The integral $\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}}dx =\frac{\pi}{2}\ln (1+\sqrt{2})$

Question

At Mathematica the numerical value of the integral $$\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}} dx$$ equals 1.3844.., which is nothing but $\frac{\pi}{2}\ln (1+\sqrt{2})=z$. Also, one of its transformed forms is evaluated to be $z$ by Mathematica. The question is: How to do it by hand?

Answer 1

Let $$I=\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}}dx =\int_{0}^{\infty} \frac{\tan^{-1}(1/\sqrt{(1+x^2})}{\sqrt{1+x^2}} dx$$ Let us use the integral representation of $$\frac{\tan^{-1}(1/\sqrt{1+x^2})}{\sqrt{1+x^2}}=\int_{0}^{1} \frac{dt}{1+x^2+t^2}$$ $$I=\int_{0}^{\infty} \int_{0}^{1} \frac {dt dx}{1+x^2+t^2}=\int_{0}^{1}\frac{dt}{\sqrt{1+t^2}} \tan^{-1} \left.\frac{x}{\sqrt{1+t^2}}\right|_{0}^{\infty} =\frac{\pi}{2} \ln (1+\sqrt{2}).$$

Answer 2

Consider the following integral: $$I(a)=\int_0^\infty \frac{\operatorname{arccot}(a\sqrt{1+x^2})}{\sqrt{1+x^2}}dx\Rightarrow I'(a)=-\int_0^\infty \frac{1}{1+a^2+a^2x^2}dx$$ $$=-\frac{1}{a\sqrt{1+a^2}}\arctan\left(\frac{ax}{\sqrt{1+a^2}}\right)\bigg|_0^\infty=-\frac{\pi}{2}\frac{1}{a\sqrt{1+a^2}} $$ Since $I(\infty)=0$ and we are looking for $I(1)$ we have: $$I(1)=-(I(\infty)-I(1))=\frac{\pi}{2}\int_1^\infty \frac{1}{a\sqrt{1+a^2}}da\overset{a=\frac{1}{x}}=\frac{\pi}{2}\int_0^1 \frac{1}{\sqrt{1+x^2}}dx$$ $$\Rightarrow \boxed{\int_0^\infty \frac{\operatorname{arccot}(\sqrt{1+x^2})}{\sqrt{1+x^2}}dx=\frac{\pi}{2}\ln(1+\sqrt 2)}$$

Answer 3

Here we will address your integral: \begin{equation} I = \int_0^\infty \frac{\cot^{-1}\left(\sqrt{1 + x^2} \right)}{\sqrt{1 + x^2}}\:dx \end{equation} We first let $x = \tan(s)$: \begin{align} I &= \int_0^\frac{\pi}{2} \frac{\cot^{-1}\left(\sqrt{1 + \tan^2(s)} \right)}{\sqrt{1 + \tan^2(s)}}\cdot \sec^2(s)\:ds = \int_0^\frac{\pi}{2} \sec(s)\cot^{-1}\left(\sec(s)\right)\:ds \nonumber \\ &= \int_0^\frac{\pi}{2} \frac{\arctan\left(\cos(s)\right)}{\cos(s)}\:ds \end{align} Here I will now employ Feynman's Trick and introduce the following function: \begin{equation} J(t) = \int_0^\frac{\pi}{2} \frac{\arctan(t\cos(s))}{\cos(s)}\:ds \end{equation} Here $I = J(1)$ and $J(0) = 0$. ,By Leibniz's Integral Rule we now differentiate under the curve with respect to $t$: \begin{align} J'(t) &= \int_0^\frac{\pi}{2} \frac{1}{t^2\cos^2(s) + 1}\cdot \cos(s) \cdot \frac{1}{\cos(s)}\:ds = \int_0^\frac{\pi}{2}\frac{1}{t^2\cos^2(s) + 1}\:ds \nonumber \\ &= \left[ \frac{1}{\sqrt{t^2 + 1}}\arctan\left(\frac{\tan(s)}{\sqrt{t^2 + 1}}\right) \right]_0^\frac{\pi}{2} = \frac{\pi}{2}\frac{1}{\sqrt{t^2 + 1}} \end{align} Thus, \begin{equation} J(t) = \frac{\pi}{2} \int \frac{1}{\sqrt{t^2 + 1}}\:dt = \frac{\pi}{2}\sinh^{-1}(t) + C \end{equation} Where $C$ is the constant of integration. To resolve $C$ we use our condition $J(0) = 0$: \begin{equation} J(0) = 0 = \frac{\pi}{2}\sinh^{-1}(0) + C \rightarrow C = 0 \end{equation} Thus, \begin{equation} J(t) = \frac{\pi}{2}\sinh^{-1}(t) \end{equation} We now may resolve $I$ \begin{equation} I = J(1) = \frac{\pi}{2}\sinh^{-1}(1) \end{equation} Noting that \begin{equation} \sinh^{-1}(t) = \ln\left|t + \sqrt{t^2 + 1}\right| \end{equation} We see that another representation for $I$ is given by: \begin{equation} I = \frac{\pi}{2}\ln(1 + \sqrt{2}) \end{equation}

Answer 4

Let $t=1/\sqrt{1+x^2}$, then $$\int_{0}^{\infty} \frac{\cot^{-1}\sqrt{1+x^2}}{\sqrt{1+x^2}}dx =\int_{0}^{1} \frac{\arctan(t)}{t\sqrt{1-t^2}}\, dt$$ Now the result follows from Definite integral $\int_0^1 \frac{\arctan x}{x\,\sqrt{1-x^2}}\,\text{d}x$ where several approaches are provided.

Answer 5

In general, for any $k>0$, $$ \begin{aligned}\int_0^{\infty} \frac{\cot ^{-1} \sqrt{x^2+k^2}}{\sqrt{x^2+k^2}} d x & =\int_0^{\infty} \frac{\tan ^{-1}\left(\frac{1}{\sqrt{x^2+k^2}}\right)}{\sqrt{x^2+k^2}} d x \\ & =\int_0^{\infty} \int_0^1 \frac{1}{y^2+\left(x^2+k^2\right)} d y d x \\ & =\int_0^1 \int_0^{\infty} \frac{1}{x^2+\left(y^2+k^2\right)} d x d y \\ & =\int_0^1 \frac{1}{\sqrt{y^2+k^2}}\left[\tan ^{-1}\left(\frac{x}{\sqrt{y^2+k^2}}\right)\right]_0^{\infty} d y \\ & =\frac{\pi}{2} \int_0^1 \frac{d y}{\sqrt{y^2+k^2}} \\ & =\frac{\pi}{2} \sinh ^{-1}\left(\frac{1}{k}\right)\\ \int_0^{\infty} \frac{\cot ^{-1} \sqrt{x^2+k^2}}{\sqrt{x^2+k^2}} d x& = \frac{\pi}{2} \ln \left(\frac{1+\sqrt{1+k^2}}{k}\right) \quad \blacksquare\\ & \end{aligned} $$ In particular, $$\int_0^{\infty} \frac{\cot ^{-1} \sqrt{x^2+1}}{\sqrt{x^2+1}} d x = \frac{\pi}{2} \ln(1+\sqrt 2)$$