1. ## Another fun integral

$\displaystyle \int_0^{\infty}\bigg[\frac{\sin^2(px)\cos^2(px)}{x^2}-\frac{\sin^4(px)}{3x^2}\bigg]~dx$

2. Originally Posted by Mathstud28
$\displaystyle \int_0^{\infty}\bigg[\frac{\sin^2(px)\cos^2(px)}{x^2}-\frac{\sin^4(px)}{3x^2}\bigg]~dx$
I got to this point...and then I'm kinda stuck...any hints?

$\displaystyle \int_0^{\infty}\frac{\sin^2(px)}{x^2}\,dx-\frac{4}{3}\int_0^{\infty}\frac{\sin^4(px)}{x^2}\, dx$

--Chris

3. Originally Posted by Chris L T521
I got to this point...and then I'm kinda stuck...any hints?

$\displaystyle \int_0^{\infty}\frac{\sin^2(px)}{x^2}\,dx-\frac{4}{3}\int_0^{\infty}\frac{\sin^4(px)}{x^2}\, dx$

--Chris
Going that route?

$\displaystyle \frac{1}{x^2}=\int_0^{\infty}ye^{-yx}~dy$

4. Originally Posted by Chris L T521
I got to this point...and then I'm kinda stuck...any hints?

$\displaystyle \int_0^{\infty}\frac{\sin^2(px)}{x^2}\,dx-\frac{4}{3}\int_0^{\infty}\frac{\sin^4(px)}{x^2}\, dx$

--Chris
I didn't just edit my quote so you would see this, I went through the calculations of this and I think it is wrong

5. Hint:

Let $\displaystyle \int_0^{\infty}\bigg[\frac{\sin^2(px)\cos^2(px)}{x^2}-\frac{\sin^4(px)}{3x^2}\bigg]~dx=f(p)$

6. Originally Posted by Mathstud28
$\displaystyle \int_0^{\infty}\bigg[\frac{\sin^2(px)\cos^2(px)}{x^2}-\frac{\sin^4(px)}{3x^2}\bigg]~dx$
No one's going to bite?

This is

$\displaystyle \frac{1}{12}\frac{d^2}{dp^2}\int_0^{\infty}\frac{\ sin^4(px)}{x^4}~dx=\frac{\pi{p}}{6}$