Closed form of $\int_0^\infty \left(\frac{\arctan x}{x}\right)^ndx$

Solution 1:

Used formulas and definitions:

$\displaystyle\Re(i^n)=\cos\left(\frac{\pi n}{2}\right)\enspace$ for integer $~n\enspace$ ; $\enspace\displaystyle\tan x = -i\frac{1-e^{-i2x}}{1+e^{-i2x}}$

$\displaystyle \sum\limits_{j=0}^{-1} a_j := \sum\limits_{j=0}^0 a_j - a_0 = 0\enspace$ ; $\enspace\displaystyle \left(\sum\limits_{k=0}^{\infty} a_k\right) \left(\sum\limits_{k=0}^{\infty} b_k\right) = \left(\sum\limits_{k=0}^{\infty} \sum\limits_{v=0}^k a_v b_{k-v}\right) $

$\displaystyle \int\limits_0^a x^m e^{zx} dx = \frac{m!}{(-z)^{m+1}} - e^{az}\sum\limits_{v=0}^m\frac{m!a^{m-v}}{(m-v)!(-z)^{v+1}} \enspace$ for $\enspace m\in\mathbb{N}_0 , \, a\in\mathbb{R} , \, z\in\mathbb{C}\setminus\{0\} $

Stirling numbers of the first kind $\displaystyle \begin{bmatrix}n\\k\end{bmatrix}$ defined by

$\hspace{4cm}\displaystyle\sum\limits_{j=0}^n\begin{bmatrix}n\\j\end{bmatrix}x^j:=\prod\limits_{k=0}^{n-1}(x+k) \enspace$ for $\enspace n\in\mathbb{N}_0 , \, x\in\mathbb{C} $

$\displaystyle \sum\limits_{j=0}^n f(j) \sum\limits_{l=0}^j g(j,l) k^l = \sum\limits_{j=0}^n k^j \sum\limits_{l=j}^n f(l)g(l,j) \enspace\enspace$ formal summation permutation

$\displaystyle \sum\limits_{v=-1}^k {\binom {n+1} {k-v}}{\binom {n+v} v} = \displaystyle \sum\limits_{v=0}^{n+1} {\binom {n+1} v}{\binom {n+k-v} n}\enspace$ for $\enspace k,n\geq 0$

$\displaystyle c_{n,j} := \frac{1}{n!} \sum\limits_{v=0}^{n+1} {\binom {n+1} {v}} \sum\limits_{l=j}^m \begin{bmatrix}n+1\\l+1\end{bmatrix} {\binom l j} (-v)^{l-j} \enspace$ for $\enspace 0\leq j\leq n\enspace$ with $\enspace 0^0:=1$

$\displaystyle b_{n,k} := \sum\limits_{v=0}^k {\binom {n+1} {k-v}} {\binom {n+v} v} = -0^{n+k} +\sum\limits_{j=0}^n k^j c_{n,j} \enspace$ for $\enspace k,n \geq 0 \,~~ ; \enspace b_{n,0 }=1$

It follows:

$\displaystyle\sum\limits_{k=1}^\infty \frac{b_{n,k}~x^k}{k^s} = \sum\limits_{j=0}^n c_{n,j}~\text{Li}_{s-j}(x) \enspace$ for $\enspace s\in\mathbb{C}\enspace$ where $\enspace \text{Li}_s(x)\enspace$ is the Polylogarithm

with the special cases $\enspace \text{Li}_s(1)\equiv \zeta(s)\,$ and

$\hspace{4.1cm}\text{Li}_s(-1)\equiv -\eta(s) = \left(2^{1-s}-1\right)\zeta(s) \enspace$ with $\enspace\eta(1)=\ln 2$

$\underline{\text{Solution:}}$

Let $\enspace\displaystyle |a| \leq \frac{\pi}{2} \, , \enspace 0\leq n\leq m \,$ .

$\displaystyle \int\limits_0^{\tan a} \frac{\arctan^m x}{x^n} dx = \int\limits_0^a \frac{x^m}{\tan^n x} dx + \int\limits_0^a \frac{x^m}{\tan^{n-2} x} dx \enspace$ for $\enspace n\geq 2$

$$\int\limits_0^a \frac{x^m}{\tan^n x}dx = i^n\frac{a^{m+1}}{m+1} + \frac{i^{n-m-1}m!}{2^{m+1}}\sum\limits_{j=0}^{n-1}~c_{n-1,j}~\text{Li}_{m+1-j}~(1) $$

$$\hspace{2.5cm} -\sum\limits_{v=0}^m \frac{i^{n-v-1}m!a^{m-v}}{(m-v)!2^{v+1}} \sum\limits_{j=0}^{n-1}~c_{n-1,j}~\text{Li}_{v+1-j}~(e^{-i2a})$$

$\displaystyle a:=\frac{\pi}{2}\,$ :

$$\int\limits_0^{\pi/2} \frac{x^m}{\tan^n x}dx = $$

$$ \cos\left(\frac{\pi n}{2}\right)\frac{(\pi/2)^{m+1}}{m+1} +\cos\left(\frac{\pi (n-m-1)}{2}\right) \frac{m!}{2^{m+1}}\sum\limits_{j=0}^{n-1}~c_{n-1,j}~\zeta(m+1-j) $$

$$\hspace{2.5cm} +\sum\limits_{v=0}^m \cos\left(\frac{\pi (n-v-1)}{2}\right)\frac{m!(\pi/2)^{m-v}}{(m-v)!2^{v+1}} \sum\limits_{j=0}^{n-1}~c_{n-1,j}~\eta(v+1-j)\hspace{1cm}$$

Finally we get:

$$\int\limits_0^\infty \left(\frac{\arctan x}{x}\right)^n dx = $$

$$\frac{1}{2^{n+1}} \sum\limits_{v=0}^n \left(\cos\frac{\pi(n-v-1)}{2}\right) \frac{n!\pi^{n-v}}{(n-v)!}~\cdot $$

$$\cdot~\left( \sum\limits_{j=0}^{n-1} c_{n-1,j}~\eta(v-j+1) - \sum\limits_{j=0}^{n-3} c_{n-3,j}~\eta(v-j+1) \right)$$

Analytical continuations $(s\in\mathbb{C})$ :

$$\begin{align} \zeta(1-s)&=\dfrac{2}{(2\pi)^s}\cos\bigg(\dfrac{\pi s}{2}\bigg)~\Gamma(s)~\zeta(s) \\\\ \eta(1-s)&=\dfrac{2^s-1}{1-2^{s-1}}~\pi^{-s}\cos\bigg(\dfrac{\pi s}{2}\bigg)~\Gamma(s)~\eta(s) \end{align} $$

Simplifications $(k\in\mathbb{N}_0)$ :

$$\begin{align} \eta(1)=\ln 2 \enspace ; \enspace \eta(1-k)~&=~\frac{2^k-1}{k}~B_k \end{align}$$

$$\begin{align} \eta(2k)~&=~(-1)^{k-1}~\frac{2^{2k-1}-1}{(2k)!}~B_{2k}~\pi^{2k} \end{align}$$

$$\begin{align} \eta(2k+1)~&=~\bigg(1-\frac{1}{2^{2k}}\bigg)\zeta(2k+1) \end{align}$$

Examples:

$\displaystyle c_{0,0} = \frac{2}{0!}(1)$

$\displaystyle (c_{1,0}~;~ c_{1,1}) = \frac{2}{1!} \left(0~;~2\right)$

$\displaystyle (c_{2,0}~;~c_{2,1}~;~c_{2,2}) = \frac{2}{2!} \left(2~;~0~;~4\right)$

$\displaystyle (c_{3,0}~;~c_{3,1}~;~c_{3,2}~;~c_{3,3}) = \frac{2}{3!} \left(0~;~16~;~0~;~8\right)$

$\displaystyle (c_{4,0}~;~c_{4,1}~;~c_{4,2}~;~c_{4,3}~;~c_{4,4}) = \frac{2}{4!} \left(24~;~0~;~80~;~0~;~16\right)$

$\displaystyle (c_{5,0}~;~c_{5,1}~;~c_{5,2}~;~c_{5,3}~;~c_{5,4}~;~c_{5,5}) = \frac{2}{5!} \left(0~;~368~;~0~;~320~;~0~;~32\right)$

$\displaystyle (c_{6,0}~;~c_{6,1}~;~c_{6,2}~;~c_{6,3}~;~c_{6,4}~;~c_{6,5}~;~c_{6,6}) = \frac{2}{6!} \left(720~;~0~;~3136~;~0~;~1120~;~0~;~64\right)$

$\displaystyle (c_{7,0}~;~c_{7,1}~;~c_{7,2}~;~c_{7,3}~;~c_{7,4}~;~c_{7,5}~;~ c_{7,6}~;~ c_{7,7}) $

$\hspace{7cm}\displaystyle =\frac{2}{7!} \left(0~;~16896~;~0~;~19712~;~0~;~3584~;~0~;~128\right)$

...

$\displaystyle \int\limits_0^\infty \left(\frac{\arctan x}{x}\right)^2 dx = \pi\ln 2$

$\displaystyle \int\limits_0^\infty \left(\frac{\arctan x}{x}\right)^3 dx = -\frac{\pi^3}{16} + \frac{3\pi}{2}\ln 2$

$\displaystyle \int\limits_0^\infty \left(\frac{\arctan x}{x}\right)^4 dx = -\frac{\pi^3}{12}(2\ln 2 + 1) + \frac{\pi}{4}(3\zeta(3) + 8\ln 2)$

$\displaystyle \int\limits_0^\infty \left(\frac{\arctan x}{x}\right)^5 dx = \frac{\pi^5}{128} - \frac{5\pi^3}{48}(8\ln 2 + 1) + \frac{5\pi}{4}(3\zeta(3) + 2\ln 2)$

$\displaystyle \int\limits_0^\infty \left(\frac{\arctan x}{x}\right)^6 dx = $

$\displaystyle\hspace{8mm} =\frac{3\pi^5}{320}(4\ln 2 + 3) - \frac{\pi^3}{16}(9\zeta(3) + 40\ln 2 + 2) + \frac{3\pi}{32}(45\zeta(5) + 120\zeta(3) + 32\ln 2)$

...

Solution 2:

It wasn't requested, but instead of exact representations the OP might want an asymptotic expression for $n \to \infty.$ This one works well with the technique of Depoissonization. Make an exponential power series and analyze it asymptotically: $$ \sum_{n=0}^\infty \frac{y^n}{n!} C_n = \int_0^\infty \exp{\big(\frac{y}{x} \text{arctan}(x) \big)} dx , \quad C_n=\int_0^\infty \Big(\frac{\text{arctan}(x)}{x}\Big)^n dx$$ Now $\text{arctan}(x)/x = 1-x^2/3+x^4/5+...$ and with $y$ large and keeping on the first term in the asymptotic expansion $$ e^{-y} \sum_{n=0}^\infty \frac{y^n}{n!} C_n \sim \int_0^\infty \exp{\big(-y\,\frac{x^2}{3}\big)} dx = \frac{1}{2} \sqrt{\frac{3\pi}{y}}.$$ By Depoissonization we can conclude that $$ C_n \sim \frac{1}{2} \sqrt{\frac{3\pi}{n}} .$$ For $n=50$ the asymptotic expression is within 2% of the value from a numerical integration.