Derivation – $\operatorname{arccot}(x)=\int\limits_{x}^ {\infty} \frac{1}{1+t^2} ~dt$

Question

For $x\in \mathbb R$ we have $$\text{arccot}(x)=\int\limits_{x}^ {\infty} \frac{1}{1+t^2} ~dt.$$ How can we prove this? for if we have definite integral then for example after assuming some hypothesis we have
$$\text{arcsin}(x)=\text{arcsin}(0)+\int\limits_{0}^ {x} \frac{1}{\sqrt{1-t^2}} ~dt.$$ By fundamental theorem of calculus. How we can use it for improper integral?
If $f:(0,\infty)\rightarrow \mathbb R$ such that for all $x$, $\int\limits_{x}^ {\infty} f(t) ~dt$ converges. And if we define$~F$ on $(0,\infty)$ as
$$ x\mapsto \int\limits_{x}^ {\infty} f(t) ~dt$$
then I proved $F'(x)=f(x)$ for all $x$.So if we have any other $G$ on $(0,\infty)$ such that $G'(x)=f(x)$. Then by mean value theorem we have$$
G(x)=C+\int\limits_{x}^ {\infty} f(t) ~dt$$ for some constant $C$.
Is there any analogous for that constant $C$ like definite integral. Is $C=G(\infty)=\lim\limits_{x\to \infty} G(x)$? Because here
$$\text{arccot}(x)=\lim\limits_{k\to \infty}\text{arccot}(k)+\int\limits_{x}^ {\infty} \frac{1}{1+t^2} ~dt.$$
So My question is about that constant $C$. Is there any way to get $C$?

Answer 1

We have: $\cot^{-1}(x) = \dfrac{\pi}{2} - \tan^{-1}(x) = \displaystyle \int_{0}^{\infty} \dfrac{1}{1+t^2}dt-\displaystyle \int_{0}^{x} \dfrac{1}{1+t^2}dt = \displaystyle \int_{x}^{\infty} \dfrac{1}{1+t^2}dt$ which is the desire formula.