Integral of a logarithmic derivative of a complex polynomial over the real line

Question

Let $f(z)$ be a complex polynomial with $n$ zeros in $y>0$, $m$ zeros in $y<0$, and no real zeros. How can we show that
$$displaystylelim_{Rto + infty}displaystyleint_{-R}^R dfrac{f'(x)}{f(x)}dx=pi i (n-m)?$$
If $R$ is large enough so that all the zeros are inside $|z|<R$, then we can apply the argument principle to get $int_{C_1} f'(z)/f(z)~dz=2pi i n$ and $int_{C_2} f'(z)/f(z)~dz=2pi i m$ where $C_1$ is the half circle $Re^{it}, 0leq tleq pi$ with $[-R,R]$, and $C_2$ is the half circle $Re^{it}, -pileq tleq 0$ with $[R,-R]$. But I think I need more information to get the result. Am I missing something? Thanks in advance

Martin R · Answer

If $C_R^+$ and $C_R^-$ are the upper and lower semicircle with radius $R$ (in positive orientation) then (as you already observed)
$$
 int_{C_R^+}  frac{f'(z)}{f(z)} , dz +  int_{-R}^R frac{f'(x)}{f(x)} , dx 
= 2pi i n \
 int_{C_R^-}  frac{f'(z)}{f(z)} , dz -  int_{-R}^R frac{f'(x)}{f(x)} , dx 
= 2pi i m 
$$
for sufficiently large $R$. It follows that
$$
2 int_{-R}^R frac{f'(x)}{f(x)} + int_{C_R^+}  frac{f'(z)}{f(z)} , dz - 
int_{C_R^-}  frac{f'(z)}{f(z)} , dz = 2 pi (n-m) , .
$$
The “missing link” is that
$$
tag{*}
 lim_{R to infty} left(int_{C_R^+}  frac{f'(z)}{f(z)} , dz - 
int_{C_R^-}  frac{f'(z)}{f(z)} , dz right) = 0
$$
and that follows from the asymptotic
$$ 
 frac{f'(z)}{f(z)} = frac{d}{z} + Oleft(frac{1}{z^2}right)
$$
for $z to infty$, where $d$ is the degree of the polynomial: Both integrals in $(*)$ are asymptotically equal to $pi i d$.

Integral of a logarithmic derivative of a complex polynomial over the real line

One Answer

Add your own answers!

Ask a Question