By the SINE INTEGRAL OF SINE PULSE, we derived in the class that
pi/2 = integral 0 to infinity ( ( sin w ) / w ) dw

now PLEASE PROVE that "" integral 0 to infinity ( (sin^2 w ) / w^2 ) dw = pi/2 "" as well.

I'll be really very thankful as it is very urgent. Thanks

Hi. I'm not sure how you're supposed to solve it but, using Fourier transform (I assumed that this could be the way since you had a function against $\omega$) I arrived to the result:

First, let's call your function $sinc(\omega) = \frac{sin(\omega)}{\omega}$ so the integral you want to solve is $\int_{0}^{\infty}sinc^2(\omega) \,d\omega$

Then, from Parseval's identity:

(1) $\int_{-\infty}^{\infty}|f(t)|^2\,dt = \frac{1}{2\pi}\int_{-\infty}^{\infty}|F(\omega)|^2\,d\omega$

We can assume that (2) $F(\omega) = \left\{\begin{array}{c l} 0 & \mbox{ if } \omega < 0 \\ sinc^2(\omega) & \mbox{ if } \omega \geq 0\end{array}\right.$

Since we have $F(\omega)$, we might want to find $f(t)$ to solve it easily. First, I'll rewrite the function so we can use a transform I've got from my table.

$sinc(\omega) = \frac{2}{2}sinc(\omega)$

The trasform is: $P_a(t) \Leftrightarrow a\cdot sinc\left(\frac{a\omega}{2}\right)$. Where P stands for square pulse and a total width.

Antitransforming:

(3) $\mathcal{F}^{-1}\left\{\frac{2 \cdot sinc(\omega)}{2}\right\} = \frac{P_2(t)}{2}$

Once we have everything, we can assume from (1), (2) and (3) that:

$2\pi \cdot \int_{0}^{\infty}\frac{P_2\,^2(t)}{2^2} \,dt = \int_{0}^{\infty}sinc^2(\omega)}\,d\omega$

Now we solve the first part (which is easier):

$2\pi \cdot \int_{0}^{\infty}\frac{P_2\,^2(t)}{4} \,dt = \frac{\pi}{2}\int_{0}^{1}\,dt = \frac{\pi}{2} \cdot \left. t \right |_0^1 = \frac{\pi}{2}$

I hope this is what you needed.