Integral $\int_{\vert z \vert = 5}{ \frac{e^{z^3}}{\sin(z)}} dz$

Question

I am currently studying complex analysis, and I came across this problem: Compute $$\int_{\vert z \vert = 5}{ \frac{e^{z^3}}{\sin(z)}} dz.$$ My first reflex was to look at singularities. For that I looked at the zeroes of $\sin(z)$. After some research I ended up finding that any $z= k\pi + 0i$ where $k \in \mathbb{Z}$, but I am only focusing on the three following : $(-\pi, 0); (0,0); (\pi, 0)$ because they are the only one in the disk of radius $5$ that we are integrating over. Then I compute the limit of $f(z)$ around these singularities and see that it goes to infinity $ \Rightarrow$ it is a pole. But now I don't know how to proceed to compute the integral, what should the next step be?

Answer 1

We have the function $$f(z)=\frac{e^{z^3}}{\sin{z}}$$ which is not analytic at $z=k\pi$, $\forall k \in \mathbb{Z}$.
In the given disk of $|z|= 5$, we have simple poles at $0, \pi, -\pi$

We can write the function $f(z)$ in the form of $$f(z)=\frac{\phi(z)}{\psi(z)}$$ where $\phi(z)$ and $\psi(z)$ have NO Common term and $\phi(z_0) \not=0 $, $\psi(z_0)=0$ and $\psi'(z_0) \not=0$ .Since $\psi'(z)$ is analytic at every point in $C$ we can expand it as $$\psi(z) = \psi(z_0)+(z-z_0)\psi'(z+0) + . . .$$ since the Residue at simple pole is calculated as $Res_{z=z_0} f(z) = lim_{z \to z_0} \left[(z-z_0)\frac{\phi(z)}{\psi(z)}\right] =\frac{\phi(z_0)}{\psi'(z_0)}$

we have $f(z)=f(z)=\frac{e^{z^3}}{\sin{z}}=\frac{\phi(z)}{\psi(z)} $ therefore, $$\frac{\phi(z)}{\psi'(z)} = \frac{e^{z^3}}{\cos{z}}$$ we have three points on which the function is not analytic so we have to calculate the Residues at three points

(i) at $z=0$,

$$Res_{z=0}f(z)=lim_{z \to 0}\left[\frac{e^{z^3}}{\cos{z}}\right] = 1 $$ (ii)at $z=\pi$,

$$Res_{z=\pi} f(z) = lim_{z \to \pi} \left[\frac{e^{z^3}}{\cos{z}}\right]=-e^{{\pi}^3}$$ (iii) at $z=-\pi$, $$Res_{z=-\pi} f(z) = lim_{z \to -\pi}\left[\frac{e^{z^3}}{\cos{z}}\right]= -e^{{-\pi}^3}$$ from the Residue theorem, $$\oint_{C} f(z)dz = 2\pi\iota\sum [Res_{z=z_0} f(z)] $$ so the integral is : $$ \oint_{|z|=5}\frac{e^{z^3}}{\sin{z}}dz = 2\pi\iota[1-e^{{\pi}^3}-e^{{-\pi}^3}]$$