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I have a problem solving those two integrals.

  1. $\int_{\partial D} \frac{z^3}{e^{z^2}-1}dz$, $D=\{z: |z|<4\}$
  2. $\int_0^2 \frac{\sqrt{x(2-x)}}{x+3}dx$

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.

D_mathematician
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1 Answers1

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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

$$J=(4-\sqrt{15})\pi$$

Mark Viola
  • 179,405
  • Ah, I've got confused since the pole of $\frac{1}{e^{z^2}-1}$ is not simple at $0$, if I'm not mistaken. Also, shouldn't the residue be equal to $-in\pi$ when $z\rightarrow e^{-i\pi/4}$, and thus the integral will be equal to 0? – D_mathematician May 10 '23 at 16:02
  • @D_mathematician Nope. The residues at $\pm i\pi/4$ are in fact equal. The $z^3/z=z^2$ term washes away the minus sign. – Mark Viola May 10 '23 at 16:11
  • But isn't $(e^{-i\pi/4})^2=e^{-i\pi/2}=-i$? – D_mathematician May 10 '23 at 16:16
  • @D_mathematician We have $z^2=\sqrt{i}$. So $z^2=e^{i\pi/2+2\ell \pi}$ and hence $z=\pm e^{i\pi/4}$. I'll fix the erratum. – Mark Viola May 10 '23 at 16:44