hello

i've been trying for ages and this one question i can't answer, show that:

u(x,y)=(y/pi) times integral , from - to + infinity, of {f(t)dt/[(x-t)^2 + y^2]}

satisfies

'second partial of u w.r.t x' + 'second partial of u w.r.t y' = 0

i've tried

du =

{-(y/pi)times integral, from -to+ inf, of {2(x-t)f(t)dt/[(x-t)^2 + y^2]^2}}dx

+

{(1/pi)times integral , from - to + infinity, of {f(t)dt/[(x-t)^2 + y^2]}

(y/pi)times integral, from -to+ infinity, of {2yf(t)dt/[(x-t)^2 + y^2]^2}}dy

and then take the second derivatives of the partials and add them but it don't work.

NOW I AM SURE i am going about this the wrong way, BUT HOPEFULLY some brighter soul out there will see what i am trying to do and any way tell me what is the right way