Show that $\text{Res}_{z = \infty}\left(f(z)\log\left(\frac{z-a}{z-b}\right)\right) = - \int_{a}^{b}f(z)\,dz$

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?

First, the residue at infinity can be written as

$$\text{Res}\left(f(z)\log\left(\frac{z-a}{z-b}\right),z=\infty\right)=\text{Res}\left(-\frac1{z^2}f(1/z)\log\left(\frac{1-az}{1-bz}\right),z=0\right)$$

Second, inasmuch as $$f$$ is entire, we can write $$f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(0)z^n}{n!}$$ whereupon integrating term by term reveals

$$\int_a^b f(z)\,dz=\sum_{n=0}^\infty \frac{f^{(n)}(0)}{(n+1)!}\left(b^{n+1}-a^{n+1}\right)\tag1$$

Third, writing $$\log(1-x)=-\sum_{n=1}^\infty \frac{x^n}n$$, we have

\begin{align} -\frac{1}{z^2}f\left(\frac1z\right)\log\left(\frac{1-az}{1-bz}\right)&=-\sum_{m=0}^\infty \frac{f^{(m)}(0)}{m!\,z^m}\sum_{n=1}^\infty \frac{b^nz^{n-2}-a^nz^{n-2}}{n}\\\\ &=-\sum_{m=0}^\infty \sum_{n=0}^\infty \frac{f^{(m)}(0)}{m!}\frac{b^{n+1}z^{n-m-1}-a^{n+1}z^{n-m-1}}{n+1}\tag2 \end{align}

The residue of $$(2)$$ at $$0$$ is the coefficient on the $$z^{-1}$$ term, which occurs when $$n=m$$. Hence, using $$(2)$$ we find that

$$\text{Res}\left(f(z)\log\left(\frac{z-a}{z-b}\right),z=\infty\right)=-\sum_{n=0}\frac{f^{(n)}(0)}{(n+1)!}\left(b^{n+1}-a^{n+1}\right)\tag3$$

Comparing $$(1)$$ and $$(3)$$ yields the coveted identity

$$\text{Res}\left(f(z)\log\left(\frac{z-a}{z-b}\right),z=\infty\right)=-\int_a^b f(z)\,dz\tag4$$

as was to be shown!

The minus sign on the integral in $$(4)$$ is a consequence of the reversal of orientation upon the transformation $$z\mapsto 1/z$$ when using the residue at infinity.

• Beautiful answer, thank you. – Nicholas Roberts Apr 2 at 4:30
• Can you check the signs in Eq (2)? $-\log(1-az)$ should yield $+ a^n z^n$, so I think the signs for $a$ and $b$ should be swapped. – Count Iblis Apr 2 at 4:51
• @CountIblis Indeed. I've edited accordingly. Thank you for catching this. – Mark Viola Apr 2 at 14:08

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.

• @NicholasRoberts This involves interchanging the order of integration allowing one to integrate over $z$ inside the $w$ integral. This interchange of the order of integration needs to be justified, in this case it will only force you to define the branch cut of the logarithm. – Count Iblis Apr 2 at 4:41
• You are missing a minus sign and this is because the integral with respect to z evaluates to $\log(w - b) - \log(w-a)$. Swapping the order of the subtraction makes the minus sign pop out. – Nicholas Roberts Apr 2 at 4:41
• I'll check again alter, but note that $1/(w-z)$ integrates to $-\log(w-z)$ inserting the limits yields $\log(w-a) - \log(w-b)$. – Count Iblis Apr 2 at 4:48
• Yes that is right, my mistake. I wonder where the minus sign went though. There is supposed to be one if you look at the integral definition of residue at infinity. – Nicholas Roberts Apr 2 at 4:49
• @NicholasRoberts Yes, that's why I suspect that there is a minus sign missing in the problem statement. The limit of the contour integral should be minus the residue at infinity, because we define that residue such that the value of a contour integral equals minus $2\pi i$ times the sum of all residues outside the contour. The sum of all residues, including the one at infinity is then by definition, equal to zero. – Count Iblis Apr 2 at 4:54

Note that $$f(z) = \frac{1}{2\pi i}\int_{C_R}\frac{f(w)}{w-z}dw$$ where we take $$C_R$$ to be the circle of radius large enough to contain the line from $$a$$ to $$b$$. Then we have $$\int_a^bf(z)dz = \int_a^b\frac{1}{2\pi i}\int_{C_R}\frac{f(w)}{w-z}dw dz$$. We can use Fubini's theorem (why?) to swap the order of integration to get $$\frac{-1}{2\pi i}\int_{C_R}f(w)(\log\frac{w-b}{w-a})dw$$, where the negative sign comes out when integrating (that part is basic calculus). This is exactly equal to the residue at infinity of the integrand. Note that we have to take a branch cut of $$\log\frac{w-b}{w-a}$$ here - in particular it is regular at infinity. To do this, we can let this branch cut be the preimage of the principal branch under the transformation $$\zeta = \frac{z-b}{z-a}$$, which happens to be the segment from $$a$$ to $$b$$. This does not interfere with the integration since we picked the cut after integrating along this segment.