Not sure how this inequality is formed - $\bigg|\int_0^{2\pi}\frac{e^{p(R+iy)}}{1+e^{R+iy}}idy\bigg| \le \frac{e^{pR}}{e^R - 1}2\pi$

Question

I have the following inequality in my notes -

$$\bigg|\int_0^{2\pi}\frac{e^{p(R+iy)}}{1+e^{R+iy}}idy\bigg| \le \frac{e^{pR}}{e^R - 1}2\pi$$

We can start as follows $$\bigg|\int_0^{2\pi}\frac{e^{p(R+iy)}}{1+e^{R+iy}}idy\bigg|$$

$$\le\int_0^{2\pi}\bigg|\frac{e^{p(R+iy)}}{1+e^{R+iy}}i\bigg|dy$$

Now I can see the magnitude of the numerator will be bounded by $e^{pR}$. But how is does the bound on the denominator, $e^R - 1$, come about? And where is the $2\pi$ coming from? And for that matter how are we 'dropping' the integration operation?

Answer 1

Notice that $|x+y| > |y| - |x| $. Apply that to the denominator. Do what you did with the numerator. Now your expression is independent of y. So, integrate and you get your 2$\pi$ factor and life works out.

Answer 2

Perhaps you're missing the equality

$$x,y\in\Bbb R\implies\left|e^{x+iy}\right|=e^x$$

Answer 3

Using $|1+e^{R+iy}|\geq|e^{R+iy}|-1=e^{R}-1$ as Rikimaru suggested, we have

$$\bigg|\int_0^{2\pi}\frac{e^{p(R+iy)}}{1+e^{R+iy}}idy\bigg|\leq\int_0^{2\pi}\frac{|e^{p(R+iy)}|}{|1+e^{R+iy}|}dy$$

$$ \leq\int_0^{2\pi}\frac{e^{pR}}{|e^{R+iy}|-1}dy=\frac{e^{pR}}{e^{R}-1}\int_0^{2\pi}dy= \frac{e^{pR}}{e^R - 1}2\pi$$