$\int_0^{2\pi} \left| \sum_{n\in \mathbb N} e^{-nt} e^{in\theta} \right| d\theta \leq C(t) C'$?

Question

Fix $t>0$, $$\int_0^{2\pi} \left| \sum_{n\in \mathbb N} e^{-nt} e^{in\theta} \right| d\theta \leq C(t) C'$$

where $C(t)$ is some constant depending on $t$ and $C'$ is some constant (independent of $t$).

My question is: Can we expect to find $C(t)$, I'm curious to know behavior of $C(t)$ at $t\to 0$ and $t \to \infty$

Answer 1

$\int_0^{2\pi} |\sum_ne^{-nt} e^{in\theta}|d\theta \leq (\int_0^{2\pi} |\sum_ne^{-nt} e^{in\theta}|^{2}d\theta)^{1/2} \sqrt {2\pi}\leq \sum_n e^{-2nt}\sqrt {2\pi}$. You can take $C'=\sqrt {2\pi}$ and $C(t)=\sum_n e^{-2nt}$. Even though $C(t)<\infty$ for all $t$ this function blows up as $t \to 0$.

Answer 2

Hint: Triangle inequality shows $$\left|\sum_{n\in \mathbb N} e^{-nt} e^{in\theta}\right|\leq \sum_{n\in \mathbb N}\left| e^{-nt} e^{in\theta}\right| = \sum_{n\in \mathbb N} e^{-nt} = \dfrac{1}{e^t-1}$$ for $t\geq0$.