1. ## Leibniz' rule

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

2. I'll do the first one (assuming everything sufficient to take the partials):

$\displaystyle \frac{\partial}{\partial x} u=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{\partial}{\partial x}\left(\frac{f(t)}{(x-t)^2+y^2}\right)dx=-\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{2(x-t)f(t)}{\left[(x-t)^2+y^2\right]^2}dt$

$\displaystyle \frac{\partial^2 u}{\partial x^2}=\frac{y}{\pi}\int_{-\infty}^{\infty} \frac{8(x-t)^2f(t)}{((x-t)^2+y^2)^3}-\frac{2f(t)}{((x-t)^2+y^2)^2}dt=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{[(x-t)^2-2y^2]f(t)}{[(x-t)^2+y^2)^3}dt$

I did check the results in Mathematica and the Laplace expression does equate to zero.

3. Originally Posted by shawsend
I'll do the first one (assuming everything sufficient to take the partials):

$\displaystyle \frac{\partial}{\partial x} u=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{\partial}{\partial x}\left(\frac{f(t)}{(x-t)^2+y^2}\right)dx=-\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{2(x-t)f(t)}{\left[(x-t)^2+y^2\right]^2}dt$

$\displaystyle \frac{\partial^2 u}{\partial x^2}=\frac{y}{\pi}\int_{-\infty}^{\infty} \frac{8(x-t)^2f(t)}{((x-t)^2+y^2)^3}-\frac{2f(t)}{((x-t)^2+y^2)^2}dt=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{[(x-t)^2-2y^2]f(t)}{[(x-t)^2+y^2)^3}dt$

I did check the results in Mathematica and the Laplace expression does equate to zero.
yes thanks, i do appreciate this . however i did get that far my self so my issue is with the second one.

p.s. don't think i am pushing to get the full answer, it simply the case and that's how it is. anyway, i do appreciate your help

4. Originally Posted by shawsend
$\displaystyle \frac{\partial^2 u}{\partial x^2}=\frac{y}{\pi}\int_{-\infty}^{\infty} \frac{8(x-t)^2f(t)}{((x-t)^2+y^2)^3}-\frac{2f(t)}{((x-t)^2+y^2)^2}dt=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{[(x-t)^2-2y^2]f(t)}{[(x-t)^2+y^2)^3}dt$
That should be:

$\displaystyle \frac{\partial^2 u}{\partial x^2}=\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{f(t)\left[6y(x-t)^2-2y^3\right]}{((x-t)^2+y^2)^3}dt$

and I think you got the other one by now. It's messy and I had problems with it too but here it is:

$\displaystyle \frac{\partial u}{\partial y}=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{-2y f(t)}{((x-t)^2+y^2)^2}+\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f(t)}{(x-t)^2+y^2}$

$\displaystyle \frac{\partial^2 u}{\partial y^2}=\frac{y}{\pi}\left[\int_{-\infty}^{\infty} \frac{f(t)\left[(x-t)^2+y^2\right]^2(-2)+8y^2\left[(x-t)^2+y^2\right]}{((x-t)^2+y^2)^4}\right]$ $\displaystyle -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{4yf(t)}{(x-t)^2+y^2)^2}$

$\displaystyle =\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{f(t)\left[-2y^3-2y(x-t)^2+8y^3-\left[4y((x-t)^2+y^2)\right]\right]}{((x-t)^2+y^2)^3}$

$\displaystyle =\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f(t)\left[2y^3-6y(x-t)^2\right]}{((x-t)^2+y^2)^3}$

which cancels the other one.