Big O in integral representation of arctan

Question

Why does this equality hold?

$$\int_{0}^{T/c}\frac{du}{1+u^2} = \int_{0}^\infty \frac{du}{1+u^2}+O(c/T)$$

Answer 1

It can be written $$ \int_{0}^{A}\frac{du}{1+u^2} = \int_{0}^\infty \frac{du}{1+u^2}+O(1/A)\quad\text{as } A \to \infty $$ and equivalently $$ \int_{A}^\infty\frac{du}{1+u^2} = O(1/A)\quad\text{as } A \to \infty $$ which is easy: for $A>0$ $$ \int_{A}^\infty\frac{du}{1+u^2} < \int_{A}^\infty\frac{du}{u^2} = \frac{1}{A} . $$

Answer 2

Cauchy's inequality gives that

$$ (u^2+1^2)(1^2+1^2)\ge(u+1)^2 $$

so we have

$$ \int_{T/c}^\infty{\mathrm du\over1+u^2}\le2\int_{T/c}^\infty{\mathrm du\over(1+u)^2}={2c\over c+T}=\mathcal O\left(\frac cT\right) $$