I'll do the first one (assuming everything sufficient to take the partials):
I did check the results in Mathematica and the Laplace expression does equate to zero.
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