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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.

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  • $\begingroup$ How should I conclude..I have seen those ....help me concluding $\endgroup$ – silent learner Jan 6 '17 at 0:11
  • $\begingroup$ You ought to check the links. Many arguments that follow your basic idea. $\endgroup$ – Simply Beautiful Art Jan 6 '17 at 0:12
  • $\begingroup$ This is a bit lazy but you can use the following result : a serie $\sum a_n$ converges if $a_n$ and $a_{n+1}$ have different sign for all $n$, $|a_n| \geq |a_{n+1}|$ and $a_n \to 0$. $\endgroup$ – user171326 Jan 6 '17 at 0:38
  • $\begingroup$ There is no need in bounding your last term, it suffices to say that what you wrote on the line above converges to $- \frac{1}{\pi} - \int_{\pi}^{\infty} \frac{\cos(x)dx}{x^2}$ as $n \to \infty$. However, as regards the aim of your exercise, it does not suffice to show that $\int_{\pi}^{n \pi} \frac{sin(x)dx}{x}$ converges in order to show that $\int_{0}^{\infty}\frac{\sin{x}}{x}dx$ converges (the integrand is not positive). You ought to replace the '$n\pi$ ' with any real positive number $A$ in your computation. $\endgroup$ – Marsan Jan 6 '17 at 0:40
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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.

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  • $\begingroup$ point taken ...but what if I have to take $n\pi$ even then it will go to zero right? $\endgroup$ – silent learner Jan 6 '17 at 0:51
  • $\begingroup$ it is not enough to take just $t=n\pi$, you really need an arbitrary $t\to \infty$. As Marsan said, your function has a sign $\endgroup$ – Gio67 Jan 6 '17 at 0:56
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$\Bigg|\bigg|\frac 1\pi\bigg|-\bigg|\frac{1}{n\pi}\bigg| \Bigg|\le\Bigg|\frac{(-1)^n}{n\pi} - \frac 1\pi \Bigg|\le \Bigg|\bigg|\frac 1\pi\bigg|+\bigg|\frac{1}{n\pi}\bigg| \Bigg|$

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

$\forall\epsilon>0, n>N\implies \Bigg|\bigg|\frac{(-1)^n}{n\pi} - \frac 1\pi\bigg| -\frac1\pi \Bigg|<\epsilon$

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  • $\begingroup$ I know but that will give me a bound and not convergence.....any thoughts?? :) $\endgroup$ – silent learner Jan 6 '17 at 0:53
  • $\begingroup$ Sure, see above. $\endgroup$ – Doug M Jan 6 '17 at 1:20

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