How to calculate this integral I have a function $u(x,y)$ defined on the domain $|x|<\infty, y>0$. I know that
$$ \frac{\partial u(x,y)}{\partial y} = \frac{y}{\pi}\int_{-\infty}^{\infty} \frac{f(w)}{y^2 + (x-w)^2}dw$$
How can I calculate $u(x,y)$? I have no idea how to tackle this integral.
 A: If $f$ is such that you can reverse the order of integration
$$u(x,y) = \frac{1}{\pi} \int_{-\infty}^{\infty} dw \, f(w) \, \int_0^{Y} dy \frac{y}{y^2+(x-w)^2} + g(x)$$
where $g$ is a function of $x$ only.  I used those integration limits over $y$ because you specified $y>0$, but the integral diverges at $\infty$.  Note that the inner integral is
$$\frac12 \log{[Y^2 + (x-w)^2]} - \log{(x-w)}$$
A: You could also do something like this if it useful...
I'll rename your $w$ as $r$ so it wont confuse things where I use $\omega$ later; your equation is then:
$${\frac {\partial }{\partial y}}u \left( x,y \right) =y\int _{-\infty }
^{\infty }\!{\frac {f \left( r \right) }{{y}^{2}+ \left( x-r \right) ^
{2}}}{dr}$$
The right hand side has the form of a convolution. Thus by the convolution theorem, if we take the Fourier transform we get the product of two functions; one is the Fourier transfrom of $f$ and the other is the Fourier transform of a Lorentzian function which is a damping exponential. In this case it is most useful to use this form of the Fourier transform:
$$U \left( \omega,y \right) =\,\dfrac{1 }{2\pi }\int _{-\infty }^{\infty }\!u
 \left( x,y \right) {{\rm e}^{-i\omega\,x}}{dx}
$$
$$u \left( x,y \right) =\int _{-\infty }^{\infty }\!U \left( \omega,y
 \right) {{\rm e}^{i\omega\,x}}{d\omega}
$$
We then get:
$${\frac {\partial }{\partial y}}U \left( \omega,y \right) =F
 \left( \omega \right)G
 \left( \omega ,y\right) $$
$$F \left( \omega \right) =\dfrac{1 }{2\pi }\int _{-\infty }^{\infty }\!f \left( x
 \right) {{\rm e}^{-i\omega\,x}}{dx}$$
$$G \left( \omega,y \right) =\dfrac{1 }{2\pi }\int _{-\infty }^{\infty }\!{\frac { \left| y
 \right| {}}{   \left| y \right|  ^
{2}+{x}^{2}}}{\rm e}^{-i\omega\,x}{dx}=1/2\,{{\rm e}^{- \left| \omega \right|  \left| y \right| }}$$
so the differential equation becomes:
$${\frac {\partial }{\partial y}}U \left( \omega,y \right) =1/2\,F
 \left( \omega \right) {{\rm e}^{- \left| \omega \right|  \left| y
 \right| }}$$
which is solvable:
$$U \left( \omega,y \right) =-1/2\,{\frac {F \left( \omega \right) {
{\rm e}^{- \left| \omega \right|  \left| y \right| }}}{ \left| \omega
 \right|  \left| y \right| }}+K(\omega)$$
for $K(\omega)$ a constant as far as $y$ is concerned, and thus:
$$u \left( x,y \right) =\int _{-\infty }^{\infty }\! \left( -1/2\,{
\frac {F \left( \omega \right) {{\rm e}^{- \left| \omega \right| 
 \left| y \right| }}}{ \left| \omega \right|  \left| y \right| }}+K(\omega)
 \right) {{\rm e}^{i\omega\,x}}{d\omega}$$
$$u \left( x,y \right) =\int _{-\infty }^{\infty }\! \left( -1/2\,{
\frac {F \left( \omega \right) {{\rm e}^{- \left| \omega \right| 
 \left| y \right| }}}{ \left| \omega \right|  \left| y \right| }}
 \right) {{\rm e}^{i\omega\,x}}{d\omega}+k(x)$$
A: You haven't said anything about the function $f(w)$ (you really should).
If $f$ is holomorphic in the complex upper half-plane, and is bounded at infinity, $f(w)=O(1)$, $|w|\to\infty$, $\Im w\geq0$, then you can use Cauchy theorem to write
$$ \frac{y}{\pi} \int_{-\infty}^\infty \frac{f(w)}{(w-x)^2+y^2}\,dw
= 2iy\,\text{Res}_{w=x+iy}\frac{f(w)}{(w-x)^2+y^2}
= f(x+iy),$$
so that your function $u(x,y)$ can be written as
$$ u(x,y) = g(x) + \int^y f(x+iy')\,dy', $$
where $g(x)$ is an arbitrary function of $x$.
Other things will also happen depending on the properties of $f(w)$.
