• Jan 7th 2013, 12:33 AM
szak1592
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
• Jan 7th 2013, 10:47 AM
russo
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.