Prove that the following function has a branch cut

Question

I am given with a function $$\zeta(z)=\int_{-\infty}^\infty\mathrm{d}x \frac{f(x)}{z-x}.$$ Any idea to prove that $\zeta(z)$ is discontinuous across real axis for $f(x)\neq 0$?

Answer 1

Let $z=x+iy\;$ with $\;y>0$, then \begin{align} \zeta(z)&=\int_{-\infty}^\infty \frac{f(t)}{z-t}dt\\ &=\int_{-\infty}^\infty \frac{(x-t)f(t)}{(x-t)^2+y^2}dt-i\int_{-\infty}^\infty \frac{yf(t)}{(x-t)^2+y^2}dt. \end{align}

We recall Poisson's integral for the half plane:
Theorem. Assume that $f(t)$ is piecewise continuous and bounded for all real $t$,$$ P_f(z)=\frac{1}{\pi}\int_{-\infty}^\infty \frac{yf(t)}{(x-t)^2+y^2}dt$$ represents a harmonic function in the upper half plane with boundary values $\lim_{y\to +0}P_f(z)=f(x)$ at points of continuity.

By this theorem we have $$ \lim_{y\to +0} \zeta(z)=I_f(x) -i\pi f(x) $$ if $f$ is continuous at $x$ and the limit $$I_f(x)=\lim_{y\to 0}\int_{-\infty}^\infty \frac{(x-t)f(t)}{(x-t)^2+y^2}dt$$ exists.

Now we consider $\lim_{y\to -0}\zeta(z)$. When $y<0$, \begin{align} \zeta(z)&=\int_{-\infty}^\infty \frac{f(t)}{z-t}dt\\ &=\int_{-\infty}^\infty \frac{(x-t)f(t)}{(x-t)^2+y^2}dt+i\int_{-\infty}^\infty \frac{(-y)f(t)}{(x-t)^2+y^2}dt\\ &\to I_f(x)+i\pi f(x) \quad (\,y<0,\, y\to 0\,), \end{align} since $$ \lim_{y\to -0} \int_{-\infty}^\infty \frac{(-y)f(t)}{(x-t)^2+y^2}dt= \lim_{\xi\to +0}\int_{-\infty}^\infty \frac{\xi f(t)}{(x-t)^2+\xi^2}dt=\pi f(x).$$ Thus we see that $$\lim_{y\to -0}\zeta(z)=\lim_{y\to +0}\zeta(z)+2i\pi f(x),$$ which implies that $\zeta (z)$ is discontinuous across the real axis.