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I want to calculate the value of $$\int_{1}^{\infty}\frac{e^{\sin x}\cos x}{x}\,dx$$

I was able to prove using Dirichlet's test that it does converge, but how can I calculate its value?

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  • $\begingroup$ Thanks, @A-Level Student I messed up my edit improvement! $\endgroup$ – amWhy Jan 31 at 17:00
  • $\begingroup$ @amWhy you're welcome, it was only a very small error :) $\endgroup$ – A-Level Student Jan 31 at 17:01
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    $\begingroup$ @ArielYael Why would you suspect that a closed form result exists? $\endgroup$ – Mark Viola Jan 31 at 17:14
  • $\begingroup$ @MarkViola I'm asking if there is, I know it converges, but I don't know the value. Perhaps there isn't, but if there is I hope someone will give me an answer $\endgroup$ – Ariel Yael Jan 31 at 18:07
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It is pretty obvious that there cannot be an elementary result for the given integral, nor one using "mild" special functions like Bessel and error functions.

So we will numerically integrate it. The problem, however, is that the integrand is oscillatory as $x\to\infty$ and only decays like $\frac1x$, which means the following approach is needed:

  • Break up $[1,\infty)$ at the integrand zeros: $\frac\pi2,\frac{3\pi}2,\frac{5\pi}2\dots$
  • Compute the integral over each segment, using series acceleration techniques to "predict" future values, then sum everything up

In mpmath this can be accomplished as
quadosc(lambda x: exp(sin(x))*cos(x)/x, [1, inf], period=2*pi) and this returns the accurate result $$-0.379597645637239886\dots$$

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We have that, integrating by parts $$ \eqalign{ & I = \int_1^\infty {e^{\,\sin x} {{\cos x} \over x}dx} = \int_1^\infty {{1 \over x}d\left( {e^{\,\sin x} } \right)} = \cr & = \left. {{{e^{\,\sin x} } \over x}} \right|_1^\infty - \int_1^\infty {e^{\,\sin x} d\left( {{1 \over x}} \right)} = - e^{\,\sin 1} + \int_1^\infty {{{e^{\,\sin x} } \over {x^{\,2} }}dx} = \cr & = - e^{\,\sin 1} + J \cr} $$ with the advantage of having replaced $1/x$ with $1/x^2$.

Then being $\exp(\sin x)$ periodic $$ \eqalign{ & J = \int_1^\infty {{{e^{\,\sin x} } \over {x^{\,2} }}dx} = \int_1^{1 + 2\pi } {{{e^{\,\sin x} } \over {x^{\,2} }}dx} + \int_{1 + 2\pi }^{1 + 4\pi } {{{e^{\,\sin x} } \over {x^{\,2} }}dx} + \cdots = \cr & = \sum\limits_{0\, \le \,k} {\int_{1 + k2\pi }^{1 + \left( {k + 1} \right)2\pi } {{{e^{\,\sin x} } \over {x^{\,2} }}dx} } = \sum\limits_{0\, \le \,k} {\int_0^{2\pi } {{{e^{\,\sin \left( {x + 1} \right)} } \over {\left( {x + 1 + k2\pi } \right)^{\,2} }}dx} } = \cr & = {1 \over {4\pi ^{\,2} }}\int_0^{2\pi } {e^{\,\sin \left( {x + 1} \right)} \left( {\sum\limits_{0\, \le \,k} {{1 \over {\left( {{{\left( {x + 1} \right)} \over {2\pi }} + k} \right)^{\,2} }}} } \right)dx} = \cr & = {1 \over {4\pi ^{\,2} }}\int_0^{2\pi } {e^{\,\sin \left( {x + 1} \right)} \zeta \left( {2,{{x + 1} \over {2\pi }}} \right)dx} = \cr & = {1 \over {4\pi ^{\,2} }}\int_0^{2\pi } {e^{\,\sin \left( {x + 1} \right)} \psi ^{\left( 1 \right)} \left( {{{x + 1} \over {2\pi }}} \right)dx} = \cr & = {1 \over {2\pi }}\int_0^{2\pi } {e^{\,\sin \left( {x + 1} \right)} \psi ^{\left( 1 \right)} \left( {{{x + 1} \over {2\pi }}} \right) d\left( {{{x + 1} \over {2\pi }}} \right)} = \cr & = {1 \over {2\pi }}\int_{{1 \over {2\pi }}}^{{1 \over {2\pi }} + 1} {e^{\,\sin \left( {2\pi t} \right)} \psi ^{\left( 1 \right)} \left( t \right)d\left( t \right)} \cr } $$ where $\psi ^{\left( 1 \right)} \left( t \right)$ denotes the Trigamma Function.

So we have reduced the improper integral to a proper integral, much more manageable by numeric integration.
My CAS gives: $$ \eqalign{ & J \approx 1 .94017 \cr & {\rm I} \approx - \,0.3796 \cr} $$

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  • $\begingroup$ @Parclytaxel : good that it checks with your results ! $\endgroup$ – G Cab Feb 1 at 23:06

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