Creating Difficult integrals by the residue theorem


Let \( f \) be analytic function in the unit circle \( |z|\leq 1 \)  such that \( f\neq 0\) . Then $$\int^{2\pi}_0f(e^{it})\,dt =2\pi \, f(0) $$


Since the function \(f \) is analytic in and on the contour we have by the Cauchy integral theorem

$$\oint_{|z|=1}\frac{f(z)}{z}\,dz = 2\pi i\,\mathrm{Res}(f/z,0)$$

Use \(z = e^{it}\) whenere \(0\leq t \leq 2\pi\)

$$i\int_{0}^{2\pi}\frac{f(e^{it})}{e^{it}}e^{it}\,dt = 2\pi i\,\mathrm{Res}(f/z,0)$$

Note that

$$\mathrm{Res}\left( \frac{f(z)}{z},0\right) = \lim_{z\to 0}z \frac{f(z)}{z} = f(0)$$


$$\int^{2\pi}_0f(e^{it})\,dt =2\pi \, f(0) $$

Example 1

Consider the function


It then follows from the theorem

$$\int^{2\pi}_0 \exp(\exp(\exp(it)))\,dt = 2\pi \, e$$

Note that

&= e^{e^{\cos t+ i\sin t}} \\
&= e^{e^{\cos t}(\cos( \sin t)+i\sin(\sin t))}\\
&= e^{e^{\cos t}\cos(\sin t)}(\cos(e^{\cos t}\sin(\sin t))+i\sin(e^{\cos t}\sin(\sin t)))\end{align}

By taking the real part

$$\int^{2\pi}_0 e^{e^{\cos t}\cos(\sin t)} \cos(e^{\cos t}\sin(\sin t))\,dt = 2\pi \,e$$

Example 2

$$f(z) = \log(2+z)$$

Where we take the principal logarithm which the branch cut located at \(y=0,x \leq -2\)

$$\int^{2\pi}_0 \log(2+e^{it}) = 2\pi \log(2)$$

Note that

\begin{align}\log(2+e^{it}) &=\log|(2+\cos t)^2+\sin^2t|+i\arctan \left( \frac{\sin t}{1+\cos t}\right) \\
&= \frac{1}{2}\log(5+4\cos t)+ i\arctan \left( \frac{\sin t}{2+\cos t}\right)

It follows then

$$\int^{2\pi}_0 \log(5+4\cos t)\,dt = 8\pi \log(2)$$

Example 3

By compining the two functions

$$f(z) = \exp(\exp(z)) \log(2+z)$$

We deduce that

$$\small \int^{2\pi}_0 \exp(e^{\cos t}\cos(\sin t)) ( \cos(e^{\cos t}\sin(\sin t))\log(5+4\cos t)-2\arctan \left( \frac{\sin t}{2+\cos t}\right)\sin(e^{\cos t}\sin(\sin t))) \,dt= 4\pi e \log(2)$$

This entry was posted in Contour Integration and tagged , , , , , , , . Bookmark the permalink.

Leave a Reply