Integral of $\frac{\sin^2(x-y)}{(x-y)^2}$

Question

Calculate the integral $$\int_{-t}^t\int_{-t}^t\frac{\sin^2(x-y)}{(x-y)^2}dxdy.$$ If we cannot get the exact value, does it equal to $2t-\frac{\log(2t)}{\pi^2}+O(1)$? Thank you!

Answer 1

Changing u=x-y we get $$\int \!{\frac { \left( \sin \left( x-y \right) \right) ^{2}}{ \left(x-y \right) ^{2}}}\,{\rm d}y=\int \!-{\frac { \left( \sin \left( u \right) \right) ^{2}}{{u}^{2}}}\,{\rm d}u$$ By parts, with f=sin(u): $$1/2\,{u}^{-1}-1/2\,{\frac {\cos \left( 2\,u \right) }{u}}-{\it Si} \left( 2\,u \right) $$ Reverting the change, and given that integrating for dx is symmetric, so we have: $$\int_{-t}^t\int_{-t}^t\frac{\sin^2(x-y)}{(x-y)^2}dxdy = 4\,{\it Si} \left( 4\,t \right) t+8\, \left( \cos \left( t \right) \right) ^{4}-8\, \left( \cos \left( t \right) \right) ^{2}-2\,\ln \left( 2 \right) -\gamma+1/2\,{\it Ci} \left( -4\,t \right) -1/2\, \ln \left( -t \right) +1/2\,{\it Ci} \left( 4\,t \right) -1/2\,\ln \left( t \right) $$

Answer 2

$$\int_{-t}^t\int_{-p}^p\frac{\sin^2(x-y)}{(x-y)^2}dxdy.$$

 (v= x-y) then upper limit is p-y and lower limit will be -p-y

If 2nd integral $p$ tends to infinity then

$$I(a)=\int_{-\infty}^{\infty}\frac{\sin^2(av)dv}{v^2}$$

$$=>\frac{dI}{da}=\int_{-\infty}^{\infty}\frac{\sin(2av)dv}{v}=2\pi a$$

$$=>I(a)=\pi a^2+c$$

because $I(0)=0$ then c= 0

$$I(a)=\pi a^2$$

$=\int_{-t}^{t}\pi a^2 dy$

$= 2\pi a^2 t $; considered a=1

I tried my best level. I hope this will be some what help full