Verify proof that $\lim_{m\rightarrow\infty} \int_0^z \frac{t^m}{1+t}dt = 0 : z\in\mathbb{C}, |z|=1.$

Question

To Prove:

$\displaystyle \lim_{m\rightarrow\infty} \int_0^z \dfrac{t^m}{1+t}dt = 0 : z\in\mathbb{C}, |z|=1.$

My Real Question: Is the following analysis valid?

My Research:
I know little about complex analysis. I (superficially) browsed some online texts and
(given $z=x+iy : x,y\in\mathbb{R}$), decided to construe

$\;\displaystyle \int_0^z \dfrac{t^m}{1+t}dt \;$ as $\;\displaystyle\left(\int_0^x \dfrac{t^m}{1+t}dt\right) \;+ i\times \left(\int_0^y \dfrac{t^m}{1+t}dt\right).$

My Work:

Since $|z|=1,\;|x|\leq 1\;$ and $\;|y|\leq 1.\;$ Further, when $\;m>0,$

$\displaystyle \left\lvert\int_0^x \dfrac{t^m}{1+t}dt\right\rvert < \left\lvert\int_0^x t^m dt\right\rvert \;= \left\lvert\;\frac{x^{m+1}}{m+1}\right\rvert$ and

$\displaystyle \left\lvert\int_0^y \dfrac{t^m}{1+t}dt\right\rvert < \left\lvert\int_0^y t^m dt\right\rvert \;= \left\lvert\;\frac{y^{m+1}}{m+1}\right\rvert.$

Since $\;|x|\leq 1\;$ and $\;|y|\leq 1,$ $\;\left\lvert\dfrac{x^{m+1}}{m+1}\right\rvert\;$ and $\;\left\lvert\dfrac{y^{m+1}}{m+1}\right\rvert\;$ both go to zero as $\;m\rightarrow\infty.$

Therefore $\;\displaystyle \left\lvert \int_0^z \frac{t^m}{1+t}dt\right\rvert\;$ goes to zero as $\;m\rightarrow\infty.$

Correction:

The above analysis assumes (for example) that as $t$ ranges from $0$ to $x,$ $\dfrac{1}{1+t} < 1.$ This is true when $x>0,$ but false when $x<0.$

However, if $-1 < x < 0,\;$ then as $t$ ranges from $0$ to $x,$ $\dfrac{1}{1+t} < \dfrac{1}{1+x}.$

This provides a remedy for the analysis except in the special cases of $x=-1$ or $y=-1.$ This forum query was originally intended as the final piece of analysis needed for a 2nd query: Request help verifying the Taylor series expansion of $\text{ln}(1 + e^{2ix})$.

In the 2nd query, $\ln(1+z)$ is disallowed when $z=-1,$ but allowed when $z=-i.$ Thus, even if my correction is valid, it still remains to prove (in the 2nd query) that $\displaystyle \ln(1-i) = \sum_{k=0}^{\infty} \frac{(-1)^k}{k+1} (-i)^{k+1}.$

This means that for marty cohen's answer in the 2nd query to apply, it must be shown that the $\displaystyle \lim_{m\rightarrow\infty} \int_0^{y} \frac{t^m}{1+t}dt = 0,$ specifically when $y=-1.$ I would love feedback here.

2nd Correction:

I have augmented the 2nd query: Request help verifying the Taylor series expansion of $\text{ln}(1 + e^{2ix})$, to manually resolve the case of $z = -i,\;$ which is the only pending case unresolved by my first Correction.

I don't know if $\int_0^z \dfrac{t^m}{1+t}dt\;$ exists when $z=-i,\;$ and would still like feedback on this. However, with respect to the 2nd query, the issue is moot.

Answer 1

For $|z|=1,z\ne -1,$ we have

$$\tag 1 \left |\int_0^z \dfrac{t^m}{1+t}dt\right|=\left | \int_0^1 \frac{(sz)^m z}{1+sz}\, ds\right | \le \int_0^1 \frac{s^m }{|1+sz|}\, ds.$$

Now for $s\in [0,1],$ $|1+sz| = |sz-(-1)|.$ The last expression is the distance from $sz$ to $-1,$ which is bounded below by the distance from the line segment $[0,z]$ to $-1.$ This last distance is a positive number I'll call $d_z.$ It follows that the right side of $(1)$ is bounded above by

$$\frac{\int_0^1s^m\,ds }{d_z}= \frac{1}{d_z(m+1)}\to 0.$$

This shows the left side of $(1)$ $\to 0,$ which is the desired result.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

I'll introduce a slight modification. Namely, $\ds{z \in \braces{\vphantom{\Large A}z \in \mathbb{C} \left.\vphantom{\large A}\right\vert \verts{z} = 1}\setminus\color{red}{\braces{-1}}}$.

Then, \begin{align} \lim_{m \to \infty}\int_{0}^{z}{t^{m} \over 1 + t}\,\dd t & \,\,\,\stackrel{t\ =\ zx}{=}\,\,\, \lim_{m \to \infty}\int_{0}^{1}{z^{m}x^{m} \over 1 + zx}\,z\,\dd x \\[5mm] & \stackrel{x\ \mapsto\ 1 - x}{=}\,\,\, \lim_{m \to \infty}\bracks{z^{m +1}\int_{0}^{1}{\pars{1 - x}^{m} \over 1 + z\pars{1 - x}}\,\dd x} \\[5mm] & = \lim_{m \to \infty}\bracks{z^{m +1}\int_{0}^{1} {\exp\pars{m\ln\pars{1 - x}} \over 1 + z\pars{1 - x}}\,\dd x} \label{1}\tag{1} \end{align}

With Laplace Method:

\begin{align} \lim_{m \to \infty}\int_{0}^{z}{t^{m} \over 1 + t}\,\dd t & = \lim_{m \to \infty}\bracks{z^{m +1}\int_{0}^{\infty} {\exp\pars{-mx} \over 1 + z}\,\dd x} = {1 \over 1 + z}\,\lim_{m \to \infty}{{z^{m +1} \over m}} \\[5mm] & = \bbx{\large 0}\quad \mbox{because}\quad\bbox[10px,#ffd]{0 < \verts{z^{m + 1} \over m} = {1 \over \verts{m}} \,\,\,\stackrel{\mrm{as}\ m\ \to\ \infty}{\LARGE\to} {\LARGE 0}} \end{align}