac{\sin(xn)}{n!}dx=\sum_{n=1}^{\infty}\frac{1}{n!}\int_0^\infty \frac{\sin(xn)}{x}dx= \frac{\pi}{2}(e-1)& \end{align}

Shouldn’t the integral converge absolutely (something like Fubini’s) for this last step to be rigorous? I also used a couple of online calculators, but they didn’t even seem convinced it converges.

Am I missing something? Is there another solution? Does it converge? Thanks so much for the help.

Тyma Gaidash
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    There is a variant on Fubini tonelli which says if both $\int_X\int_Y$ and $\int_Y\int_X$ exist then they are the same, roughly speaking. – FShrike Aug 28 '22 at 17:56
  • https://math.stackexchange.com/questions/2574916/definite-integral-possible-evaluation-using-real-methods?noredirect=1 and https://math.stackexchange.com/questions/2667866/show-that-int-limits-0-infty-frac-dtte-cos-t-sin-sin-t-frac-pi2e?noredirect=1 – Svyatoslav Aug 28 '22 at 18:29

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