$\int_{0}^{\infty}\frac{\sin x}{x}dx $ converges I want to show that $\int_{0}^{\infty}\frac{\sin x}{x}dx $ converges. I am facing difficluty in my last step
Although there are many proofs regarding this but doubt is hasnt been addressed.
My attempt
Now  $\int_{0}^{\infty}\frac{\sin x}{x}dx  =\int_{0}^{\pi}\frac{\sin x}{x}dx + \int_{\pi}^{\infty}\frac{\sin x}{x}dx $ The first integral  converges as $\frac{\sin x}{x}$ is continuous in $[ 0, \pi]$ and $$\int_{\pi}^{n\pi}\frac{\sin x}{x}dx = -\frac{\cos x}{x}|_{\pi}^{n \pi} - \int_{\pi}^{n\pi}\frac{\cos x}{x^2}dx  \\= \frac{(-1)^n}{n\pi} - 1/\pi - \int_{\pi}^{n\pi}\frac{\cos x}{x^2}dx \\
\leq \Bigg|\frac{(-1)^n}{n\pi} - 1/\pi \Bigg| + \int_{\pi}^{n\pi}\frac{1}{x^2}dx $$
Since $\int_{\pi}^{n\pi}\frac{1}{x^2}dx$ is improperly integrable then so is $\int_{\pi}^{n\pi}\frac{\cos x}{x^2}dx$. But there there is a $(-1)^n$ which may cause trouble in convergence. 
I feel that I am almost through but I need a good mathematical argument to conclude.
 A: you should not integrate between $\pi$ and $n\pi$ but between $\pi$ and a generic $t$. Then using your calculations you are going to get
$$-\frac{\cos t}{t}-1/\pi-\int_\pi^t\frac{\cos x}{x^2} dx.$$
Then $-\frac{\cos t}{t}\to 0$ as $t\to\infty$, while $\int_\pi^\infty\frac{\cos x}{x^2} dx$ exists because as you said $\int_\pi^\infty\frac{1}{x^2} dx$ exists.
ADDED: As Marsan said when you are computing $\int_a^\infty f(x)\,dx$, if $f\ge 0$, then you know that the limit $\lim_{t\to\infty} \int_a^t f(x) dx$ exists because the function $g(t)=\int_a^t f(x) dx$ is increasing. In that case (and only in  that case) you can take $t$ to be any sequence going to $\infty$. But if $f$ changes sign, you cannot do that. Along a sequence $t_n\to\infty$ you could have that $\int_a^{t_n} f(x) dx$ goes to a limit and along another sequence $s_n\to\infty$ you could have that $\int_a^{s_n} f(x) dx$ might go to a completely different limit. 
A: $\Bigg|\bigg|\frac 1\pi\bigg|-\bigg|\frac{1}{n\pi}\bigg|  \Bigg|\le\Bigg|\frac{(-1)^n}{n\pi} - \frac 1\pi \Bigg|\le \Bigg|\bigg|\frac 1\pi\bigg|+\bigg|\frac{1}{n\pi}\bigg| \Bigg|$
Let $N = \frac 1{\pi\epsilon}$
When $n>N, \frac 1\pi-\epsilon \le \Bigg|\frac{(-1)^n}{n\pi} - \frac 1\pi \Bigg|\le\frac 1\pi+\epsilon$
or
$\forall\epsilon>0, n>N\implies \Bigg|\bigg|\frac{(-1)^n}{n\pi} - \frac 1\pi\bigg| -\frac1\pi \Bigg|<\epsilon$
