Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
    
Wolfram Alpha doesn't seem to be able to do it... –  Daniel Freedman Dec 14 '11 at 0:46
    
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 1:15
6  
The portion of the integrands with the sines is not periodic, so the prognosis is not great. –  ncmathsadist Dec 14 '11 at 2:47
    
Maple gives the result $\frac{2\pi}{4+\pi^4}$. –  Jon Dec 14 '11 at 8:32
2  
@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 8:40
show 9 more comments

2 Answers

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.

share|improve this answer
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. –  Antonio Vargas Apr 14 '12 at 23:29
1  
Not sure what this serves; OP is asking for a closed form and not mere existence... –  J. M. Apr 15 '12 at 16:16
add comment

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$\ldots$

share|improve this answer
4  
"I wish you luck..." LOL –  Pedro Tamaroff Apr 16 '12 at 0:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.