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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.

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    $\begingroup$ How did you get $\frac{2 \pi}{m+1}$ or $0$ as the solution on the first line. Why not just take the limit of that? I'm confused by the return of the integral and sum in the 2nd line. Isn't letting the sum be finite to get rid of those? $\endgroup$
    – futurebird
    Apr 8 '20 at 10:37
  • $\begingroup$ @futurebird When I tried to calculate it first I just supposed that we can intertwine the infinite sum with the integral. If you intertwine them, you will se that if $m<n$ all integrals are zeros. If $m=n$, the integral $\int_{0}^{2\pi}d\theta=2\pi$ , the same happens when $m>n$ and you find $2\pi$. $\endgroup$
    – Student
    Apr 8 '20 at 10:58
  • $\begingroup$ I will edit it now and remove the result from the first line. $\endgroup$
    – Student
    Apr 8 '20 at 11:02
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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.

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