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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.

silent learner
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  • Related: http://math.stackexchange.com/questions/1089495/convergence-of-int-1-infty-frac-sin-xx-alphadx?rq=1 http://math.stackexchange.com/questions/1310536/find-whether-int-1-infty-fracsinxx-dx-is-converging-or-not?noredirect=1&lq=1 http://math.stackexchange.com/questions/5248/solving-the-integral-int-0-infty-frac-sinxx-dx-frac-pi2?noredirect=1&lq=1 – Simply Beautiful Art Jan 06 '17 at 00:10
  • How should I conclude..I have seen those ....help me concluding – silent learner Jan 06 '17 at 00:11
  • You ought to check the links. Many arguments that follow your basic idea. – Simply Beautiful Art Jan 06 '17 at 00:12
  • 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 $a_n \to 0$. –  Jan 06 '17 at 00:38
  • 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. – Marsan Jan 06 '17 at 00:40
  • ooops, I did not see your message Marsan and replied the exact same thing... – Gio67 Jan 06 '17 at 00:43
  • @Marsan for $\int_{0}^{\pi}\frac{\sin x}{x}$ the integrand is continuous on that interval hence I was left with the other part. Isn't that correct?? – silent learner Jan 06 '17 at 01:01
  • silentlearner yes this is correct! But for the rest see the detailed answer of @Gio67 – Marsan Jan 06 '17 at 01:15

2 Answers2

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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.

Gio67
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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

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

Doug M
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