Evaluate $\int\limits_0^{\infty}\frac{\sin x}{x}\, dx$ How can I evaluate the following improper integral:   
$$ \int\limits_0^{\infty}\frac{\sin x}{x}\, dx$$  
I have tried to evaluate this by integration by parts but failed. Please help
 A: The integral is convergent by Dirichlet's test. We may notice that 
$$ I=\frac{1}{2}\int_{-\infty}^{+\infty}\frac{\sin x}{x}\,dx = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\sin(x)\cdot\color{green}{\left[\frac{1}{x}+\sum_{n\geq 1}(-1)^n\left(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right)\right]}dx\tag{1}$$
hence it is enough to understand what kind of function is the green function.
By the Weierstrass product for the sine function:
$$ \sin(x) = x\cdot\prod_{n\geq 1}\left(1-\frac{x^2}{n^2\pi^2}\right)\tag{2}$$
and by considering the logarithmic derivative of both sides:
$$ \cot(x) = \frac{1}{x}+\sum_{n\geq 1}\left(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right) \tag{3} $$
that leads to:
$$ \frac{1}{\sin(x)}=\frac{1}{x}+\sum_{n\geq 1}(-1)^n\left(\frac{1}{x+n\pi}+\frac{1}{x-n\pi}\right)\tag{4} $$
that is exactly the green function. It follows that:
$$ I = \frac{1}{2}\int_{-\pi/2}^{\pi/2}\frac{\sin x}{\sin x}\,dx = \color{red}{\frac{\pi}{2}}.\tag{5} $$
This is an explanation of qoqosz' answer here.
A: you define the function
$$F(t)=\int_0^\infty \frac{\sin x}{x}e^{-xt}dx$$
you show that $F$ satisfies a simple differential equation.
you find $F(t)$
and $F(0)$ gives you the result $\frac{\pi}{2}$.
