Closed form for improper definite integral involving trig functions and exponentials? Is it possible to calculate the following definite integral in a closed form?
$$ \int_0^\infty \left| \sin x \cdot \sin (\pi x) \right| e^{-x} \, dx$$
 A: 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.
A: 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$\ldots$
