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

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

Gerry Myerson
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Vladimir Reshetnikov
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  • Wolfram Alpha doesn't seem to be able to do it... – Daniel Freedman Dec 14 '11 at 00:46
  • 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. – dls Dec 14 '11 at 01:15
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    The portion of the integrands with the sines is not periodic, so the prognosis is not great. – ncmathsadist Dec 14 '11 at 02:47
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    Maple gives the result $\frac{2\pi}{4+\pi^4}$. – Jon Dec 14 '11 at 08:32
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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 $2\pi/(4+\pi^4) < 0.1$. – Greg Martin Dec 14 '11 at 08:40
  • @daniel: what does it give for {0,4}, which is the first time the $\sin x$ term changes sign? – Greg Martin Dec 14 '11 at 08:43
  • Mathematica 8 is unable to compute. Possibly, there is no closed form. – Jon Dec 14 '11 at 21:11
  • 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. – Jon Dec 16 '11 at 07:37
  • @GM:n=4--(1/(4+Pi^4)(e^(-4-Pi)(Pie^Pi (-2cos[4] +2e (2cos[3] +e (2cos[2] + (e+2cos[1])(e+Pisin[1])) Pi^2sin[3])- Pi^2sin[4])+ 2e^4 (-2Pi*cos[Pi^2]+(Pi^2-2)sin[Pi^2]))). – daniel Dec 16 '11 at 11:54
  • If it is integrable then it would be a 'constant' which is a closed form? – FiniteA Mar 16 '12 at 08:53
  • @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! – Mariano Suárez-Álvarez Mar 20 '12 at 00:39
  • Wolfram Alpha gives an answer $\approx 0.3587091$ – Kirthi Raman Apr 11 '12 at 12:27
  • It's tricky to compute this integral even numerically, isn't $0.336549871;;;$ a better estimation? Has anyone tried to compute it? – Yrogirg Apr 26 '12 at 18:38
  • MAybe something in the following direction helps: If $\frac ab\approx \pi$ is a good approximation, then the intergral should be more or less $\approx \int_0^{b\pi}|\sin(x)\cdot \sin(\frac ab x)|\sum_{k=0}^\infty e^{-x-kb\pi},dx$. – Hagen von Eitzen Nov 02 '12 at 19:33

2 Answers2

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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$\ldots$

Christian Blatter
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4

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

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

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

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

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

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

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

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

pavybez
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