6
$\begingroup$

I got a homework and I've trying to do this problem about 2 days, but I "lost my fight". So I turn to you. I have to prove that $$\int _0^\infty \frac{\sin (x)}{x} \, dx = \frac{\pi}{2}.$$ I can't use complex numbers, double integral, laplace transforms.

I found Mr. Berry's proof, which is looks good, but I don't understand any steps so I need some explanation. Proof on the picture. Thank you!

enter image description here

$\endgroup$
  • 1
    $\begingroup$ At which point do you start having problems ? $\endgroup$ – Claude Leibovici Jan 10 '15 at 8:31
  • 1
    $\begingroup$ I dont understand the second line on the picture. What did he do? Integral has changed and the fraction denominator, it get a (-1)^i, but how? And the penultimate sum equal 1/sin(x), but why? (sry I cant use latex form on the page) $\endgroup$ – Zarrev Jan 10 '15 at 8:51
  • 2
    $\begingroup$ I think because he changed the integration bounds from $[i\pi,(i+1)\pi]$ to $[0,\pi]$ by the change of variables $x=y+i\pi$. I agree that it would have been better to change the name of the variable. The last summation is the development of $\csc(x)$. Please, tell me if this helps. Cheers. $\endgroup$ – Claude Leibovici Jan 10 '15 at 8:56
  • $\begingroup$ I'm so sorry, but I found a mistake in the equation, I've repaired it. Your help is good, but I still don't understand how get it. $\endgroup$ – Zarrev Jan 10 '15 at 9:14
  • 1
    $\begingroup$ math.stackexchange.com/questions/5248/… $\endgroup$ – Guy Fsone Jan 2 '18 at 8:50
5
$\begingroup$

Note that

$$\underbrace{\int_{i \pi}^{(i+1) \pi} dx \frac{\sin{x}}{x}}_{x=u+i \pi} = \int_0^{\pi} du \frac{\sin{(u+i \pi)}}{u+i \pi} = (-1)^i \int_0^{\pi} du \frac{\sin{u}}{u+i \pi}$$

Then

$$\sum_{i=-\infty}^{\infty} \int_{i \pi}^{(i+1) \pi} dx \frac{\sin{x}}{x} = \sum_{i=-\infty}^{\infty} (-1)^i \int_0^{\pi} du \frac{\sin{u}}{u+i \pi} = \sum_{i=-\infty}^{\infty} (-1)^i \int_0^{\pi} du \frac{\sin{u}}{u-i \pi}$$

We may reverse the order of summation and integration, so that the integral is equal to

$$\frac12 \int_0^{\pi} du \, \sin{u} \sum_{i=-\infty}^{\infty} \frac{(-1)^i}{u-i \pi} $$

This last sum is $\csc{u}$; one may show this using the Residue Theorem.

$\endgroup$
  • $\begingroup$ Residue Theorem is useing complex analysis, is it? Because I cant use complex so I can't use it :/ $\endgroup$ – Zarrev Jan 10 '15 at 9:31
  • 1
    $\begingroup$ @Zarrev: there are other ways to evaluate the sum that do not use complex analysis. $\endgroup$ – Ron Gordon Jan 10 '15 at 9:46
  • $\begingroup$ Thanks to you :) I hope I will find it ;) $\endgroup$ – Zarrev Jan 10 '15 at 10:02
2
$\begingroup$

Since you cannot use complex numbers, double integrals, or Laplace transforms, here is a method that comes to mind. After showing that the improper integral exists, let $A = \int_0^\infty \frac{\sin x}{x}\, dx$. Then $$A = \lim_{n\to \infty} \int_0^{(2n+1)\frac{\pi}{2}} \frac{\sin x}{x}\, dx = \lim_{n\to \infty} \int_0^\pi \frac{\sin \frac{(2n+1)}{2}x}{x}\, dx = \lim_{n\to \infty} \int_0^\pi f(x)g_n(x)\, dx,$$

where $f(x) = \frac{\sin x/2}{x/2}$ and $g_n(x) = \frac{\sin (2n+1)x/2}{2\sin x/2}$. Since $$g_n(x) = \frac{1}{2} + \sum_{k = 1}^n \cos kx$$ for $0 < x < \pi$, we have $$\int_0^\pi g_n(x)\, dx = \frac{\pi}{2}.$$ Therefore

$$\int_0^\pi f(x)g_n(x)\, dx = \frac{\pi}{2} + \int_0^\pi h(x)\sin \frac{(2n+1)x}{2}\, dx,$$

where $h(x) = (f(x) - 1)/(2\sin x/2)$. Not only is $h$ continuous on $(0,\pi)$, but $h$ also has a right-hand limit at $0$. Indeed, $$\lim_{x \to 0^+} h(x) = \lim_{x\to 0^+} \frac{f(x) - 1}{x} \lim_{x\to 0^+} \frac{x/2}{\sin x/2} = \lim_{x\to 0^+} \frac{f(x) - 1}{x} = \lim_{x\to 0^+} \frac{2\sin \frac{x}{2} - x}{x^2},$$ and

$$\lim_{x\to 0^+} \frac{2\sin \frac{x}{2} - x}{x^2} = \lim_{x\to 0^+} \frac{O(x^3)}{x^2} = \lim_{x\to 0^+} O(x) = 0.$$ Thus $h(x)$ is piecewise continuous on $(0,\pi)$. Hence, by the Riemann-Lebesgue lemma, $\lim_{n\to \infty} \int_0^\pi h(x)\sin \frac{(2n+1)x}{2}\, dx = 0$. Now we deduce $$\lim_{n\to \infty} \int_0^\pi f(x)g_n(x)\, dx = \frac{\pi}{2},$$ in other words, $A = \frac{\pi}{2}$.

$\endgroup$
  • $\begingroup$ Thank you so much, but I don't know what is O(x^3). Can you tell me? $\endgroup$ – Zarrev Jan 10 '15 at 13:29
  • 1
    $\begingroup$ It's big-O notation. If you're uncomfortable with it, you can use L'hospital's rule to find $\lim_{x\to 0^+} (2\sin(x/2) - x)/x^2$. $\endgroup$ – kobe Jan 10 '15 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.