Evaluate $\int_{0}^{\infty}\frac{dx}{(x+\sqrt{1+x^2})^{2n}}\cdot \frac{1}{1+x^2}$

Question

$$F(n)=\large \int_{0}^{\infty}\frac{\mathrm dx}{(x+\sqrt{1+x^2})^{2n}}\cdot \frac{1}{1+x^2}$$

$\large x=\tan u$

$\large \mathrm dx=\sec^2 u\mathrm du$

$$F(n)=\large \int_{0}^{\pi/2}\frac{\mathrm du}{(\tan u+\sec u)^{2n}}$$

$\large \tan u+\sec u=\frac{2\tan(u/2)}{1-\tan^2(u/2)}+\frac{1+\tan^2(u/2)}{1-\tan^2(u/2)}$

$$F(n)=\large \int_{0}^{\pi/2}\frac{\mathrm du}{\left(\frac{2\tan(u/2)}{1-\tan^2(u/2)}+\frac{1+\tan^2(u/2)}{1-\tan^2(u/2)}\right)^{2n}}$$

$\large t=\tan(u/2)$

$\large \mathrm du=\frac{2}{\sec^2(u/2)}\mathrm dt=\frac{2}{1+t^2}\mathrm dt $

$$F(n)=\large 2\int_{0}^{1}\left(\frac{t-1}{t+1}\right)^{2n}\cdot \frac{\mathrm dt}{1+t^2}$$

I trying to evaluate $F(n)$, but I got stuck, how can I continue?

Answer 1

Alternatively: $$F(n)=\int_{0}^{\infty}\frac{dx}{(x+\sqrt{1+x^2})^{2n}}\cdot \frac{1}{1+x^2}= \int_{0}^{\infty}\frac{(x-\sqrt{1+x^2})^{2n}}{1+x^2}dx$$ Change: $x-\sqrt{1+x^2}=t \Rightarrow x=\frac{1-t^2}{2t}, dx=-\frac{1+t^2}{2t^2}dt$, then: $$F(n)= \int_{-1}^{0}\frac{t^{2n}}{\frac{(1+t^2)^2}{4t^2}}\cdot \left(-\frac{1+t^2}{2t^2}\right)dt=-2 \int_{-1}^{0}\frac{t^{2n}}{1+t^2}dt=\\ -2 \int_{-1}^{0}\frac{t^{2n}+t^{2n-2}-t^{2n-2}}{1+t^2}dt=\\ -2\int_{-1}^{0}t^{2n-2}dt+2\int_{-1}^{0}\frac{t^{2n-2}}{1+t^2}dt=\\ -\frac{2t^{2n-1}}{2n-1}\big{|}_{-1}^0-F(n-1).$$

Answer 2

I would say that the most natural substitution here is $x=\sinh u$, leading to $$ F(n) = \int_{0}^{+\infty}e^{-2nu}\frac{du}{\cosh u} $$ such that $F(n)$ is given by the Laplace transform of the hyperbolic secant.
By letting $u=\log v$ we get $$ F(n) = \int_{1}^{+\infty}\frac{1}{v^{2n}}\cdot \frac{2}{v^2+1}\,dv = \int_{0}^{1}\frac{2z^{2n}}{z^2+1}\,dz$$ and the last integral is clearly related to the tails of Gregory's series: $$ F(n) = 2\sum_{k\geq 0}\frac{(-1)^k}{2n+2k+1}. $$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \mrm{F}\pars{n} & \equiv \int_{0}^{\infty}{\dd x \over \pars{x + \root{1 + x^{2}}}^{2n}}\,{1 \over 1+x^2} \\[5mm] & \stackrel{x\ =\ \pars{1/t - t}/2}{=}\,\,\, 2\int_{0}^{1}{t^{2n} \over 1 + t^{2}}\,\dd t \\[5mm] & = 2\int_{0}^{1}{t^{2n} - t^{2n + 2} \over 1 - t^{4}}\,\dd t = {1 \over 2}\int_{0}^{1}{t^{n/2 - 3/4} - t^{n/2 -1/4} \over 1 - t}\,\dd t \\[5mm] & = {1 \over 2}\bracks{% \int_{0}^{1}{1 - t^{n/2 -1/4} \over 1 - t}\,\dd t - \int_{0}^{1}{1 - t^{n/2 -3/4} \over 1 - t}\,\dd t} \\[5mm] & = \bbx{{1 \over 2}\bracks{\Psi\pars{{n \over 2} + {3 \over 4}} - \Psi\pars{{n \over 2} + {1 \over 4}}}} \end{align}

where $\ds{\Psi}$ is the Digamma Function.

Note that a induced recurrence is provided by the Digamma Recurrence Formula $\ds{\Psi\pars{z + 1} = \Psi\pars{z} + {1 \over z}}$.