Why does the following inequality hold ? $\left|\int_0^1 e^{ire^{\pi i t}} \, dt\right|\leq \int_0^1 e^{-r\sin(\pi t)}dt$

Question

In one of your complex analysis exercise, we had to find an integral using the residue theorem, and at one moment, to prove that one section of our domain border tends to $0$ as the radius tends toward infinity, our professor used the following inequality :

$$\left|\int_0^1 e^{ire^{\pi i t}} \, dt\right| \leq \int_0^1 e^{-r\sin(\pi t)} \, dt$$

But I have some troubles figuring out why this is true. How do you prove this inequality, where does the $-r\sin(\pi t)$ come from ?

I searched on the internet but couldn't find more information about this inequality.

Thanks for your help !

Answer 1

Since $$e^{ire^{\pi it}}=e^{ir(\cos\pi t+i\sin\pi t)}=e^{-r\sin\pi t+ir\cos\pi t},$$ we can see that $$|e^{ire^{\pi it}}|=e^{-r\sin\pi t}.$$ Now you can use triangle inequality.

Answer 2

First you have this: $$ \left|\int_0^1 e^{ire^{\pi i t}} \, dt\right| \le \int_0^1 \left|e^{ire^{\pi it}} \right| \, dt $$ The proof of this is a bit more involved than if we had had only real-valued functions. At this point I'll suggest that if you don't know how to do this part, then say so in comments below.

Next: $$ \left| e^{ire^{\pi it}} \right| = \left| e^{ir(\cos (\pi t) + i\sin(\pi t))} \right| = \left| e^{ir\cos(\pi t) -r\sin(\pi t)} \right| = e^{-r\sin(\pi t)} \left| e^{ir\cos(\pi t)} \right| = e^{-r \sin(\pi t)} $$ That last equality holds because if $a,b$ are real then \begin{align} \left| e^{a+ib} \right| = {} & e^a \left| e^{ib} \right| \\ & \text{since $e^a$ is a positive real number} \\[12pt] = {} & e^a \left| \cos b + i\sin b \right| = e^a \sqrt{(\cos b)^2 + (\sin b)^2} = e^a. \end{align}