Okay, I'm trying to compute the following integral:

$\displaystyle I=\int_0^{\infty}\frac{\sin x}{x}\ dx$

But first, I will try to prove that $\displaystyle I$ converge.

$\displaystyle I=\int_0^{\infty}\frac{\sin x}{x}\ dx =\int_0^{1}\frac{\sin x}{x}\ dx +\int_1^{\infty}\frac{\sin x}{x}\ dx$.

The integral $\displaystyle \int_1^{\infty}\frac{\sin x}{x}\ dx <\infty$ (Dirichlet's test)

But what about $\displaystyle \int_0^{1}\frac{\sin x}{x}\ dx$ ?

I know that $\displaystyle x=0$ is ourproblematicpoint and also I know that $\displaystyle \lim_{x\to 0} \frac{\sin x}{x}=1$.

Can I say that because that limit exists the function is integrable?

Now to the the second part-computation.

It's from a book byG. M. Fikhtengoltz,The Fundamentals of Mathematical Analysis.

(I'm translating this from Russian into Hebrew and then into English.)

We will represent I in form of sum:

$\displaystyle I=\sum_{\nu =0}^{\infty}\int^{(\nu+1)\frac{\pi}{2}}_{\nu\frac{ \pi}{2}}$.

Putting $\displaystyle \nu=2\mu$ or $\displaystyle \nu=2\mu-1$ and the he says the following:

we substituting $\displaystyle x=\mu\pi+t$ or $\displaystyle x=\mu\pi-t$, and we get:

$\displaystyle \int_{2\mu\frac{\pi}{2}}^{(2\mu+1)\frac{\pi}{2}} = (-1)^{\mu}\int_{0}^{\frac{\pi}{2}}\frac{\sin t}{\mu\pi+t}$

and

$\displaystyle \int_{(2\mu-1)\frac{\pi}{2}}^{2\mu\frac{\pi}{2}} =(-1)^{\mu - 1}\int_{0}^{\frac{\pi}{2}}\frac{\sin t}{\mu\pi-t}$

Question:

Why we need the both substitutions, $\displaystyle x=\mu\pi+t$$\displaystyle x=\mu\pi-t$ ?or

Thank you.

To be continued...