Evaluate $\int_0^{2\pi}\frac{i}{e^{it}+2}dt$

Question

I was doing a contour integration over a circle and, after a few steps, got stuck here:

$$\int_0^{2\pi}\frac{i}{e^{it}+2}dt=-\frac{1}{2}\log(1+2e^{-it})\Big\vert_0^{2\pi}=0$$ which is clearly not correct.

I know this has to do with the fact that the logarithm function can't be simultaneously defined at $0$ and $2\pi$. How could I formalize the argument?

BTW I know this can be done using the Cauchy Integral formula. But we are not there yet. I'd appreciate a more elementary approach.

Answer 1

Your integral is equal to$$\int_{\lvert z\rvert=1}\frac1{z(z+2)}\,\mathrm dz$$and, by Cauchy's integral formula, this integral is equal to$$\frac{2\pi i}{0+2}=\pi i.$$

Answer 2

It equals $\int_{\mid z\mid=1}\frac1{z(z+2)}\operatorname dz=\frac12\int_{\mid z\mid=1} (\frac1z-\frac 1{z+2})\operatorname dz=\frac12(\int_0^{2\pi}\frac1{e^{it}}ie^{it}\operatorname dt-\int_{\mid z\mid=1}\frac{\operatorname dz}{z+2})=\pi i $ (where the second integral is zero by Cauchy's theorem).

Answer 3

Separate the real and imaginary parts; the integral of the real part is easily seen to be $0$; the integral of the imaginary part is, with the Weierstrass substitution, \begin{align} i\int_0^{2\pi}\frac{\cos t+2}{5+4\cos t}\,dt &=i\int_{-\pi}^{\pi}\frac{\cos t+2}{5+4\cos t}\,dt \\[4px] &=2i\int_{0}^{\pi}\frac{\cos t+2}{5+4\cos t}\,dt \\[4px] &=2i\int_0^{\infty}\frac{1-u^2+2+2u^2}{5+5u^2+4-4u^2}\frac{2}{1+u^2}\,du \\[4px] &=4i\int_0^\infty\frac{u^2+3}{(u^2+9)(u^2+1)}\,du \end{align} With partial fractions this becomes $$ i\int_0^{\infty}\left(\frac{3}{u^2+9}+\frac{1}{u^2+1}\right)\,du =2i\int_0^\infty\frac{1}{u^2+1}=2i\frac{\pi}{2}=\pi i $$

Answer 4

Since you asked for a solution not using Cauchy's formula, we need to develop $e^{it}$.

$\displaystyle \int_0^{2\pi}\dfrac{i}{e^{it}+2}\mathop{dt}=\int_0^{2\pi}\dfrac{i}{\cos(t)+2+i\sin(t)}\mathop{dt}=\int_0^{2\pi}\dfrac{i\left(\cos(t)+2-i\sin(t)\right)}{4\cos(t)+5}\mathop{dt}=I_1+I_2+I_3$

$\displaystyle I_1=\int_{0}^{2\pi}\dfrac i4\mathop{dt}=\dfrac{i\pi}2$

$\displaystyle I_3=\int_{0}^{2\pi}\dfrac {\sin(t)}{4\cos(t)+5}\mathop{dt}=\bigg[-\frac 14\ln(4\cos(t)+5)\bigg]_0^{2\pi}=0$

For the last one we have to be careful, because we need a substitution in $\tan(\frac t2)$ which is not bijective on the given interval $[0,2\pi]$

$\displaystyle I_2=\int_{0}^{2\pi}\dfrac {3i/4}{4\cos(t)+5}\mathop{dt}$

If we do not pay attention this can result in $\arctan(\tan(\pi))-\arctan(\tan(0))=0$ instead of $\pi$, this is the same problem you get with the complex logarithm in your attempt.

By substitution $t'=2\pi-t$ it is immediate that $\displaystyle I_2=2\int_{0}^{\pi}\dfrac {3i/4}{4\cos(t)+5}\mathop{dt}$

Now we can substitute $u=\tan(\frac t2)$ this gives $\displaystyle I_2=\int_0^{\infty}\dfrac{3i}{u^2+9}\mathop{du}=\bigg[i\arctan\left(\frac u3\right)\bigg]_0^{\infty}=\dfrac{i\pi}2$

And the result $I=I_1+I_2+I_3=\dfrac{i\pi}{2}+\dfrac{i\pi}{2}+0=i\pi$