Calculate the integral $\int_0^{2\pi}\sum_{k=n}^{\infty}\frac{e^{i(k-m)\theta}}{k+1}d\theta$

Question

I would like to calculate this integral:

$$\int_0^{2\pi}\sum_{k=n}^{\infty}\frac{e^{i(k-m)\theta}}{k+1}d\theta$$ where $n$ and $m$ belong to $\mathbb{N}$.

My attempt:

Instead of considering an infinite sum, I will consider a finite one.

Let $l>n$, we have:

\begin{alignat*}{2} \int_0^{2\pi}\sum_{k=n}^{l}\frac{e^{i(k-m)\theta}}{k+1}d\theta=\sum_{k=n}^{l}\frac{1}{k+1}\int_0^{2\pi}e^{i(k-m)\theta}d\theta \end{alignat*}

Taking the limit when $l$ tends to $\infty$, we have:

$$\lim_{l\to\infty}\int_0^{2\pi}\sum_{k=n}^{l}\frac{e^{i(k-m)\theta}}{k+1}d\theta=\left \{ \begin{array}{l c r} \frac{2\pi}{m+1} & \text{if} & m\geq n \\ 0 & \text{if} & m<n \end{array} \right. $$

In this case, do we have: $$\lim_{l\to\infty}\int_0^{2\pi}\sum_{k=n}^{l}\frac{e^{i(k-m)\theta}}{k+1}d\theta=\int_0^{2\pi}\sum_{k=n}^{\infty}\frac{e^{i(k-m)\theta}}{k+1}d\theta\quad?$$ Many thank's in advance.

Answer 1

Consider a branch of logarithm $$f(z):=-\ln(1-z)$$ for $z\in\mathbb{C}\setminus\mathbb{R}_{\geq 1}$ that coincides with $\displaystyle\sum_{k=1}^\infty\,\frac{z^k}{k}$ when $|z|<1$. Define also $$f_r(z):=\sum_{k=1}^{r}\,\frac{z^k}{k}$$ for all $z\in\mathbb{C}$ and $r\in\mathbb{Z}_{\geq0}$. It is known that $\lim\limits_{r\to\infty}\,f_r(z)=f(z)$ for all $z\in\mathbb{C}$ such that $|z|\leq1$ and $z\neq 1$.

We now define $$g_r(\theta):=f_r\big(\text{e}^{\text{i}\theta}\big)=\sum_{k=1}^{r}\,\frac{\text{e}^{\text{i}k\theta}}{k}$$ and $$g(\theta):=f\big(\text{e}^{\text{i}\theta}\big)=-\ln\big(1-\text{e}^{\text{i}\theta}\big)$$ for all $\theta\in(0,2\pi)$ and $r\in\mathbb{Z}_{\geq 0}$. Thus, $g_r\to g$ pointwise as $r\to\infty$. For each compact subset $K$ of $(0,2\pi)$, we want to show that the family $\Big(g_r\big|_K\Big)_{r=0}^\infty$ is uniformly equicontinuous. This proves that $g_r\big|_K\to g\big|_K$ uniformly as $r\to\infty$, and thereby, proving the OP's assertion that $$\lim_{l\to\infty}\,\int_0^{2\pi}\,\frac{g_{l+1}(\theta)-g_n(\theta)}{\text{e}^{\text{i}(m+1)\theta}}\,\text{d}\theta=\int_0^{2\pi}\,\frac{g(\theta)-g_n(\theta)}{\text{e}^{\text{i}(m+1)\theta}}\,\text{d}\theta\,.$$

Here is the proof of the claim above. Let $K$ be a compact subset of $(0,2\pi)$. Fix $\epsilon>0$ and $\phi\in K$. We want to find $\delta>0$ such that $$\big|g_r(\theta)-g_r(\phi)\big|<\epsilon\tag{*}$$ for every $r\in\mathbb{Z}_{\geq 0}$ and every $\theta\in K$ such that $|\theta-\phi|<\delta$. However, $$g'_r(\theta)=\text{i}\text{e}^{\text{i}\theta}\,f_r'\left(\text{e}^{\text{i}\theta}\right)$$ so that $$\left|g'_r(\theta)\right|=\Big|f_r'\left(\text{e}^{\text{i}\theta}\right)\Big|=\left|\frac{1-\text{e}^{\text{i}r\theta}}{1-\text{e}^{\text{i}\theta}}\right|\leq \frac{2}{\big|1-\text{e}^{\text{i}\theta}\big|}=\frac{1}{\Big|\sin\left(\frac{\theta}{2}\right)\Big|}\,.$$ Let $M$ be the supremum of $\dfrac{1}{\Big|\sin\left(\frac{\theta}{2}\right)\Big|}$ for $\theta\in K$. Then, each $g_r$ is $M$-Lipschitz. Therefore, for $\delta:=\dfrac{\epsilon}{M}$, we see that, whenever $\theta\in K$ satisfies $|\theta-\phi|<\delta$, (*) holds. Since $\delta$ does not depend on $\phi$, the family $\Big(g_r\big|_K\Big)_{r=0}^\infty$ is uniformly equicontinuous.