I would like to show that $\text{Res}_{z = \infty}\left(f(z)\log\frac{z-a}{z-b} \right)= – \int_{a}^{b}f(z)\,dz$, where $f(z)$ is an entire function and for $\log\left(\frac{z-a}{z-b}\right)$ we take any branch that is regular at $z = \infty$.
I have tried using both definitions of the residue at infinity but have had no luck.
First I tried the integral definition:
$$\text{Res}_{z = \infty}\left(f(z)\log\frac{z-a}{z-b}\right) = \frac{-1}{2\pi i}\int_{C_r}f(z)\log\left(\frac{z-a}{z-b}\right) $$
But now I am unsure how to compute this integral. Breaking the logarithm up as a difference of two logs did not seem to help. I also tried to parametrize the circle of radius r and write this as a real integral. This got really messy.
Next, I tried the definition which uses the residue at $0$
$$\text{Res}_{z = \infty}\left(f(z)\log\left(\frac{z-a}{z-b}\right) \right)$$
$$ = – \text{Res}_{z = 0}\left(\frac{1}{z^2}f\left(\frac{1}{z}\right)\log\left(\frac{z^{-1}-a}{z^{-1}-b}\right)\right) $$
$$ = – \lim_{z \to 0} \left(\frac{1}{z}f\left(\frac{1}{z}\right)\log\left(\frac{1-az}{1-bz} \right)\right)$$
The above limit computation corresponds to there being a simple pole at $0$ which I'm not even sure is true.
And yet again, I am stuck. Any tips?
Best Answer
Cauchy's integral formula says:
$$f(z) = \frac{1}{2\pi i}\oint_{C}\frac{f(w)dw}{w-z}$$
where $C$ is a counterclockwise contour encircling the point $z$. Integrating both sides from $z = a$ to $b$, assuming that the contour $C$ contains the entire interval gives:
$$\int_a^b f (z)dz = \frac{1}{2\pi i}\oint_{C}\log\left(\frac{w-a}{w-b}\right)f(w)dw$$
The r.h.s. is minus the sum of all residues outside of the contour of the integrand, since $f(z)$ is assumed to be analytic, this is therefore equal to minus the residue at infinity.
One can use this formula (or a generalized version involving a weight function) to derive quadrature rules without having to go through the formalism involving orthogonal polynomials. I've explained this here.