one complex variables (integration)

Question

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 $R>0$

Answer 1

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.