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

Question

I know that for $n=1$ the integral is divergent and that for $n=2$ the integral has a closed form. However, I wonder if the general expression has a closed form.

My attempt: $$\int_0^\infty \left(\frac{\arctan(x)}{x}\right)^ndx=\frac{n}{1-n}\int_0^\infty\frac{\arctan^{n-1}(x)}{x^{n-1}(x^2+1)}dx=\frac{n}{1-n}\int_0^{(\frac{\pi}{2})^{n-1}} u^{n-1}\cot^{n-1}\left(u^{1/(n-1)}\right)du$$

I don't know if I'm on the right track here or not but I do not know through what methods to evaluate the last integral. Any help is appreciated.

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

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

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

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

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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 = $

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

...

Answer 2

$$\int_{0}^{+\infty}\left(\frac{\arctan x}{x}\right)^3\,dx \stackrel{x\mapsto\tan\theta}{=}\int_{0}^{\pi/2}\frac{\theta^3\,d\theta}{\tan^3\theta\cos^2\theta}\stackrel{\text{IBP}}{=}-\frac{\pi^3}{16}+\frac{3}{2}\int_{0}^{\pi/2}\frac{\theta^2\,d\theta}{\sin^2\theta}$$ and in general the computation of the wanted integrals boils down to the computation of $\int_{0}^{\pi/2}\left(\frac{\theta}{\sin\theta}\right)^m\,d\theta$ or the computation of $\oint_\gamma \frac{\log^m z}{z\left(z-\frac{1}{z}\right)^m}\,dz$ where $\gamma$ is the quarter circle joining $1$ and $i$. The integration of $\frac{\log^m z}{z\left(z-\frac{1}{z}\right)^m}$ along the segments joining $0$ and $1$ or $0$ and $i$ can be simply performed through Maclaurin series; in particular the wanted integrals can be always expressed in terms of standard or alternating Euler sums.

Answer 3

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.

Answer 4

Here is a general answer: Let $m \geq n \geq 2$ be integers, and define

$$ \mathcal{I}(m,n) = \int_{0}^{\infty} \frac{\arctan^m x}{x^n} \, dx. $$

Our aim is to obtain a manageable formula for $\mathcal{I}(m,n)$.

Proposition. $\mathcal{I}(m,n)$ equlas $$ \begin{cases} \displaystyle (-1)^{\frac{m-n,}{2}} \int_{0}^{\infty} \frac{\sinh^{n-2} x}{\cosh^n x} \cdot \operatorname{Im} \left( x + \frac{i\pi}{2} \right)^{m} \, dx, & m+n \ \text{even} \\ \displaystyle (-1)^{\frac{m-n-1}{2}} \int_{0}^{\infty} \frac{\sinh^{n-2} x}{\cosh^n x} \cdot \frac{2 \log \tanh x}{\pi} \cdot \operatorname{Im} \left( x + \frac{i\pi}{2} \right)^{m} \, dx, & m+n \ \text{odd} \end{cases} $$

When $m+n$ is even, this integral can be written as a linear combination of values of the Dirichlet $\eta$-function using the well-known formula $\int_{0}^{\infty} x^{s-1}/(e^x + 1) \, dx = \Gamma(s)\eta(s)$.

Here is a Mathematica code for numerical verification:

{m, n} = {7, 5};
NIntegrate[ArcTan[x]^m/x^n, {x, 0, Infinity}, WorkingPrecision -> 20]
(-1)^((m - n)/2) Im[NIntegrate[ Tanh[x]^(n - 2) Sech[x]^2 (x + (I Pi)/2)^m, {x, 0, Infinity}, WorkingPrecision -> 20]]
Clear[m, n];

---

Proof. We only prove the case where $m+n$ is even. (The proof for the other case goes almost the same.) We begin by noticing that

$$ \arctan x = \int_{0}^{1} \frac{x \, ds}{1+x^2s^2} = \lim_{\delta \to 0^+} \int_{i\delta}^{1+i\delta} \frac{x \, ds}{1+x^2s^2}. $$

We temporarily fix $0 < \delta_1 < \cdots < \delta_m$ and consider the line segment $L_k$ beginning at $i\delta_k$ and ending at $1+i\delta_k$. Writing $\vec{\delta} = (\delta_1, \cdots, \delta_k)$, we consider the following perturbed version of $\mathcal{I}(m,n)$.

$$ \mathcal{I}_{\vec{\delta}}(m,n) := \int_{0}^{\infty} x^{m-n} \prod_{k=1}^{m} \left( \int_{L_k} \frac{dz_k}{1+x^2 z_k^2} \right) \, dx. $$

As $\vec{\delta} \to 0$, this converges to the original integral $\mathcal{I}(m,n)$. Now, we invoke the following lemma.

Lemma. If $\alpha_1, \cdots, \alpha_n$ are distinct complex numbers and $m \in \{0, \cdots,n-1\}$, then $$ \frac{x^m}{(1-\alpha_1 x)\cdots (1-\alpha_n x)} = \sum_{k=1}^{n} \frac{\alpha_k^{n-1-m}}{1-\alpha_k x} \prod_{l \neq k} \frac{1}{\alpha_k - \alpha_l}. $$

Since $0 \leq m-n < 2m$, we can apply Lemma to write

$$ \frac{x^{m-n}}{\prod_{k=1}^{m} (1+z_k^2 x^2)} = \frac{x^{m-n}}{\prod_{k=1}^{m} (1 - i z_k x)(1 + i z_k x)} = (-1)^{\frac{m-n}{2}} \sum_{k=1}^{m} \frac{z_k^{m+n-2}}{1+z_k^2 x^2} \prod_{l \neq k} \frac{1}{z_k^2 - z_l^2}. $$

(At this step the assumption that $m+n$ is even is utilized.) Plugging this back to $\mathcal{I}_{\vec{\delta}}(m,n)$ and interchanging the order of integration, we obtain

$$ \mathcal{I}_{\vec{\delta}}(m,n) := (-1)^{\frac{m-n}{2}} \frac{\pi}{2} \sum_{k=1}^{m} \int_{L_1} dz_1 \cdots \int_{L_n} dz_n \left( z_k^{m+n-3} \prod_{l \neq k} \frac{1}{z_k^2 - z_l^2} \right). $$

Now for $l \neq k$, writing $z_k = s_k + i\delta_k$ shows

\begin{align*} \int_{L_l} \frac{dz_l}{z_k^2 - z_l^2} &= \frac{1}{2z_k}\left( \log(z_k + 1 + i\delta_l) - \log(z_k + i\delta_l) - \log(z_k - 1 - i\delta_l) + \log(z_k - i\delta_l) \right) \\ &\xrightarrow[\vec{\delta} \to 0]{} \frac{1}{2s_k}\left( \log\left(\frac{1 + s_k}{1 - s_k} \right) + i\pi\operatorname{sign}(l-k) \right) \end{align*}

Plugging this back,

\begin{align*} \mathcal{I}(m,n) &= (-1)^{\frac{m-n}{2}} \frac{\pi}{2} \sum_{k=1}^{m} \int_{0}^{1} s^{n-2} \left( \operatorname{artanh}(s) - \frac{i\pi}{2} \right)^{k-1}\left( \operatorname{artanh}(s) + \frac{i\pi}{2} \right)^{n-k} \, ds \\ &= (-1)^{\frac{m-n}{2}} \operatorname{Im} \int_{0}^{1} s^{n-2} \left( \operatorname{artanh}(s) + \frac{i\pi}{2} \right)^{m} \, ds \\ &= (-1)^{\frac{m-n}{2}} \operatorname{Im} \int_{0}^{\infty} \frac{\sinh^{n-2} x}{\cosh^n x} \left( x + \frac{i\pi}{2} \right)^{m} \, dx, \end{align*}

completing the proof.

Answer 5

As user90369 doesn't have the time apparently I thought I repesent my own solution explicitly for those interested.

Starting with \begin{align} \int_{-\pi/2}^{\pi/2} \frac{x^n \left(\cos x\right)^{n-2}}{\left(\sin x\right)^n} \, {\rm d}x &\stackrel{y=2x}{=} 2^{-n+1} \, i^n \int_{-\pi}^{\pi} y^n \, e^{-iy} \, \frac{\left(1+e^{-iy}\right)^{n-2}}{\left(1-e^{-iy}\right)^n} \, {\rm d}y \\ &\stackrel{z=e^{iy}}{=} -i \, 2^{-n+1} \int_\gamma \left(\log z\right)^n \, \frac{(z+1)^{n-2}}{(z-1)^n} \, {\rm d}z \end{align} where $\gamma$ is the unit circle in positive direction. The integrand is holomorph off the negative real line and the contour can be deformed to only enclose the cut $[-1,0]$.

Then \begin{align} &=i \, 2^{-n+1} (-1)^n \int_0^1 \frac{(1-z)^{n-2}}{(1+z)^n} \, \left\{ \left(\log z + i\pi\right)^n - \left(\log z - i\pi\right)^n \right\} {\rm d}z \\ &=2^{-n+2} (-1)^{n+1} \sum_{k=0}^n \sum_{p=0}^{n-2} \sum_{q=0}^\infty \begin{pmatrix} n \\ k \end{pmatrix} \begin{pmatrix} n-2 \\ p \end{pmatrix} \begin{pmatrix} -n \\ q \end{pmatrix} (-1)^p \pi^{n-k} \sin\left(\frac{\pi}{2}(n-k)\right) \\ &\qquad \times \int_0^1 z^{p+q} \, \left( \log z \right)^k \, {\rm d}z \end{align} and using the two formulas \begin{align} \int_0^1 x^n \, \left(\log x\right)^m \, {\rm d}x &= \frac{(-1)^m \, m!}{(n+1)^{m+1}} \tag{1} \\ \begin{pmatrix} -n \\ q \end{pmatrix} &= (-1)^q \begin{pmatrix} n+q-1 \\ q \end{pmatrix} = (-1)^q \, \frac{(q+1)\cdots(q+n-1)}{(n-1)!} \\ &= \frac{(-1)^q}{(n-1)!} \sum_{m=0}^{n-1} \left[ {n-1 \atop m} \right] (q+1)^m \tag{2} \end{align} with the Stirling numbers of the first kind we arrive at \begin{align} =2^{-n+2} (-1)^{n+1} \sum_{k=0}^n \sum_{p=0}^{n-2} \sum_{m=0}^{n-1} \sum_{q=0}^\infty \begin{pmatrix} n-2 \\ p \end{pmatrix} \left[ {n-1 \atop m} \right] \frac{(q+1)^m (-1)^{k+p+q} \pi^{n-k} n! \sin\left(\frac{\pi}{2}(n-k)\right)}{(p+q+1)^{k+1} (n-1)! (n-k)!} \, . \end{align} Next we reverse the summation $k \rightarrow n-k$ and decompose another time the factor $(q+1)^{m}=(p+q+1-p)^{m}$ in the nominator \begin{align} &=-2^{-n+2}\,n \sum_{k=0}^n \sum_{p=0}^{n-2} \sum_{m=0}^{n-1} \sum_{l=0}^m \begin{pmatrix} n-2 \\ p \end{pmatrix} \left[ {n-1 \atop m} \right] \begin{pmatrix} m \\ l \end{pmatrix} (-p)^{m-l} \\ &\qquad \frac{(-\pi)^{k} \sin\left(\frac{\pi}{2}k\right)}{k!} \sum_{q=0}^\infty \frac{(-1)^{q+p}}{(p+q+1)^{n-k-l+1}} \\ &=-2^{-n+2} \, n \sum_{k=0}^n \sum_{p=0}^{n-2} \sum_{l=0}^{n-1} \sum_{m=l}^{n-1} \begin{pmatrix} n-2 \\ p \end{pmatrix} \left[ {n-1 \atop m} \right] \begin{pmatrix} m \\ l \end{pmatrix} (-p)^{m-l}\\ &\qquad \frac{(-\pi)^{k} \sin\left(\frac{\pi}{2}k\right)}{k!} \left\{ \eta\left(n-k-l+1\right) -\sum_{q=1}^p \frac{(-1)^{q+p}}{(p-q+1)^{n-k-l+1}} \right\} \tag{3} \end{align} where $\eta(s)$ is the Dirichlet Eta-Function.

It seems as if \begin{align} &\quad \sum_{p=0}^{n-2} \sum_{q=1}^{p} \sum_{l=0}^{n-1} \sum_{m=l}^{n-1} \begin{pmatrix} n-2 \\ p \end{pmatrix} \left[ {n-1 \atop m} \right] \begin{pmatrix} m \\ l \end{pmatrix} \frac{(-p)^{m-l} (-1)^{q+p}}{(p-q+1)^{n-k-l+1}} \\ &=\sum_{p=0}^{n-2} \sum_{q=1}^{p} \begin{pmatrix} n-2 \\ p \end{pmatrix} \frac{(-1)^{q+p}}{(p-q+1)^{n-k+1}} \sum_{m=0}^{n-1} \left[ {n-1 \atop m} \right](-q+1)^m \\ &=(-1)^{n-1}(n-1)! \sum_{p=0}^{n-2} \sum_{q=1}^{p} \begin{pmatrix} n-2 \\ p \end{pmatrix} \begin{pmatrix} q-1 \\ n-1 \end{pmatrix} \frac{(-1)^{q+p}}{(p-q+1)^{n-k+1}} \\ &= 0 \, , \end{align} since $q\leq n-2$ and $\begin{pmatrix} n-3 \\ n-1 \end{pmatrix}=0$, so the second term in the curly bracket does not contribute.

Here is a Maple Code of the result (3) for verification:

restart; 
n := 5; 
eta := proc (s) options operator, arrow; limit((1-2^(1-S))*Zeta(S), S = s) end proc; 
r1 := simplify(-2^(-n+2)*n*add(add(add(add(binomial(n-2, p)*Stirling1(n-1, m)*(-1)^(n-1-m)*binomial(m, l)*(-p)^(m-l)*(-Pi)^k*sin((1/2)*Pi*k)*eta(n-k-l+1)/factorial(k), m = l .. n-1), l = 0 .. n-1), p = 0 .. n-2), k = 0 .. n)); 
r2 := simplify(int(x^n*cos(x)^(n-2)/sin(x)^n, x = -(1/2)*Pi .. (1/2)*Pi)); 
evalf([r1, r2])

Answer 6

A complete asymptotic expansion for large $n$ follows from Laplace's method: \begin{align*} \int_0^{ + \infty } {\left( {\frac{{\arctan t}}{t}} \right)^n \mathrm{d}t} & = \int_0^{ + \infty } {\exp \left( { - n\log \left( {\frac{t}{{\arctan t}}} \right)} \right)\mathrm{d}t} \sim \frac{1}{2}\sqrt {\frac{{3\pi }}{n}} \sum\limits_{k = 0}^\infty {\frac{{a_k }}{{n^k }}} \\ & = \frac{1}{2}\sqrt {\frac{{3\pi }}{n}} \left( {1 + \frac{{39}}{{40n}} + \frac{{4763}}{{4480n^2 }} + \frac{{25401}}{{25600n^3 }} + \frac{{13025883}}{{12615680n^4 }} + \ldots } \right), \end{align*} where $$ a_k = \left( {\frac{3}{4}} \right)^k \frac{1}{{k!}}\left[ {\frac{{\mathrm{d}^{2k} }}{{\mathrm{d}t^{2k} }}\left( {\frac{{t^2 }}{{3\log \left( {\frac{t}{{\arctan t}}} \right)}}} \right)^{k + 1/2} } \right]_{t = 0} . $$