Hi!
I need the following to complete a proof.
$\displaystyle \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
$\displaystyle \chi$ $\displaystyle \sigma$
Here is a third method using the Laplace transform
Consider the function
$\displaystyle f(t)=\int_{0}^{\infty}\frac{\sin^2(tx)}{x^2}dx$
Note that
$\displaystyle f(1)=\int_{0}^{\infty}\frac{\sin^2(x)}{x^2}dx$
Is the integral you are looking for
Using a trig Identity we get
$\displaystyle f(t)=\int_{0}^{\infty}\frac{\sin^2(tx)}{x^2}dx= \int_{0}^{\infty} \frac{1-\cos(2tx)}{2x^2}$
Now if we take the Laplace transform with respect to t we get
$\displaystyle \mathcal{L}(f)=\frac{1}{2}\int_{0}^{\infty} \left( \frac{1}{x^2s}-\frac{s}{x^2(s^2+4x^2)} \right) dx$
Now by partial fractions on the 2nd term we get
$\displaystyle \frac{s}{x^2(s^2+4x^2)}=\frac{1}{sx^2}-\frac{4}{s(s^2+4x^2)}$
Plugging this back in gives
$\displaystyle \mathcal{L}(f)=\frac{2}{s}\int_{0}^{\infty} \frac{1}{(s^2+4x^2)} dx=\frac{1}{s^2} \cdot \tan^{-1}\left( \frac{2x}{s}\right) \bigg|_{0}^{\infty}$
$\displaystyle \mathcal{L}(f)=\frac{\pi }{2s^2}$
Now finally take the inverse transform to get
$\displaystyle f(t)=\frac{\pi}{2}t \implies f(1)=\frac{\pi}{2}$
Another way: using integration by parts with $\displaystyle u=\sin^2 x,\; dv=dx/x^2$ :
$\displaystyle \displaystyle\int_0^{+\infty}\dfrac{\sin ^2 x}{x^2}dx=\left[\dfrac{\sin ^2 x}{x}\right]_0^{+\infty}+\displaystyle\int_0^{+\infty}\dfrac{\ sin 2x}{x}dx=\displaystyle\int_0^{+\infty}\dfrac{\sin 2 x}{x}dx$
Using the substitution $\displaystyle t=2x$ we obtain the Dirichlet's integral:
$\displaystyle \displaystyle\int_0^{+\infty}\dfrac{\sin 2 x}{x}dx=\displaystyle\int_0^{+\infty}\dfrac{\sin t}{t}dt=\dfrac{\pi}{2}$