Integral inequality $I_2(z) = \mathcal{O}\left(\frac{1}{(1+z)^2}\right)$

Question

I want to prove the inequality of an integral like this:

$$I_2(z) = \int_1^{+\infty} \frac{\rho(t)}{(z+t)^2} d t = \mathcal{O}\left(\frac{1}{(1+z)^2}\right)$$ , when $z \to +\infty$

Below is my current attempt：

$$\begin{aligned} I_2(z) & =\int_1^{+\infty} \frac{\rho(t)}{(z+t)^2} d t \\ & =\frac{R(t)}{(z+t)^2}\Big|_1 ^{+\infty}+2 \int_1^{+\infty} \frac{R(t)}{(z+t)^3} d t \\ &\leq \int_{0}^{+\infty} \frac{1}{|(z+1)+t|^3} d t \end{aligned}$$

Where，

$$\begin{gathered} \rho(t)=\{t\}-\frac{1}{2},|\rho(t)| \leq \frac{1}{2}, \\ R(t)=\int_1^t \rho(s) d s, \quad|R(t)| \leq \frac{1}{2} . \end{gathered}$$

The last integral with absolute value I don't know how to handle it properly.

Answer 1

For $z=x+iy$ with $x > 0$ is, as you already calculated, $$ I_2(z) = \int_1^\infty \frac{2R(t)}{(z+t)^3} \, dt $$ and $|2 R(t)|\le 1$. Let us first assume that $y \ne 0$. Then $$ \begin{align} |I_2(z)| &\le \int_1^\infty \frac{1}{(x+t)^2+y^2)^{3/2}} \, dt \\ &= \left[ \frac {x+t}{y^2 \sqrt{(x+t)^2+y^2}}\right]_{t=1}^{t=\infty} \\ &= \frac{1}{y^2} \left( 1 - \frac{x+1}{\sqrt{(x+1)^2+y^2}}\right) \\ &= \frac{1}{\sqrt{(x+1)^2+y^2} \cdot (\sqrt{(x+1)^2+y^2} + (x+1))} \\ &\le \frac{1}{(x+1)^2+y^2} \\ &= \frac{1}{|z+1|^2} \, . \end{align} $$ By continuity it follows that the same estimates holds also if $y=0$.