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

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

If $F:\Bbb R \to \Bbb R$ is continuously differentiable then $$ \tag{1} F(y) - F(x) = \int_x^y F'(t) \, dt $$ for all $x, y \in \Bbb R$ due to the fundamental theorem of calculus. (The conditions can be relaxed, e.g. that $F$ is differentiable almost everywhere and $F'$ is locally Lebesgue-integrable).

If, in addition, $C = \lim_{x \to \infty} F(x)$ exists then we can take the limit $y \to \infty$ in $(1)$ and conclude that $$ \tag{2} F(x) = C - \int_x^\infty F'(t)\, dt $$

For $F(x) = \operatorname{arccot}(x)$ is $F'(x) = -\frac{1}{1+x^2}$ and $C = \lim_{x \to \infty} F(x) = 0$, so that $$ \operatorname{arccot}(x) = \int_x^\infty \frac{1}{1+t^2} \, dt $$

Answer 2

YES there is a way to find $C$. Hint $\lim\limits_{x\to 0^{-}} \cot (x) =\infty$

Answer 3

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.