$\displaystyle \int_0^{\infty}\bigg[\frac{\sin^2(px)\cos^2(px)}{x^2}-\frac{\sin^4(px)}{3x^2}\bigg]~dx$
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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
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? $\displaystyle \frac{1}{x^2}=\int_0^{\infty}ye^{-yx}~dy$
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
Hint: Let $\displaystyle \int_0^{\infty}\bigg[\frac{\sin^2(px)\cos^2(px)}{x^2}-\frac{\sin^4(px)}{3x^2}\bigg]~dx=f(p)$
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 $\displaystyle \frac{1}{12}\frac{d^2}{dp^2}\int_0^{\infty}\frac{\ sin^4(px)}{x^4}~dx=\frac{\pi{p}}{6}$
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