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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
  • 0
    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
  • 1
    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

2 Answers 2

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One could give the following a try: Develop $|\sin x|$ into a Fourier series. You get $$|\sin x|={2\over\pi}-{4\over\pi}\sum_{k=1}^\infty {1\over 4k^2 -1}\cos(2kx)\ .$$ Similarly $$|\sin (\pi x)|={2\over\pi}-{4\over\pi}\sum_{k=1}^\infty {1\over 4k^2 -1}\cos(2\pi kx)\ .$$ Since the two series are absolutely convergent you can multiply them, obtaining a double series of the form $$\sum_{k,l} 2c_{k,l}\cos(2kx)\cos(2\pi l x)=\sum_{k,l} c_{k,l}\bigl(\cos \bigl((2(k+\pi l)x\bigr)+\cos\bigl(2(k-\pi l) x\bigr)\bigr)\ .$$ Now $$\int_0^\infty \cos(q x)e^{-x}\ dx={1\over 1+q^2}\ ;$$ therefore you will end up with a huge double series containing terms of the form $${c\over (4k^2-1)(4l^2-1)\bigl(1+4(k\pm \pi l)^2\bigr)}\ .$$ I wish you luck$\ldots$

  • 6
    "I wish you luck..." LOL2012-04-16
4

Let $f(x) = e^{-x} \sin(x)\sin(\pi x)$

Let $A=\{x : e^{-x}sin(x)\sin(\pi x) > 0\}$
Let $B=\{x : e^{-x}sin(x)\sin(\pi x) < 0\}$

A and B are disjoint and hence $\int_{0}^{\infty}f(x)=\int_A f \,du + \int_B f\,du$

Range of $f(x)=0$ to $\kappa =\max(f(x))$

Split the range of $f(x)$ into n intervals, $n\rightarrow \infty$ such that

$\displaystyle \int_A f \,du = \lim_{n\to\infty} \sum_{j=1}^{n} \left (\left(j+1\right)\frac{\kappa }{n}-j\frac{\kappa }{n} \right ) \int I_{A_j}$

$\displaystyle \int_B f \,du = \lim_{n\to\infty} \sum_{j=1}^{n} \left (\left(j+1\right)\frac{\kappa }{n}-j\frac{\kappa }{n} \right ) \int I_{B_j}$

$\displaystyle \int_{A+B} f \,du = \lim_{n\to\infty} \frac{\kappa }{n} \sum_{j=1}^{n} \int I_{A_j} + I_{B_j}$

$\displaystyle A_j =\left (\frac{j\kappa }{n} < f(x) < \frac{(j+1)\kappa }{n} \right )$
$\displaystyle B_j =\left (\frac{j\kappa }{n} < -f(x) < \frac{(j+1)\kappa }{n} \right )$

$\displaystyle h(a,x,b) = \begin{cases} 1 &\text{if } |a| < |x| < |b|, \\ 0 &\text{if } otherwise. \end{cases} $

$\displaystyle I_{A_j} =\frac{1}{2}h\left(j\frac{\kappa }{n},f(x),\left(j+1\right)\frac{\kappa }{n} \right ) \left(1 + \frac{\left|f(x)\right|}{f(x)}\right)$

$\displaystyle I_{B_j} =\frac{1}{2}h\left(-1\left(j+1\right)\frac{\kappa }{n},-f(x),-j\frac{\kappa }{n} \right ) \left(1 - \frac{\left|f(x)\right|}{f(x)}\right)$

working on it.

  • 3
    Could you fill in a few of the details? As a spectator I'm interested in seeing a complete solution, not necessarily doing it myself.2012-04-14
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    Not sure what this serves; OP is asking for a closed form and not mere existence...2012-04-15