# Differentiating Integral with Green's function

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1. Nov 19, 2015

### RUber

Hello, I am having trouble finding the proper justification for being able to pass the derivative through the integral in the following:
$u(x,y) = \frac{\partial}{\partial y} \int_0^\infty\int_{-\infty}^\infty f(x') K_0( \sqrt{ (x - x')^2 + (y-y')^2 } \, dx' dy'$
$K_0$ is the Modified Bessel function of the second kind with properties:
1. $K_0(z) \approx - \ln z \text{ as } z \to 0 .$
2. $\mathop {\lim }\limits_{|z| \to \infty} K_\nu (z) = 0 .$
3. $K_\nu(z)$ is real and positive for $\nu > -1$ and $z \in \mathbb{R} >0 .$
4. $K_{-\nu}(z) = K_\nu(z) .$
5. $\frac{\partial K_\nu (z) }{\partial z } = -\frac 12 \left( K_{\nu -1 }(z) + K_{\nu+1}(z)\right).$

I have seen it done, but my advisor asked me to justify the step.
The problem occurs at the singularity $x= x', y = y'.$

I also know that $K_0$ has an equivalent representation in the Fourier transform domain of:
$c \int_{-\infty}^\infty \frac { e ^{ - \sqrt{ \xi^2+\alpha^2 } |y- y'|}}{\sqrt{ \xi^2+\alpha^2 }} e^{i (x-x') \xi } d\xi$
for some scaling constant c.
In a few articles, they discuss the properties of pseudo-differential operators...which may be a hint for justifying the derivative.

I am really stumped on this one...any insight would be helpful. I feel like there is something intuitive that I am simply missing.
Thanks.

2. Nov 20, 2015

### geoffrey159

It looks difficult !
I don't know the answer, but have you tried this: assuming that you've shown integrability of your double integral on $\mathbb{R}_+\times \mathbb{R}$, would a polar change of variable transform this complicated double integral to a single, parameter dependent integral ? Then, would the usual theorems on inversion under the integral sign apply ?