Sine squared integral

liyi · #1 $Old$ November 14th, 2007, 06:15 PM

Evaluate $\int_0^\infty\left(\frac{\sin x}x\right)^2\,dx$

ThePerfectHacker · #2 $Old$ November 14th, 2007, 07:50 PM

Use complex analysis, the answer is $\frac{\pi}{2}$ .

By considering the function $f(z) = \frac{\frac{1}{2i}(z-z^{-1})}{z}$ .

This is Mine 77

th Post!!!

Krizalid · #3 $Old$ November 15th, 2007, 02:11 PM

Quote:

Originally Posted by liyi $View Post$

Evaluate $\int_0^\infty\left(\frac{\sin x}x\right)^2\,dx$

$\int_0^\infty {\left( {\frac{{\sin x}}{x}} \right)^2 \,dx} = \int_0^\infty {\frac{{1 - \cos 2x}}{{2x^2 }}\,dx} .$

Use the following parameter: $\frac{1}{{x^2 }} = \int_0^\infty {ue^{ - ux} \,du} .$

(The only reason that I'd ever create a double integral is that I can reverse the integration order.)

From there you can solve the rest, without problems.

Krizalid · #4 $Old$ November 16th, 2007, 07:19 AM

Or we can define

$I(\varphi ) = \int_0^\infty {\frac{{\sin ^2 (\varphi x)}}{{x^2 }}\,dx} ,\,\varphi > 0.$

Then you can apply the Leibniz's Rule for differentiation under the integral sign. You'd get

$I(\varphi ) = \frac{\pi}{2}\varphi .$

The conclusion follows.

ThePerfectHacker · #5 $Old$ November 16th, 2007, 11:02 AM

$\int_{0}^{\infty} \frac{\cos ax - \cos bx}{x^2} dx = \int_{0}^{\infty} \int_a^b \frac{\sin tx}{x}dt~dx = \int_a^b \int_0^{\infty}\frac{\sin tx}{x}dx~dt$ .

But $\int_0^{\infty} \frac{\sin tx}{x} dx = \int_a^b \frac{\pi}{2} dt = \frac{\pi}{2}(b-a)$ .

That means,
$\int_0^{\infty} \frac{\sin^2 x}{x^2} dx = \frac{1}{2}\int_0^{\infty}\frac{1-\cos 2x}{x^2} dx = \frac{1}{2} \int_0^{\infty} \frac{\cos 0x - \cos 2x}{x^2} dx = \frac{1}{2}(2\pi ) = \pi$ .

Where did I make a mistake? Anyways, you get the idea I am too lazy to check this right now.