Unable to think why integral should be convergent

Question

I am unable to think why an integral must converge while studying a research paper in analytic number theory.

Integral is $\to I_n(z)$

where $n,r,a$ are integers satisfying $1\leq r<a/2$.

Here author writes that integral must converge for any complex $z$ such that $|z|>1$.

Can someone please give hints.

Answer 1

My answer ended up being fairly long so I was unsure about how to frame it as a hint.

I've kept all the details below. You can take the nonhidden bit as a hint (it is quite easy to work out the rest from it).

We'll start off with a relevant fact.

Fact: For any $a \in [0,1]$ and any $n\in \mathbb Z^+$, $|a^n-a^{n+1}+a|\leq 1$.

Proof: Proceed by induction. If $n=1$, the expression becomes $|a^1-a^{1+1}+a|=|2a-a^2|=|1-(a-1)^2|$, which, for $a\in [0,1]$ is clearly bounded above by $1$.

Suppose we have proven the Fact statement for the case of $n=k\geq 1$. Then, when $n=k+1$, we have for any $a\in [0,1]$: $$a^{k+1}-a^{k+2}+a=a(a^k-a^{k+1}+a)+(a-a^2)\leq a+(a-a^2)=2a-a^2\leq 1$$ And also, trivially: $$a^{k+1}-a^{k+2}+a=a^{k+1}(1-a)+a\geq a\geq 0$$

So we conclude that the statement holds when $n=k+1$ as well, completing the proof by induction.

Now, apply the fact to $|x_l^k-x_l^{k+1}+x|$ to bound $|x_l^k(1-x_l)|$.

The Problem: Define $F:[0,1]^{a+1}\rightarrow \mathbb R_{\geq 0}$ so that: $$F\left (x_1,x_2,\dots,x_{a+1}\right )=\left |\frac{\prod_{l=1}^{a+1} x_l^r(1-x_l)}{\left(z-x_1x_2\cdots x_{a+1}\right)^{2r+1}}\right |$$
We'll show that $F\left (x_1,x_2,\dots,x_{a+1}\right)<1$ always.
Let $\left (y_1,y_2,\dots,y_{a+1}\right)$ any given point in $[0,1]^{a+1}$. Each term in the numerator of the expression for $F\left (y_1,y_2,\dots,y_{a+1}\right)$, $|y_l^r(1-y_l)|$, is $\leq 1$, so the numerator of $F\left (y_1,y_2,\dots,y_{a+1}\right)$ as a whole has magnitude $\leq 1$.
So, if $|z-y_1y_2\cdots y_{a+1}|>1$, then it directly follows that $F\left (y_1,y_2,\dots,y_{a+1}\right)<1$. If $|z-y_1y_2\cdots y_{a+1}|\leq 1$, then we employ some manipulation to bound $F\left (y_1,y_2,\dots,y_{a+1}\right)$ (using the fact that $r$ and $a$ are integers for which $1\leq r<\frac{a}{2}$).
Using the Fact proven at the start, we have for all $1\leq l\leq a+1$ that $|y_l^r-y_l^{r+1}+y_l|\leq 1$. But we also have, by assumption, that $|z|>1$.
So, for any $l$: $$|y_l^r-y_l^{r+1}+y_l|<|z|$$
Since $y_l^r-y_l^{r+1}$ and $y_l$ are both positive terms: $$|y_l^r-y_l^{r+1}+y_l|=x_l^r-y_l^{r+1}+y_l=|y_l^r-y_l^{r+1}|+|y_l|<|z|$$ $$\implies|y_l^r-y_l^{r+1}|<|z|-|y_l|\leq |z|-|y_l|\left(|y_1y_2\cdots y_{l-1}y_{l+1}\cdots y_{a+1}|\right)\leq |z-y_1y_2\cdots y_{a+1}|$$ $$\implies|y_l^r(1-y_l)|<|z-y_1y_2\cdots y_{a+1}|$$
Since the inequality holds for arbitrary $l$, we then get: $$\prod_{l=1}^{a+1}|y_l^r(1-y_l)|<|z-y_1y_2\cdots y_{a+1}|^{a+1}$$
Now, given that we are looking at the case where $|z-y_1y_2\cdots y_{a+1}|<1$, and $r<\frac{a}{2}$ by assumption, we have: $$\prod_{l=1}^{a+1}|y_l^r(1-y_l)|<|z-y_1y_2\cdots y_{a+1}|^{a+1}<|z-y_1y_2\cdots y_{a+1}|^{2r+1}$$
This finally gives us: $F\left (y_1,y_2,\dots,y_{a+1}\right)=\left |\frac{\prod_{l=1}^{a+1} y_l^r(1-y_l)}{\left(z-y_1y_2\cdots y_{a+1}\right)^{2r+1}}\right|<1$. Since $\left (y_1,y_2,\dots,y_{a+1}\right)$ was an arbitrary element of $[0,1]^{a+1}$, $F<1$.
But, note that $F$ is a continuous function (it can be broken up into compositions and products of continuous functions) and $F$ is defined on $[0,1]^{a+1}$, which is compact.
This means $F$ attains a maximum somewhere in $[0,1]^{a+1}$. Since $|F|<1$, this then means that there exists some $0\leq \epsilon< 1$ s.t. $|F|\leq \epsilon$ everywhere in $[0,1]^{a+1}$.
Now, applying this to the integral at hand: $$|I_n(z)|=\left|\int_{[0,1]^{a+1}}\left(\frac{\prod_{l=1}^{a+1} x_l^r(1-x_l)}{\left(z-x_1x_2\cdots x_{a+1}\right)^{2r+1}}\right)^n\frac{dx_1dx_2\cdots dx_{a+1}}{\left(z-x_1x_2\cdots x_{a+1}\right)^2}\right|\leq \int_{[0,1]^{a+1}} \left(\left|\frac{\prod_{l=1}^{a+1} x_l^r(1-x_l)}{\left(z-x_1x_2\cdots x_{a+1}\right)^{2r+1}}\right|\right)^n\frac{dx_1dx_2\cdots dx_{a+1}}{\left|z-x_1x_2\cdots x_{a+1}\right|^2}\leq \int_{[0,1]^{a+1}} |F\left (x_1,x_2,\dots,x_{a+1}\right)|^n\frac{dx_1dx_2\cdots dx_{a+1}}{\left(|z|-|1|\right)^2}\leq \int_{[0,1]^{a+1}} \epsilon^n \frac{dx_1dx_2\cdots dx_{a+1}}{\left(|z|-1\right)^2}=\frac{\epsilon^n}{\left(|z|-1\right)^2}$$
It is clear now that, for any $r,a,$ and $z$ satisfying the conditions, $\lim_{n\rightarrow \infty} I_n(z)$ exists and is $0$.