Differentiating this integral,

Question

I want to show that $u_{xx} + u_{yy} = 0$ for the integral given below, so I think I want to differentiate under the integral with respect to both $x$ and $y$. The goal is to show that $u$ is harmonic. This is one part of an old complex analysis exam question, but I think this part of the problem is really an advanced calculus problem, dealing with the interchange of derivatives and integral, so I have added both the real analysis tag and complex analysis tag for anyone who's interested in answering.

So I've reviewed several versions of the Leibniz integral rule for differentiating under the integral sign, however, all of the versions have variable limits, whereas the integral below has constants in the lower and upper limits of integration, say, the constant limit $-R$ and $+R$, and letting $R \to \infty$ to turn it into an integral over the extended real line.

So, since there are no variable limits, is $u_x$ and $u_y$ just...immediately equal to $0$? And so there's not even a need to check the second derivatives of $u$, and we can conclude that $u$ is harmonic in the upper half plane. Of course, I am almost certain that this is not correct, but if I just follow what I reviewed regarding Leibniz's rule, this would seem to be the case.

What do you think? Thanks,

The problem statement is:

If $f(x) = \sum_j c_j \sin(a_jx)$ is a finite trigonometric sum given on the real axis $y=0$, show that:

$$u(x,y) = \frac{|y|}{\pi}\int_{-\infty}^{\infty}\frac{f(x+t)}{t^2+y^2}dt$$

is harmonic both on the upper half plane as well as the lower half plane, and that

$$\lim_{y \to 0} u(x,y) = f(x) $$

from either side.

Any hints or suggestions on how to get started on this problem are welcome.

Thanks,

Answer 1

It is not necessary to differentiate under the integral sign. The given $f$ is a finite linear combination of functions of the form $t\mapsto e^{i\omega t}$. Use residue calculus to compute once and for all $$\int_{-\infty}^\infty{e^{i\omega t}\over t^2+y^2}\>dt={\pi\over y}\>e^{-y|\omega|}\qquad(y>0)\ .$$

Answer 2

For the upper half plane, rewrite as

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

Then note

$$\tag 1 \frac{y}{(t-x)^2+y^2} = - \text {Im } \frac{1}{z-t}.$$

Thus, for fixed $t,$ $(1)$ is harmonic in $z=x+iy$ (no calculation necessary!). It follows that if $\Delta_z$ is the Laplacian in $z,$ then

$$\Delta_z u(z) = \frac{1}{\pi}\int_{-\infty}^{\infty}\Delta_z \left [\frac{y}{(t-x)^2+y^2}\right]\,f(t)\,dt = \frac{1}{\pi}\int_{-\infty}^{\infty}0\cdot\,f(t)\,dt =0.$$

This is all provided that differentiation across the integral sign is legit, which for your bounded function $f$ will be standard.