TITLE: Evaluating $\oint_{\gamma(t)} \frac{e^z}{(z+1)^3}dz$ where $\gamma(t)$ is the circle with radius $2$ centered at $z=0$
QUESTION [1 upvotes]: I have to evaluate the following integral $\oint_{\gamma(t)} \frac{e^z}{(z+1)^3}dz$ where $\gamma(t)$ is the circle with radius $2$ centered at $z=0$.
I know that $\int f(z)dz = \int f(\gamma(t))\dot{\gamma}(t)dt$ and I know that the path for a circle with radius $2$ around $z = 0$ is $\gamma(t) = 2e^{it}$. Then, $\dot{\gamma}(t) = 2ie^{it}$. Inserting this into the integral I get $\oint_{\gamma(t)} \frac{e^z}{(z+1)^3}dz = \int_0^{2\pi} \frac{e^{2e^{it}}}{{(2e^{it} + 1})^3} 2ie^{it}dt$. For solving it, I can make the substitution $u = 2e^{it} ; \ dt = \frac{du}{2ie^{it}} = \frac{du}{iu}$ and then I get $\int \frac{{e^{u}}}{(u + 1)^3}\frac{iu}{iu}du = \int \frac{{e^{u}}}{(u + 1)^3} du$.
But here I'm stuck.

Did I got it right up to the last integral with the substitution?
Any tips for how to continue?

REPLY [2 votes]: $\newcommand{\d}{\mathrm{d}}$Unfortunately, the majority of complex integrals (or just the majority of all integrals) that can be easily written down can't be easily found if you only use techniques of Riemann integration. In general, it is good advice to completely avoid substitutions $z=\gamma(t)$! Complex analysis has produced many beautiful theorems involving integration of complex contours.
The one I will present to you now is the Cauchy Integral Formula:

$$f^{(n)}(s)=\frac{n!}{2\pi i}\oint_\gamma\frac{f(z)}{(z-s)^{n+1}}\,\d z$$

If $f$ is a holomorphic function in some neighbourhood of $s\in\Bbb C$, and $\gamma$ is a closed contour that is contained in this neighbourhood which also winds counterclockwise - once - about the point $s$.
Here, $s=-1$ and the circle of radius $2$ - I assume a counterclockwise winding - is a contour of the correct type as it encloses $-1$. $f(z)=e^z$ is of course holomorphic in this neighbourhood of $-1$. It follows that: $$\frac{\d^2}{\d z^2}[e^z]\Big |_{z=-1}=\frac{2!}{2\pi i}\oint_{\gamma}\frac{e^z}{(z+1)^3}\,\d z$$But that just means to say your integral equals $\frac{\pi i}{e}$.
{"set_name": "stack_exchange", "score": 1, "question_id": 4553631}