Evaluate $\int_{0}^{\infty}\frac{dx}{(x+\sqrt{1+x^2})^{2n}}\cdot \frac{1}{1+x^2}$ $$F(n)=\large \int_{0}^{\infty}\frac{\mathrm dx}{(x+\sqrt{1+x^2})^{2n}}\cdot \frac{1}{1+x^2}$$
$\large x=\tan u$
$\large \mathrm dx=\sec^2 u\mathrm du$
$$F(n)=\large \int_{0}^{\pi/2}\frac{\mathrm du}{(\tan u+\sec u)^{2n}}$$
$\large \tan u+\sec u=\frac{2\tan(u/2)}{1-\tan^2(u/2)}+\frac{1+\tan^2(u/2)}{1-\tan^2(u/2)}$
$$F(n)=\large \int_{0}^{\pi/2}\frac{\mathrm du}{\left(\frac{2\tan(u/2)}{1-\tan^2(u/2)}+\frac{1+\tan^2(u/2)}{1-\tan^2(u/2)}\right)^{2n}}$$
$\large t=\tan(u/2)$
$\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?
 A: 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:
$$ F(n) = 2\sum_{k\geq 0}\frac{(-1)^k}{2n+2k+1}. $$
A: 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:
$$F(n)= \int_{-1}^{0}\frac{t^{2n}}{\frac{(1+t^2)^2}{4t^2}}\cdot \left(-\frac{1+t^2}{2t^2}\right)dt=-2 \int_{-1}^{0}\frac{t^{2n}}{1+t^2}dt=\\
-2 \int_{-1}^{0}\frac{t^{2n}+t^{2n-2}-t^{2n-2}}{1+t^2}dt=\\
-2\int_{-1}^{0}t^{2n-2}dt+2\int_{-1}^{0}\frac{t^{2n-2}}{1+t^2}dt=\\
-\frac{2t^{2n-1}}{2n-1}\big{|}_{-1}^0-F(n-1).$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}}}
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\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
$\ds{\Psi\pars{z + 1} = \Psi\pars{z} + {1 \over z}}$.
