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how to prove $\int_{C_R}\frac{\log^3(z)}{(1+z^2)^2}\,dz$ goes to $0$ as $R$ goes to $\infty$, with $C_R=Re^{it}$ for $0<t<\pi$, and $R>0$

Kyle Miller
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Abdallah
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  • Your contour appears to be the upper half of the unit circle. Do you mean $z=Re^{it}$, $0\le t\le \pi$? If so, then have a look at the denominator. It is of order $R^4$ for large $R$. – Mark Viola Apr 24 '15 at 23:09
  • @Dr.MV I think the principal branch of $\log$ isn't defined on the negative real axis so the path can't be defined at $z = Re^{i\pi}$ hence the strict inequalities? – Ilham Apr 25 '15 at 00:16
  • @Ilham Actually, one can choose to cut the plane in a variety of ways. So, the negative real axis may be included. But that was not the point of the question. – Mark Viola Apr 25 '15 at 02:01
  • @Dr.MV Yeah, I know that, but since he hasn't said the branch I'm assuming it's the one where the log isn't defined on the negative real axis. Anyway it doesn't matter, if you read his history he has a tendency not to reply to feedback. – Ilham Apr 25 '15 at 02:14

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See: http://en.wikipedia.org/wiki/Estimation_lemma

An upper bound for your integral is the path length multiplied by the maximum of the modulus of your integrand along the path). Then show that this bound converges to $0$ as $R$ tends to infinity.

Let $z \in C_R$. Then $|\log(z)| = |\ln|z|+i\text{Arg}(z) |\leq |\ln|z|| + |\text{Arg}(z)| < \ln R + \pi $

Also, note that $|z|^2 − 1=|z^2| − 1=|z^2| − | − 1| \leq |z^2 − (−1)| = |z^2 + 1|$.

And if $|z|\geq \sqrt{2}$, then $|z|^2-1 \geq \frac{|z|^2}{2}$.

So if $|z|\geq \sqrt{2}$, then $\frac{1}{|z^2 + 1|} \leq \frac{1}{|z|^2 − 1} \leq \frac{2}{|z|^2}$ so $\frac{1}{|z^2 + 1|^2} \leq \frac{4}{|z|^4}$.

So on $C_R$, $$|\frac{\log^3(z)}{(1+z^2)^2} |= \frac{|\log^3(z)|}{|1+z^2|^2} \leq\frac{(\ln(R)+\pi)^3}{|1+z^2|^2}\leq\frac{4(\ln(R)+\pi)^3}{|z|^4}=\frac{4(\ln(R)+\pi)^3}{R^4}$$.

Now show that this bound converges to $0$ as $R$ goes to infinity by L'hopital's rule.

So since by the estimation lemma $$|\int_{C_R}\frac{\log^3(z)}{(1+z^2)^2}\,dz| \leq \pi . \max_\limits{z \in C_R} |\frac{\log^3(z)}{(1+z^2)^2} | \leq \pi . \frac{4(\ln(R)+\pi)^3}{R^4}$$ and the RHS converges to $0$, your integral converges to $0$ as $R$ tends to infinity.

Ilham
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