an integral

**mgarson** · May 16th 2011, 01:42 AM

Hi!

I need the following to complete a proof.

$\int _0^\infty \frac{sin^2(x)}{x^2}dx = \frac{\pi}{2}$

I've checked it in Maple and it's true. All I have left is to actually show it (without referring to Maple ofc).

Thanks!

**Prove It** · May 16th 2011, 01:53 AM

I expect you would have to use the residue theorem...

**chisigma** · May 16th 2011, 07:42 AM

Originally Posted by mgarson

Hi!

I need the following to complete a proof.

$\int _0^\infty \frac{sin^2(x)}{x^2}dx = \frac{\pi}{2}$

I've checked it in Maple and it's true. All I have left is to actually show it (without referring to Maple ofc).

Thanks!

A solution of the integral based on the Laplace Tranform is illustrated here...

http://www.mathhelpforum.com/math-he...rm-170211.html

Kind regards

$\chi$ $\sigma$

**TheEmptySet** · May 16th 2011, 08:32 AM

Here is a third method using the Laplace transform

Consider the function

$f(t)=\int_{0}^{\infty}\frac{\sin^2(tx)}{x^2}dx$

Note that

$f(1)=\int_{0}^{\infty}\frac{\sin^2(x)}{x^2}dx$

Is the integral you are looking for

Using a trig Identity we get

$f(t)=\int_{0}^{\infty}\frac{\sin^2(tx)}{x^2}dx= \int_{0}^{\infty} \frac{1-\cos(2tx)}{2x^2}$

Now if we take the Laplace transform with respect to t we get

$\mathcal{L}(f)=\frac{1}{2}\int_{0}^{\infty} \left( \frac{1}{x^2s}-\frac{s}{x^2(s^2+4x^2)} \right) dx$

Now by partial fractions on the 2nd term we get

$\frac{s}{x^2(s^2+4x^2)}=\frac{1}{sx^2}-\frac{4}{s(s^2+4x^2)}$

Plugging this back in gives

$\mathcal{L}(f)=\frac{2}{s}\int_{0}^{\infty} \frac{1}{(s^2+4x^2)} dx=\frac{1}{s^2} \cdot \tan^{-1}\left( \frac{2x}{s}\right) \bigg|_{0}^{\infty}$

$\mathcal{L}(f)=\frac{\pi }{2s^2}$

Now finally take the inverse transform to get

$f(t)=\frac{\pi}{2}t \implies f(1)=\frac{\pi}{2}$

**FernandoRevilla** · May 16th 2011, 09:35 AM

Another way: using integration by parts with $u=\sin^2 x,\; dv=dx/x^2$ :

$\displaystyle\int_0^{+\infty}\dfrac{\sin ^2 x}{x^2}dx=\left[\dfrac{\sin ^2 x}{x}\right]_0^{+\infty}+\displaystyle\int_0^{+\infty}\dfrac{\ sin 2x}{x}dx=\displaystyle\int_0^{+\infty}\dfrac{\sin 2 x}{x}dx$

Using the substitution $t=2x$ we obtain the Dirichlet's integral:

$\displaystyle\int_0^{+\infty}\dfrac{\sin 2 x}{x}dx=\displaystyle\int_0^{+\infty}\dfrac{\sin t}{t}dt=\dfrac{\pi}{2}$

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