Show that if $a < b$:
$$ \int_0^\infty \frac{\sin ax \sin bx}{x^2} \, dx = \frac{\pi a}{2} $$
I could see solution by Fourier analysis or by Contour Integration, but why does the larger frequency $b$ not matter?
Asked
Active
Viewed 557 times
1
-
Please have a look at my answer to this very similar question: http://math.stackexchange.com/questions/1769766/complex-how-to-approach-improper-integral/1769860 – Jack D'Aurizio May 03 '16 at 16:33
-
The reason is essentially that $\mathbb{1}{(-a,a)}(x)\cdot \mathbb{1}{(-b,b)}(x) = \mathbb{1}_{(-\min(a,b),\min(a,b))}(x)$ if $a,b>0$. – Jack D'Aurizio May 03 '16 at 16:34
-
my question is a duplicate: http://math.stackexchange.com/questions/441106/how-do-i-show-that-int-0-infty-sinax-sinbx-x2-pi-mina-b-2?rq=1 – cactus314 May 03 '16 at 17:08