Need help to generalise $\int_0^\infty\left(\arctan \frac1x\right)^2 \mathrm d x$.

Question

The integral posted six years ago contains 10 solutions for proving that $$\int_0^\infty\left(\arctan \frac1x\right)^2 \mathrm d x = \pi\ln 2,$$ I then want to go further with the general case

$$I_n=\int_0^\infty\left(\arctan \frac1x\right)^n \mathrm d x$$

with the usual substitution $y=\arctan \frac{1}{x}$ and obtain

$$ \begin{aligned} I_n & =-\int_0^{\frac{\pi}{2}} y^n d(\cot y) \\ & =n \int_0^{\frac{\pi}{2}} y^{n-1} \cot y d y \\ & =n \int_0^{\frac{\pi}{2}} y^{n-1} d(\ln (\sin y)) \\ & =-n(n-1) \int_0^{\frac{\pi}{2}} y^{n-2} \ln (\sin y) d y \end{aligned} $$ Let’s start with the smaller one $n=3$. Then the Fourier series of $\ln(\sin y)$ come to my mind

$$ \ln(\sin y)=-\ln 2-\sum_{n=1}^{\infty} \frac{\cos (2 k y)}{k} $$

and arrived at $$ \begin{aligned} \int_0^{\frac{\pi}{2}} y \ln (\sin y) d y & =-\int_0^\frac \pi 2 y\left(\ln 2+\sum_{k=1}^\infty \frac{\cos (2ky)}{k}\right) d y\\&=-\frac{\pi^2}{8} \ln 2-\sum_{k=1}^{\infty} \frac{1}{k} \int_0^{\frac{\pi}{2}} y \cos (2 k y) d y \\&=-\frac{\pi^2}{8} \ln 2-\sum_{k=1}^{\infty} \frac{1}{k} \cdot \frac{\cos (\pi k)-1}{4 k^2} \\ & =-\frac{\pi^2}{8} \ln 2-\frac{1}{4} \sum_{k=1}^{\infty} \frac{1}{k^3}\left[(-1)^k-1\right] \\ & =-\frac{\pi^2}{8} \ln 2+\frac{1}{4} \sum_{k=1}^{\infty} \frac{2}{(2 k+1)^3} \\ & =-\frac{\pi^2}{8} \ln 2+\frac{1}{2}\left[\sum_{k=1}^{\infty} \frac{1}{k^3}-\sum_{k=1}^{\infty} \frac{1}{(2 k)^3}\right] \\ & =-\frac{\pi^2}{8} \ln 2+\frac{7}{16} \zeta(3) \end{aligned} $$

Plugging back yields $$I_3=\frac{3}{4} \pi^2 \ln 2-\frac{21}{8} \zeta(3)$$

For $n=4$, we have

$$ \begin{aligned} \int_0^{\frac{\pi}{2}} y^3 \ln (\sin y) d y = & -\frac{\pi^3}{24} \ln 2-\sum_{k=1}^{\infty} \frac{1}{k} \int_0^{\frac{\pi}{2}} y^2 \cos (2 k y) d y \\ = & -\frac{\pi^3}{24} \ln 2-\frac{\pi}{4} \sum_{k=1}^{\infty} \frac{(-1)^k}{k^3} \\ = & -\frac{\pi^3}{24} \ln 2+\frac{\pi}{4}\left[\sum_{k=1}^{\infty} \frac{1}{k^3}-2 \sum_{k=1}^{\infty} \frac{1}{(2 k)^3}\right] \\ = & -\frac{\pi^3}{24} \ln 2+\frac{\pi}{4}\left(1-\frac{1}{4}\right) \zeta(3) \\ = & -\frac{\pi^3}{24} \ln 2+\frac{3 \pi}{16} \zeta(3) \end{aligned} $$ Hence

$$ I_4=\frac{\pi^3}{2} \ln 2-\frac{9 \pi}{4} \zeta(3) $$

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However, in general, $$ \int_0^{\frac{\pi}{2}} y^n \ln (\sin y) d y = -\frac{\ln 2}{n+1}\left(\frac{\pi}{2}\right)^{n+1}-\sum_{k=1}^{\infty} \frac{1}{k} \int_0^{\frac{\pi}{2}} y^n \cos (2 k y) d y $$ where the last integral becomes complicated for higher powers $n$.

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My Question:

Do we have any solution for the first or last integral?

Answer 1

\begin{align} I_n &=\int_0^\infty \arctan^n \frac1x \ {d x}\>\>\>\>\>\>\> (y=\cot x )\\ &= \int_0^{\frac{\pi}{2}} y^{n} \csc^2 y \overset{t=e^{2i y}} {d y}=\frac1{i^{n+1}2^{n-1}}\int_{-1}^1 \frac{\ln^n t}{(1-t)^2}\ dt\\ & =\frac1{i^{n+1}2^{n-1}}\bigg[ \int_{0}^1 \frac{\ln^n t}{(1-t)^2}\ dt + \int_{-1}^0 \frac{\ln^n t}{(1-t)^2}\overset{t\to -t}{dt }\bigg]\\ & =\frac1{i^{n+1}2^{n-1}} \int_{0}^1 \frac{\ln^n t}{(1-t)^2}+ \frac{(i\pi +\ln t)^n}{(1+t)^2}\ dt \end{align} Utilize $\frac1{1-{2^{1-k}}} \int_{0}^1 \frac{\ln^k t}{(1+t)^2}dt =\int_{0}^1 \frac{\ln^k t}{(1-t)^2}dt= (-1)^k k! \zeta(k)=J_k $ \begin{align} I_n & =\frac1{i^{n+1}2^{n-1}} \bigg[ J_n +\frac{(i\pi)^n}2 - n (i\pi)^{n-1}\ln2 \\ &\hspace{25mm} + \sum_{k=2}^n (i\pi)^{n-k}\binom n k \left( 1-\frac1{2^{k-1}}\right)J_k \bigg] \end{align} Retain the real terms only to express the result as \begin{align} I_n&=(\frac\pi2)^{n-1}n\ln2 - \frac{n!}{2^{n-1}} \bigg[\sin^2\frac{n\pi}2 \ \zeta(n)\\ & \hspace {20mm} -\sum_{k=1}^{[\frac{n-1}2]}\frac{(-1)^k {\pi}^{n-2k-1}}{(n-2k-1)!} \left( 1-\frac1{2^{2k}}\right)\zeta(2k+1)\bigg] \end{align}

Answer 2

This answer evaluates the integral in the form provided by @Lai:

\begin{align*} I_n = (\frac{\pi}{2})^{n-1} n\ln 2 - n(n-1) \sum_{k=1}^\infty \frac{1}{k} \int_0^{\pi/2} y^{n-2} \cos (2ky) dy \end{align*}

Taking $k$ to be a continuous variable to start with. First $n$ even:

\begin{align*} & \int_0^{\pi/2} y^{2l} \cos (2ky) dy \nonumber \\ & = \frac{(-1)^l}{2^{2l}} \frac{d^{2l}}{dk^{2l}} \int_0^{\pi/2} \cos (2ky) dy \nonumber \\ & = \frac{(-1)^l}{2^{2l}} \frac{d^{2l}}{dk^{2l}} \frac{\sin (k \pi)}{2k} \nonumber \\ & = \frac{(-1)^l}{2^{2l+1}} \sum_{m=0}^{2l} \frac{(2l)!}{m! (2l-m)!} \left( \frac{d^{m}}{dk^{m}} \frac{1}{k} \right) \frac{d^{2l-m}}{dk^{2l-m}} \sin (k \pi) \nonumber \\ & = \frac{(-1)^l}{2^{2l}} \sum_{r=0}^l \frac{(2l)!}{(2l-2r-1)!} \left( -\frac{1}{k^{2r+2}} \right) \frac{d^{2l-2r-1}}{dk^{2l-2r-1}} \sin (k \pi) + \cdots \nonumber \\ & = \frac{(-1)^l}{2^{2l}} \sum_{r=0}^l \frac{(2l)!}{(2l-2r-1)!} \left( - \frac{1}{k^{2r+2}} \right) (-1)^{l-r+1} \pi^{2l-2r-1} \cos (k \pi) + \cdots \nonumber \\ & = \frac{1}{2^{2l}} \sum_{r=0}^{l} \frac{(2l)!}{(2l-2r-1)!} \left( \frac{-\cos (k \pi)}{k^{2r+2}} \right) (-1)^{r+1} \pi^{2l-2r-1} + \cdots \qquad (*) \end{align*}

where $ + \cdots$ indicates additional terms that vanish when $k$ is taken to be an integer.

For $n$ odd:

\begin{align*} & \int_0^{\pi/2} y^{2l+1} \cos (2ky) dy \nonumber \\ & = \frac{(-1)^l}{2^{2l+1}} \frac{d^{2l+1}}{dk^{2l+1}} \int_0^{\pi/2} \sin (2ky) dy \nonumber \\ & = \frac{(-1)^l}{2^{2l+1}} \frac{d^{2l+1}}{dk^{2l+1}} \frac{\cos (k \pi) - 1}{2k} \nonumber \\ & = \frac{(-1)^l}{2^{2l+2}} \sum_{m=0}^{2l+1} \frac{(2l+1)!}{m! (2l-m+1)!} \left( \frac{d^m}{dk^m} \frac{1}{k} \right) \frac{d^{2l-m+1}}{dk^{2l-m+1}} (\cos (k \pi)-1) \nonumber \\ & = \frac{(-1)^l}{2^{2l+2}} \sum_{r=0}^{l} \frac{(2l+1)!}{(2l-2r)!} \left( - \frac{1}{k^{2r+2}} \right) \frac{d^{2l-2r}}{dk^{2l-2r}} (\cos (k \pi)-1) + \cdots \nonumber \\ & = \frac{(-1)^{l+1}}{2^{2l+2}} (2l+1)! \left( - \frac{1}{k^{2l+2}} \right) + \frac{(-1)^l}{2^{2l+2}} \sum_{r=0}^l \frac{(2l+1)!}{(2l-2r)!} \left( - \frac{1}{k^{2r+2}} \right) \frac{d^{2l-2r}}{dk^{2l-2r}} \cos (k \pi) + \cdots \nonumber \\ & = \frac{(-1)^l}{2^{2l+2}} (2l+1)! \frac{1}{k^{2l+2}} + \frac{(-1)^l}{2^{2l+2}} \sum_{r=0}^l \frac{(2l+1)!}{(2l-2r)!} \left( - \frac{1}{k^{2r+2}} \right) \pi^{2l-2r} (-1)^{l-r} \cos (k \pi) + \cdots \nonumber \\ & = \frac{(-1)^l}{2^{2l+2}} (2l+1)! \frac{1}{k^{2l+2}} + \frac{1}{2^{2l+2}} \sum_{r=0}^l \frac{(2l+1)!}{(2l-2r)!} \left( \frac{-\cos (k \pi)}{k^{2r+2}} \right) \pi^{2l-2r} (-1)^r + \cdots \qquad (**) \end{align*}

where $ + \cdots$ indicates additional terms that vanish when $k$ is taken to be an integer.

For $n$ even, using $(*)$ with $n-2=2l$:

\begin{align*} & (\frac{\pi}{2})^{n-1} n\ln 2 - n(n-1) \sum_{k=1}^\infty \frac{1}{k} \int_0^{\pi/2} y^{n-2} \cos (2ky) dy \nonumber \\ & = (\frac{\pi}{2})^{n-1} n\ln 2 -n(n-1) \sum_{k=1}^\infty \frac{1}{k} \frac{1}{2^{n-2}} \sum_{r=0}^{l} \frac{(n-2)!}{(n-2r-3)!} \left( \frac{(-1)^{k+1}}{k^{2r+2}} \right) (-1)^{r+1} \pi^{n-2r-3} \nonumber \\ & = (\frac{\pi}{2})^{n-1} n\ln 2 + n! \frac{1}{2^{n-2}} \sum_{r=1}^{\frac{n-2}{2}} \frac{(-1)^r \pi^{n-2r-1}}{(n-2r-1)!} \left( 1-\frac{1}{2^{2r}} \right) \zeta (2r+1) \end{align*}

In agreement with @Quanto's result.

For $n$ odd, using $(**)$ with $n-2=2l+1$:

\begin{align*} & (\frac{\pi}{2})^{n-1} n\ln 2 - n(n-1) \sum_{k=1}^\infty \frac{1}{k} \int_0^{\pi/2} y^{n-2} \cos (2ky) dy \nonumber \\ & = (\frac{\pi}{2})^{n-1} n\ln 2 - n(n-1) \sum_{k=1}^\infty \frac{1}{k} \left[ \frac{(-1)^\frac{n-3}{2}}{2^{n-1}} (n-2)! \frac{1}{k^{n-1}} + \frac{1}{2^{n-1}} \sum_{r=0}^\frac{n-3}{2} \frac{(n-2)!}{(n-2r-3)!} \left( \frac{(-1)^{k+1}}{k^{2r+2}} \right) \pi^{n-2r-3} (-1)^r \right] \nonumber \\ & = (\frac{\pi}{2})^{n-1} n\ln 2 + \frac{(-1)^\frac{n-1}{2}}{2^{n-1}} n! \sum_{k=1}^\infty \frac{1}{k^n} + \frac{n!}{2^{n-1}} \sum_{r=1}^{\frac{n-1}{2}} \frac{1}{(n-2r-1)!} \left( \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^{2r+1}} \right) (-1)^r \pi^{n-2r-1} \nonumber \\ & = (\frac{\pi}{2})^{n-1} n\ln 2 + \sin \frac{n \pi}{2} \frac{n!}{2^{n-1}} \zeta (n) + \frac{n!}{2^{n-1}} \sum_{r=1}^{\frac{n-1}{2}} \frac{(-1)^r \pi^{n-2r-1}}{(n-2r-1)!} \left( 1-\frac{1}{2^{2r}} \right) \zeta (2r+1) \end{align*}

This is nearly in agreement with @Quanto's result, though I suspect there may be a typo in their expression.

Answer 3

Another possible generalisation is $$I(a,b)=\int_0^{\infty}\arctan\left(\frac{a}{x}\right)\arctan\left(\frac{b}{x}\right)\mathrm{d}x=\frac{\pi}{2}\ln\left(\frac{(a+b)^{a+b}}{a^a b^b}\right)$$ for $a,b>0$. If we have $a,b>0$, then \begin{align*} I(a,b) &= \int_0^{\infty}\frac{\arctan(ax)\arctan(bx)}{x^2}\mathrm{d}x,\quad \left[x\mapsto \frac{1}{x}\right]\\ &= \int_0^{\infty}\left(\int_0^a \frac{1}{y^2x^2+1}\mathrm{d}y\right)\left(\int_0^b \frac{1}{z^2x^2+1}\mathrm{d}z\right)\mathrm{d}x,\\ &= \int_0^b\int_0^a \frac{1}{y^2-z^2}\left[\int_0^{\infty} \left(\frac{y^2}{y^2x^2+1}-\frac{z^2}{z^2x^2+1}\right)\mathrm{d}x\right]\mathrm{d}y\,\mathrm{d}z,\\ &= \int_0^b\int_0^a \frac{1}{y^2-z^2} [y\tan^{-1}(yx)-z\tan^{-1}(zx)]_0^{\infty} \mathrm{d}y\,\mathrm{d}z,\\ &= \frac{\pi}{2}\int_0^b\int_0^a \frac{1}{y+z}\mathrm{d}y\,\mathrm{d}z,\\ &= \frac{\pi}{2}\int_0^b [\ln(y+z)]_{y=0}^a\mathrm{d}z,\\ &= \frac{\pi}{2}\int_0^b (\ln(z+a)-\ln(z))\,\mathrm{d}z,\\ &= \frac{\pi}{2}[(z+a)\ln(z+a)-z\ln(z)]_0^b,\\ &= \frac{\pi}{2}[(a+b)\ln(a+b)-b\ln(b)-a\ln(a)],\\ &= \frac{\pi}{2}\ln\left(\frac{(a+b)^{a+b}}{a^a b^b}\right).\end{align*}

Answer 4

For your last integral $$I_n=\int_0^{\frac{\pi}{2}} y^n \cos (2 k y)\, dy=\Re\Bigg(\int_0^{\frac{\pi}{2}} y^n e^{ 2 k i y}\, dy\Bigg)$$ $$J_n=\int_0^{\frac{\pi}{2}} y^n e^{ 2 k i y}\, dy=(-i)^{3 (n+1)}\,(2k)^{-(n+1)} \Big(\Gamma (n+1)-\Gamma (n+1,-i k \pi )\Big)$$

$\Re(J_n)$ reduces to the hypergeometric function @Caesar.tcl provided in comments.