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Is it possible to calculate the following definite integral in a closed form?

$$ \int_0^\infty \left| \sin x \cdot \sin (\pi x) \right| e^{-x} \, dx$$

  • 0
    Wolfram Alpha doesn't seem to be able to do it...2011-12-14
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    You could evaluate the indefinite integral, compute the definite integral over each "period" and then write as a sum. Not sure if this gets you anywhere.2011-12-14
  • 6
    The portion of the integrands with the sines is not periodic, so the prognosis is not great.2011-12-14
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    Maple gives the result $\frac{2\pi}{4+\pi^4}$.2011-12-14
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    @Jon: that's the result if the integral didn't have absolute value signs in it. Numerically the OP's integral is a little less than 0.359, while $2\pi/(4+\pi^4) < 0.1$.2011-12-14
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    @daniel: what does it give for {0,4}, which is the first time the $\sin x$ term changes sign?2011-12-14
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    Mathematica 8 is unable to compute. Possibly, there is no closed form.2011-12-14
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    The following holds $$\int_{0}^{\infty}|\sin x\sin(\pi x)|e^{-x}dx=\frac{1}{2}\int_0^\infty\sqrt{(1-\cos(2x))(1-\cos(2\pi x))}e^{-x}dx$$ that just removes the absolute value but does not seem to make things simpler.2011-12-16
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    @GM:n=4--(1/(4+Pi^4)(e^(-4-Pi)(Pi*e^Pi (-2cos[4] +2e (2cos[3] +e (2cos[2] + (e+2cos[1])(e+Pi*sin[1])) Pi^2*sin[3])- Pi^2*sin[4])+ 2e^4 (-2Pi*cos[Pi^2]+(Pi^2-2)sin[Pi^2]))).2011-12-16
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    If it is integrable then it would be a 'constant' which is a closed form?2012-03-16
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    @FiniteA, that's quite not the meaning of 'closed form' in the context of integration. Indeed, if *that* were the meaning, *all* integrable functions would be integrable in closed form, and then clearly the notion of being integrable in closed form would be completely useless!2012-03-20
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    Wolfram Alpha gives an answer $\approx 0.3587091$2012-04-11
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    It's tricky to compute this integral even numerically, isn't $0.336549871\;\;\;$ a better estimation? Has anyone tried to compute it?2012-04-26
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    MAybe something in the following direction helps: If $\frac ab\approx \pi$ is a good approximation, then the intergral should be more or less $\approx \int_0^{b\pi}|\sin(x)\cdot \sin(\frac ab x)|\sum_{k=0}^\infty e^{-x-kb\pi}\,dx$.2012-11-02

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