Proof verification: $\int _0 ^{2π} \frac {2e^{it}} {1+(2e^{it})^2}dt=0$

Question

I wanted to compute $\int _0 ^{2π} \frac {2e^{it}} {1+(2e^{it})^2}dt$.

Here is what I tried. $\int _0 ^{2π} \frac {2e^{it}} {1+(2e^{it})^2}dt=\frac {1} i \int _0 ^{2π} \frac{1} {1+(2e^{it})^2}2ie^{it}dt=\frac 1 i [arctan(2e^{it})]_{t=0}^{2π}=0$

However, I've been feeling a little skeptical over my assumption that $\int _0 ^{2π} \frac{1} {1+(2e^{it})^2}2ie^{it}dt=[arctan(2e^{it})]_{t=0}^{2π}$. Nevertheless, is what I've done correct? Why or why not?

Answer 1

Although this doesn't quite answer your question, here's a none complex way to do this. Observe \begin{align} \frac{2e^{it}}{1+4e^{2it}}= \frac{10\cos(t)}{9\sin^2(t)+25\cos^2(t)}-i\frac{6\sin(t)}{9\sin^2(t)+25\cos^2(t)} \end{align} then we see that \begin{align} \int^{2\pi}_0 \frac{10\cos(t)}{9\sin^2(t)+25\cos^2(t)}\ dt= 0 \end{align} by symmetry about $t=\pi$, and likewise for the other integral.

Answer 2

Now, what region is $\arctan z$ defined on in the complex plane? Its derivative $\frac1{1+z^2}$ has poles at $z=\pm i$, and we can't go around one of those poles continuously - we'll have to jump to another branch if we do. The usual branch of $\arctan$ is defined to be nice on $\mathbb{R}$, and when we extend to $\mathbb{C}$, it'll work everywhere except two branch cuts - the ray along the $y$ axis from $i$ to $\infty i$ and the ray from $-i$ to $-\infty i$.

As we go around the circle $|z|=2$ that $2e^{it}$ parametrizes, we cross both of those branch cuts. We can't simply evaluate the antiderivative at both ends, because the antiderivative isn't a continuous function along the path. The argument fails.

OK, how do we patch it? The first approach is the residue theorem; we substitute $z=2e^{it}$ to transform the integral into the line integral $\frac1i \int_C \frac1{1+z^2}\,dz$, where $C$ is the circle of radius $2$. The theorem then tells us that $\int_C \frac1{1+z^2}\,dz= 2\pi i\sum_j r_j$, where the $r_j$ are the residues of the function we're integrating at the singularities inside $C$. Here, those singularities are the poles $\pm i$, with corresponding residues $\pm\frac1{2i}$. Sum them, and we get zero. The integral is zero. Huh - that's the same answer we'd get if there was a continuous antiderivative. Maybe...

Alternatively, we can find an antiderivative that's actually continuous all the way around - a different branch of the arctangent. Note that $\arctan z$ is analytic for $|z|<1$ - if we can invert that and find a branch that's analytic for $|z|>1$, we'll be good.
We can. Let $g(z)=\frac{\pi}{2}-\arctan\left(\frac 1z\right)$ (where $\arctan$ represents the usual branch). We have $g'(z)=-\frac{-1/z^2}{1+1/z^2}=\frac{1}{z^2+1}$, so it's an antiderivative of $\frac1{1+z^2}$ just like $\arctan$. Its definition is naturally analytic for $|z|>1$, and it even agrees with $\arctan$ on the positive reals.

Using this antiderivative, we can then write $$\int_0^{2\pi}\frac{2e^{it}}{1+(2e^{it})^2}\,dt = \frac1i\left[g(2e^{it})\right]_0^{2\pi}=\frac1i(g(2)-g(2))=0$$