How to do this integral $\int_0^{2\pi}\frac{1}{e^{it}+x}ie^{it}dt$?

Question

For context, I wish to evaluate the below contour integral

$$\int_\Gamma \frac1z dz$$

where $\Gamma$ is a circle of radius $1$ centered at distance $x$ away from origin on positive x-axis. In particular, I want to show that

$$\int_\Gamma \frac1z dz=\int_{\Gamma_0} \frac1z dz=2\pi i$$

when $0\le x < 1$. Here, $\Gamma_0$ is the unit circle. So, I am trying to explicitly evaluate the integral to show that

$$\int_0^{2\pi}\frac{1}{e^{it}+x}ie^{it}dt=\begin{cases}2\pi i & \text{if $0\le x<1$} \\ 0 & \text{if $x\ge 1$} \end{cases}$$

I started with substitution $u=e^{it}+x$ to get $dt=-ie^{-it}du$ which results in

$$\int_{\frac32}^{\frac32}du=1-1=0$$

I am not sure where I did wrong.

Answer 1

If you want a real-analytic approach, you may invoke the geometric series formula

$$ \frac{1}{1-r} = \sum_{n=0}^{\infty} r^n $$

which converges uniformly on any compact subset of the unit disk $\mathbb{D} = \{z\in\mathbb{C}:|z|<1\}$.

Case 1. If $|x|<1$, then

\begin{align*} \int_{0}^{2\pi} \frac{ie^{it}}{e^{it} + x} \, \mathrm{d}t &= \int_{0}^{2\pi} \frac{i}{1 + xe^{-it}} \, \mathrm{d}t \\ &= \int_{0}^{2\pi} i \sum_{n=0}^{\infty} (-xe^{-it})^n \, \mathrm{d}t \\ &= i \sum_{n=0}^{\infty} (-x)^n \int_{0}^{2\pi} e^{-int}\, \mathrm{d}t \\ &= i \sum_{n=0}^{\infty} (-x)^n \cdot 2\pi \mathbf{1}_{\{n = 0\}} \\ &= 2\pi i. \end{align*}

Case 2. If $|x|>1$, then

\begin{align*} \int_{0}^{2\pi} \frac{ie^{it}}{e^{it} + x} \, \mathrm{d}t &= \int_{0}^{2\pi} \frac{ix^{-1}e^{it}}{1 + x^{-1}e^{it}} \, \mathrm{d}t \\ &= \int_{0}^{2\pi} i \sum_{n=0}^{\infty} (-1)^n (x^{-1}e^{it})^{n+1} \, \mathrm{d}t \\ &= i \sum_{n=0}^{\infty} \frac{(-1)^n}{x^{n+1}} \int_{0}^{2\pi} e^{i(n+1)t} \, \mathrm{d}t \\ &= 0. \end{align*}

Answer 2

Yours is ultimately a contour integral $$ \int_{\Gamma_0} \frac{dz}{z+x},\qquad \Gamma_0=\{z\in \Bbb C\ :\ |z|=1\}. $$ By the main theorems of complex integration, the value of the integral depends on whether the real parameter $x\geq 0$ is contained within the region bounded by the contour $\Gamma_0$, i.e. the open unit disk $\{|z|<1\}$, or lies outside it. In the latter case, that is, when $x>1$, you may use Cauchy's theorem to deduce that the integral vanishes. In the former case ($x<1$), the residue theorem tells you that the integral evaluates to $$2\pi i \operatorname{Res}_{z=-x} \frac{1}{z+x}=2\pi i.$$

(The integral as written is undefined when $x=1$, although one may certainly calculate its Cauchy principal value.)

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Edit. (The user has updated their question.) Alternatively, you may "avoid" the residue theorem as follows. A variant of Cauchy's theorem – a corollary, or a lemma, depending on how you want to go about the proofs – tells you that if two contours $\Gamma_0,\Gamma$ lie within the region of holomorphicity $H$ of the integrand $f(z)$, and they are homotopic (i.e. one is continuously deformable into the other without exiting $H$), then $$ \int_{\Gamma}f(z)\ d{z} = \int_{\Gamma_0}f(z)\ d{z}.$$ In your case, take $f(z) := 1/z$, $H:=\Bbb C\setminus\{0\}$, $\Gamma_0 := \{|z|=1\}$, $\Gamma := \{|z+x|=1\}$ to reduce the LHS to a very famous complex integral, which you may compute by substituting $z=e^{it}$.

Answer 3

Putting $z=e^{it}$ converts the integral into a contour integral $$ \int_0^{2 \pi} \frac{1}{e^{i t}+x} i e^{i t} d t= \oint_{|z|=1} \frac{d z}{z+x} $$ A. For $x\ge 1$, $$z+x=0 \Rightarrow z=-x \textrm{ and hence }|z|=x\ge 1 \textrm{ and there is no singularity inside } |z|=1. $$

$\quad $ By Cauchy’s Theorem,$ \displaystyle \qquad \oint_{|z|=1} \frac{d z}{z+x} =0.$

B. For $x<1$, $$z+x=0 \Rightarrow z=-x \textrm{ and hence }|z|=x<1 \textrm{ and there is one singularity}-x \textrm{ inside } |z|=1. $$

$\quad $ By Cauchy’s Theorem,$ \displaystyle \qquad \oint_{|z|=1} \frac{d z}{z+x} =2\pi i \lim _{z \rightarrow-x}(z+x) \frac{1}{z+x}=2 \pi i.$

We can now conclude that

$$ \int_0^{2 \pi} \frac{1}{e^{i t}+x} i e^{i t} d t =\begin{cases}0 & \text { if } x \geqslant 1 \\ 2 \pi i & \text { if } x<1\end{cases} $$