I have a problem solving those two integrals.
- $\int_{\partial D} \frac{z^3}{e^{z^2}-1}dz$, $D=\{z: |z|<4\}$
- $\int_0^2 \frac{\sqrt{x(2-x)}}{x+3}dx$
Since I found them in the old complex analysis course under the title "TRAINING SET FOR RESIDUES AND INTEGRALS", I believe I should solve them using the residue theorem.
As for the first integral, I see that the singularity at $0$ is removable, but we are still left with singularities at $\pm \sqrt{2\pi i}$ and $\pm \sqrt{4\pi i}$. Unfortunately, I don't know how to find residues at those points. I can Taylor expand $e^{z^2}-1$ around any point, but since I divide by this, I don't see how it helps (or maybe is there any easy way to find a Laurent expansion of 1/f, when we have a Taylor expansion of f?)
As for the second one, I'm totally lost. Usually, when I have an integral over a real axis, I find a suitable contour (semi-circle, keyhole) where the integral vanishes over a semicircle etc. Here, however, the integral is over a finite interval, and the numerator is "big", so even if I integrate over a semicircle, I don't think the upper part (over the circle) will vanish (one of my ideas was to substitute $x=u^2$, look at the integral over whole real axis, since the one I'm interested in should be just a real part of that, and integrate over a semicircle – it will be just using a residue theorem with residue at $\sqrt{3}i$, but I don't think the integral over semicircle will vanish (not mentioning the problems with a branch of a square root…).
Any help/techniques would be appreciated.
$$
\begin{align}
&\int_{-1}^{-r}\frac1z\,\mathrm{d}z+\int_{\pi}^0\overbrace{\frac1{re^{i\theta}}}^{1/z}\,\overbrace{\vphantom{\frac1z}ire^{i\theta}\mathrm{d}\theta}^{\mathrm{d}z}+\int_r^1\frac1z\,\mathrm{d}z\\
&=\log(r)+\int_\pi^0i\,\mathrm{d}\theta-\log(r)\\
&=-\pi i
\end{align}
$$
which is $-\pi i$ times the residue of $\frac1z$ at $0$ since the contour takes $\frac12$ turn clockwise around $0$.
Best Answer
Let $I$ be the contour integral
$$I=\oint_{|z|=4} \frac{z^3}{e^{z^2}-1}\,dz$$
There are four simple poles at $z=\pm e^{i\pi/4}\sqrt{2\pi}$ and $z=\pm e^{\ i \pi/4}\sqrt{4\pi}$ contained in $|z|<4$. We can find the residues using L'Hospital's Rule. Proceeding we find that the residues are in general
$$\begin{align} \text{Res}\left(\frac{z^3}{e^{z^2}-1}, z=\pm e^{ \pi/4 }\sqrt{2n\pi}\right)&=\lim_{z\to \pm e^{ i\pi/4}\sqrt{2n\pi}}\frac{(z-e^{\pm \pi/4 }\sqrt{2n\pi})z^3}{e^{z^2}-1}\\\\ &=\lim_{z\to \pm e^{\ i\pi/4}\sqrt{2n\pi}}\frac{3z^2(z-e^{\pm \pi/4 }\sqrt{2n\pi})+z^3}{2ze^{z^2}}\\\\ &=in\pi \end{align}$$
Hence, we find that
$$I=2\pi i (i\pi+i\pi+i2\pi+i2\pi)=-12\pi^2$$
See THIS ANSWER for a detailed explanation of integration around a dog bone contour.
Let $J$ be the real integral given by
$$J=\int_0^2 \frac{\sqrt{x(2-x)}}{x+3}\,dx$$
We can evaluate $J$ be analyzing the contour integral $K$ as given by
$$K=\oint_{C_{DB}}\frac{\sqrt{z(2-z)}}{z+3}\,dz$$
where $C_{DB}$ is the classical dog bone contour around the branch cut from $z=0$ to $z=2$. We can evaluate $K$ by deforming the contour and evaluating the residues at $z=-3$ and at infinity. Proceeding, we have
$$\text{Res}\left(\frac{\sqrt{z(2-z)}}{z+3}, z=-3\right)=i\sqrt{15}$$
$$\text{Res}\left(\frac{\sqrt{z(2-z)}}{z+3}, z=\infty\right)=i4$$
Therefore, we find that
$$K=8\pi-2\sqrt{15}\pi$$
Dividing by $2$ reveals that
$$J=(4-\sqrt{15})\pi$$