(Complex) integration by substitution

Question

Consider $\gamma\colon [0,2\pi]\longrightarrow \mathbb{C}, \ \gamma(t)=3e^{it}$

I want to compute $\int_\gamma \frac{1}{(z+2)^n} \, \mathrm{d}z$.

I am stuck with $\int_0^{2\pi}\frac{3ie^{it}}{(3e^{it}+2)^n} \, \mathrm{d}t$ where it is tempting to substitute, but this can not be done that easily, as the substitute would also require some sort of path. If I was just naively substituting the result would be wrong. How can this issue be resolved, and this is important, without using Cauchy formular nor any advanced theorem?

Answer 1

You still can use the fundamental theorem of calculus. $\int_\Gamma g(z)dz$ for some path $\Gamma$ is equal to the change of $f(z)$ along the path, where $f(z)$ is a primitive of $g(z)$, i.e. $f'(z)=g(z)$. In your case, assuming $n$ is a positive integer:

$$f(z) = \begin{cases} \frac{-1}{(n-1)(z+2)^{n-1}}, &\text{if $n>1$}\\ \ln(z+2), &\text{if $n=1$} \end{cases}\;\Longrightarrow\; f'(z)=\frac{1}{(z+2)^n}\,. $$

Because $f(z)$ is single valued for $n>1$, and your path is closed, the integral is zero if $n>1$ ($f(z)$ does not change after the path is done). For $n=1$ the change of $\ln(z+2)$ along you path is $i$ times the change of the phase of $z+2$ along the loop, which is $2\pi$, so the integral has a value of $2\pi i$. The big difference in the $n=1$ case is that $f(z)$ is multivalued, so when you return to the original point in the loop $f(z)$ has increased by $2\pi i$.

Answer 2

The problem is easy to solve if $n>1$. In fact, since, under this assumption, the function which is being integrated has a primitive (which is $\frac{-1}{(n-1)(z+2)^{n-1}}$) and since $\gamma$ is a loop, the integral is $0$.

Things are more complicated when $n=1$. In this case, one wants to compute\begin{align}\int_0^{2\pi}\frac{3ie^{it}}{3e^{it}+2}\,\mathrm dt&=\int_0^{2\pi}\frac{-6\sin t}{13+12\cos t}+\frac{9+6\cos t}{13+12\cos t}i\,\mathrm dt\\&=\int_0^{2\pi}\frac{-6\sin t}{13+12\cos t}\,\mathrm dt+i\int_0^{2\pi}\frac{9+6\cos t}{13+12\cos t}\,\mathrm dt.\end{align}

The first of these two last integrals is easy to compute: since the function being integrated is odd and periodic with period $2\pi$,\begin{align}\int_0^{2\pi}\frac{-6\sin t}{13+12\cos t}\,\mathrm dt&=\int_0^\pi\frac{-6\sin t}{13+12\cos t}\,\mathrm dt+\int_\pi^{2\pi}\frac{-6\sin t}{13+12\cos t}\,\mathrm dt\\&=\int_0^\pi\frac{-6\sin t}{13+12\cos t}\,\mathrm dt+\int_{-\pi}^0\frac{-6\sin t}{13+12\cos t}\,\mathrm dt\\&=\int_{-\pi}^\pi\frac{-6\sin t}{13+12\cos t}\,\mathrm dt=0.\end{align}

The second integral is trickier. Formally, a primitive of $\frac{9+6\cos t}{13+12\cos t}$ is $\frac t2+\arctan\left(5\cot\left(\frac t2\right)\right)$, but this is undefined when $t\in\pi\mathbb Z$. So, let us define $F\colon[0,2\pi]\longrightarrow\mathbb R$ by\begin{align}F(t)&=\begin{cases}\frac t2+\arctan\left(5\cot\left(\frac t2\right)\right)&\text{ if }t\in(0,2\pi)\\\lim_{t\to0^+}\frac t2+\arctan\left(5\cot\left(\frac t2\right)\right)&\text{ if }t=0\\\lim_{t\to2\pi^-}\frac t2+\arctan\left(5\cot\left(\frac t2\right)\right)&\text{ if }t=2\pi\end{cases}\\[10pt]&=\begin{cases}\frac t2+\arctan\left(5\cot\left(\frac t2\right)\right)&\text{ if }t\in(0,2\pi)\\-\frac\pi2&\text{ if }t=0\\\frac{3\pi}2&\text{ if }t=2\pi.\end{cases}\end{align}Then $F$ is a primitive of $\frac{9+6\cos t}{13+12\cos t}$. Therefore,$$\int_0^{2\pi}\frac{9+6\cos t}{13+12\cos t}\,\mathrm dt=\frac{3\pi}2-\left(-\frac\pi2\right)=2\pi,$$and so$$\int_0^{2\pi}\frac{3ie^{it}}{3e^{it}+2}\,\mathrm dt=2\pi i.$$