Log-tangent integral using exponential generating function $\int_0^{\pi/2}x\log^n(\tan x)\,dx$

Question

$$I_n=\int_0^{\pi/2}x\log^n(\tan x)\ dx,\quad n\in\mathbb{Z}$$ I want to evaluate this log-tangent integral using exponential generating functions. My attempt is below.

---

Define, $$G(t)=\sum_{n=0}^\infty\frac{I_n}{n!}t^n=\int_0^{\pi/2}x\sum_{n=0}^\infty\frac{(t\log(\tan x))^n}{n!}\ dx=\int_0^{\pi/2}x\tan^{t}(x)\ dx$$ rewriting $x$ as $\arcsin(\sin(x))$ and expanding by its Maclaurin series, $$\int_0^{\pi/2}x\tan^{t}(x)\ dx=\int_0^{\pi/2}\arcsin(\sin(x))\frac{\sin^t(x)}{\cos^{t}(x)}\ dx=\int_0^{\pi/2}\sum_{n=0}^\infty\binom{2n}{n}\frac{\sin^{2n+1}(x)}{4^n(2n+1)}\frac{\sin^t(x)}{\cos^{t}(x)}\ dx\\ =\sum_{n=0}^\infty\binom{2n}{n}\frac{4^{-n}}{(2n+1)}\int_0^{\pi/2}\sin^{2n+t+1}(x)\cos^{-t}(x)\ dx=\sum_{n=0}^\infty\binom{2n}{n}\frac{4^{-n}}{2(2n+1)}B\left(\frac{2n+t+2}{2},\frac{1-t}{2}\right)$$ where $B(m,n)$ is the Beta function, $$B(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$ hence, $$G(t)=\sum_{n=0}^\infty\binom{2n}{n}\frac{4^{-n}}{2(2n+1)}\frac{\Gamma\left(n+t/2+1\right)\Gamma(1/2-t/2)}{\Gamma(n+3/2)}.$$ Now I need to extract the coefficient of $t^n/n!$ as, $$I_n=[t^n/n!]G(t)$$ but I am having difficulties doing so. If I can get a closed form for $G(t)$ I may be able to take its logarithmic derivative and use the Cauchy product, but I don't have a closed form either.

Any ideas? Thanks in advance.

Answer 1

$$I_n=\int_0^{\pi/2}x\log^n(\tan x))\ dx\overset{t=\tan x}{=}\int_0^\infty\frac{\arctan t}{1+t^2}\ln^nt\,dt=\frac{\partial^n}{\partial s^n}J(s)\,\bigg|_{s=0}$$ where $$J(s)=\int_0^\infty\frac{\arctan t}{1+t^2}t^sdt=\int_0^1\frac{dx}{x^2}\int_0^\infty\frac{t^{s+1}}{(1+t^2)(\frac1{x^2}+t^2)}dt$$ To evaluate first integral we use the keyhole contour: $$(1-e^{2\pi is})\int_0^\infty\frac{t^{s+1}}{(1+t^2)(\frac1{x^2}+t^2)}dt=\pi i\frac{x^2}{1-x^2}(e^\frac{3\pi is}2+e^\frac{\pi is}2)(1-x^{-s})$$ Therefore, $$J(s)=\frac\pi{2\sin\frac{\pi s}2}\int_0^1\frac{x^{-s}-1}{1-x^2}dx=\frac\pi{4\sin\frac{\pi s}2}\left(\psi\Big(\frac12\Big)-\psi\Big(\frac12-\frac s2\Big)\right)$$ and $$\boxed{\,\,I_n=\frac{\partial^n}{\partial s^n}J(s)\,\bigg|_{s=0}=\frac\pi{2^{2+n}}\frac{d^n}{dx^n}\,\frac{\psi\big(\frac12\big)-\psi\big(\frac12-x\big)}{\sin\pi x}\,\bigg|_{x=0}\,\,}$$ Quick check: at $n=0$ we get $$I_0=\frac14\psi^{(1)}\Big(\frac12\Big)=\frac{\pi^2}8=\int_0^\infty\frac{\arctan t}{1+t^2}dt$$ as it should be.

For example, for $n=1$ we get $$I_1=-\frac{\psi^{(2)}\big(\frac12\big)}{16}$$ The derivatives of digamma-function can be expressed via zeta-function.

Answer 2

Changing $u=\tan x$ in the integral representation of the generating function ($-2<t<1$), \begin{align} G(t)&=\int_0^{\pi/2}x\tan^{t}(x)\,dx\\ &=\int_0^\infty \frac{\arctan u}{1+u^2}u^t\,dt\\ &=-\frac t2\int_0^\infty (\arctan u)^2 u^{t-1}\,dt \end{align} (the latter is obtained after IBP). This integral is evaluated by a CAS as \begin{equation} G(t)=\frac{\pi^{2}}{4} \sec \left(\frac{\pi t}{2}\right)-\frac{\pi} {4}\left[\gamma +2 \ln \left(2\right)+\psi \left(\tfrac{1}{2}+\tfrac{t}{2}\right)\right] \csc \left(\frac{\pi t}{2}\right) \end{equation} where $\gamma$ is Euler's constant and $\psi$ is the digamma function. Alternatively, it can be obtained by considering that the integral is a Mellin ($u\to t$) transform of the square of $\arctan u$. The multiplication formula for the Mellin transform (Parseval theorem) can be used with (Erdelyi TI 6.5.46, p.322) \begin{equation} \int_0^\infty\arctan u\,u^{t-1}du=-\frac{\pi}{2t}\sec\left( \frac{\pi t}{2} \right) \end{equation} The contour integral can be computed with the residue theorem and the result is expressed as a series which can be identified with the above expression.

The functional relation for the digamma function reads \begin{equation} \psi\left(2z\right)=\tfrac{1}{2}\left(\psi\left(z\right)+\psi\left(z+\tfrac{1} {2}\right)\right)+\ln 2 \end{equation} with $z=(1+t)/2$, it comes \begin{equation} \psi \left(\tfrac{1}{2}+\tfrac{t}{2}\right) =2\psi(1+t)-2\ln2-\psi\left( 1+\tfrac t2 \right) \end{equation} Introducing the series expansion \begin{equation} \psi\left(1+z\right)=-\gamma+\sum_{k=2}^{\infty}(-1)^{k}\zeta\left(k\right)z^{% k-1}\text{ for }|z|<1 \end{equation} to write for $|t|<1$ \begin{equation} \psi \left(\tfrac{1}{2}+\tfrac{t}{2}\right) =-\gamma-2\ln2+2\sum_{k=2}^{\infty}(-1)^{k}\zeta\left(k\right)(1-2^{-k})t^{k-1} \end{equation} Thus \begin{equation} G(t)=\frac{\pi^{2}}{4} \sec \left(\frac{\pi t}{2}\right)-\frac{\pi} {2}\csc \left(\frac{\pi t}{2}\right)\sum_{k=2}^{\infty}(-1)^{k}\zeta\left(k\right)(1-2^{-k})t^{k-1} \end{equation} Now, we have the expansions DLMF: \begin{align} \sec z&=1+\frac{z^{2}}{2}+\frac{5}{24}z^{4}+\frac{61}{720}z^{6}+\cdots+\frac{(-1)^{n}E_{2n}}{(2n)!}z^{2n}+\cdots\\ \csc z&=\frac{1}{z}+\frac{z}{6}+\frac{7}{360}z^{3}+\frac{31}{15120}z^{5}+\cdots +\frac{(-1)^{n-1}2(2^{2n-1}-1)B_{2n}}{(2n)!}z^{2n-1}+\cdots \end{align} where the Euler and Bernoulli numbers appear. The coefficients of the $t-$expansion of the generating function are then obtained using the Cauchy product formula: \begin{align} G(t)=\sum_{s=0}^\infty&\frac{(-1)^{s}E_{2s}}{(2s)!}\left( \frac{\pi}{2} \right) ^{2s+2}t^{2s}+\\ & +2\sum_{s=0}^\infty(-1)^st^s\sum_{m=0}^{\lfloor s/2\rfloor}(-1)^m\frac{(2^{2m-1}-1)(1-2^{-s+2m-2})}{(2m)!}B_{2m}\zeta(s-2m+2)\left(\frac\pi2\right)^{2m} \end{align}

Answer 3

For the even case \begin{align} I_{2m}=& \int_0^{\pi/2}x\ln^{2m}(\tan x)\ \overset{t=\tan x}{dx}\\ =& \int_0^{\infty}\frac{\tan^{-1}t\ln^{2m}t}{1+t^2}\ \overset{t\to 1/t}{dt} =\int_0^{\infty}\frac{(\frac\pi2-\tan^{-1}t)\ln^{2m}t}{1+t^2}\ dt\\ =&\ \frac\pi4 \int_0^{\infty}\frac{\ln^{2m}t}{1+t^2}\ dt =\frac{(-1)^m E_{2m}}{2}\left(\frac\pi2\right)^{2(m+1)} \end{align} where $E_{2m}=(-1,5,-61,1385,\cdots)$are the Euler numbers. In particular $$I_2= \frac{\pi^4}{32},\> I_4 = \frac{5\pi^6}{128}, \> I_6 =\frac{61\pi^8}{512},\cdots$$