$ \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx$

I'm currently working on a definite integral and am hoping to find alternative methods to evaluate. Here I will to address the integral: \begin{equation} I_n = \int_0^\frac{\pi}{2}\ln^n\left(\tan(x)\right)\:dx \end{equation} Where $n \in \mathbb{N}$. We first observe that when $n = 2k + 1$ ($k\in \mathbb{Z}, k \geq 0$) that, \begin{equation} I_{2k + 1} = \int_0^\frac{\pi}{2}\ln^{2k + 1}\left(\tan(x)\right)\:dx = 0 \end{equation} This can be easily shown by noticing that the integrand is odd over the region of integration about $x = \frac{\pi}{4}$. Thus, we need only resolve the cases when $n = 2k$, i.e. \begin{equation} I_{2k} = \int_0^\frac{\pi}{2}\ln^{2k}\left(\tan(x)\right)\:dx \end{equation} Here I have isolated two methods.

Method 1:

Let $u = \tan(x)$: \begin{equation} I_{2k} = \int_0^\infty\ln^{2k}\left(u\right) \cdot \frac{1}{u^2 + 1}\:du = \int_0^\infty \frac{\ln^{2k}\left(u\right)}{u^2 + 1}\:du \end{equation} We note that: \begin{equation} \ln^{2k}(u) = \frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0} \end{equation} By Leibniz's Integral Rule: \begin{align} I_{2k} &= \int_0^\infty \frac{\frac{d^{2k}}{dy^{2k}}\big[u^y\big]_{y = 0}}{u^2 + 1}\:du = \frac{d^{2k}}{dy^{2k}} \left[ \int_0^\infty \frac{u^y}{u^2 + 1} \right]_{y = 0} \nonumber \\ &= \frac{d^{2k}}{dy^{2k}} \left[ \frac{1}{2}B\left(1 - \frac{y + 1}{2}, \frac{y + 1}{2} \right) \right]_{y = 0} =\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \Gamma\left(1 - \frac{y + 1}{2}\right)\Gamma\left( \frac{y + 1}{2} \right) \right]_{y = 0} \nonumber \\ &=\frac{1}{2}\frac{d^{2k}}{dy^{2k}} \left[ \frac{\pi}{\sin\left(\pi\left(\frac{y + 1}{2}\right)\right)} \right]_{y = 0} = \frac{\pi}{2}\frac{d^{2k}}{dy^{2k}} \left[\operatorname{cosec}\left(\frac{\pi}{2}\left(y + 1\right)\right) \right]_{y = 0} \end{align}

Method 2:

We first observe that: \begin{align} \ln^{2k}\left(\tan(x)\right) &= \big[\ln\left(\sin(x)\right) - \ln\left(\cos(x)\right) \big]^{2k} \nonumber \\ &= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right) \end{align} By the linearity property of proper integrals we observe: \begin{align} I_{2k} &= \int_0^\frac{\pi}{2} \left[ \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right) \right]\:dx \nonumber \\ &= \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \int_0^\frac{\pi}{2} \ln^j\left(\cos(x)\right)\ln^{2k - j}\left(\sin(x)\right)\:dx \nonumber \\ & = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j F_{n,m}(0,0) \end{align} Where \begin{equation} F_{n,m}(a,b) = \int_0^\frac{\pi}{2} \ln^n\left(\cos(x)\right)\ln^{m}\left(\sin(x)\right)\:dx \end{equation} Utilising the same identity given before, this becomes: \begin{align} F_{n,m}(a,b) &= \int_0^\frac{\pi}{2} \frac{d^n}{da^n}\big[\sin^a(x) \big] \cdot \frac{d^m}{db^m}\big[\cos^b(x) \big]\big|\:dx \nonumber \\ &= \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[ \int_0^\frac{\pi}{2} \sin^a(x)\cos^b(x)\:dx\right] = \frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{1}{2} B\left(\frac{a + 1}{2}, \frac{b + 1}{2} \right)\right] \nonumber \\ &= \frac{1}{2}\frac{\partial^{n + m}}{\partial a^n \partial b^m}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right] \end{align} Thus, \begin{equation} I_{2k} = \sum_{j = 0}^{2k} { 2k \choose j}(-1)^j \frac{1}{2}\frac{\partial^{2k }}{\partial a^j \partial b^{2k - j}}\left[\frac{\Gamma\left(\frac{a + 1}{2}\right)\Gamma\left(\frac{b + 1}{2}\right)}{\Gamma\left(\frac{a + b}{2} + 1\right)}\right]_{(a,b) = (0,0)} \end{equation}

So, I'm curious, are there any other Real Based Methods to evaluate this definite integral?

Solution 1:

$$\begin{split} I_{2k} &= \int_0^\infty\frac{\ln^{2k}u}{u^2 + 1}du \\ &= \int_0^1\frac{\ln^{2k}u}{u^2 + 1}du +\int_1^{+\infty}\frac{\ln^{2k}\left(u\right)}{u^2 + 1}du \\ &=\int_0^1\frac{\ln^{2k}u}{u^2 + 1}du +\int_0^{1}\frac{\ln^{2k}t}{t^2 + 1}dt \,\,\,\left(\text{by } u\rightarrow \frac 1 t\right)\\ &=2\int_0^1\frac{\ln^{2k}u}{u^2 + 1}du\\ &=2\sum_{n\in\mathbb N}(-1)^n\int_0^1u^{2n}\ln^{2k} (u) du \end{split}$$ Now, let $$J_{p,q}=\int_0^1u^p\ln^q (u)du$$ If $q\geq 1$, by integration by parts, $$J_{p, q}=\left. \frac{u^{p+1}}{p+1}\ln^q u\right]_0^1-\int_0^1\frac{u^{p+1}}{p+1}q\ln^{q-1}(u)\frac{du}u=-\frac q{p+1}J_{p,q-1}$$ Consequently, if $q\geq 1$, $$J_{p,q} = (-1)^{q}\frac{q!}{(p+1)^{q+1}}$$ We conclude that $$I_{2k}=2\cdot (2k)!\sum_{n\in\mathbb N}\frac{(-1)^n}{(2n+1)^{2k+1}}$$ Following Zachy's suggestion, the last sum is known as the Dirichlet Beta function $$I_{2k}=2\cdot (2k)!\beta(2k+1)$$ Finally, values of $\beta$ at odd numbers are known in terms of Euler's numbers and we get $$\boxed{I_{2k}=2\frac{(-1)^kE_{2k}\pi^{2k+1}}{4^{k+1}}}$$

Solution 2:

Make an exponential generating function with $I_k$, $$I_k: = \frac{1}{2} (1 + (-1)^k) \int_0^\infty \frac{ \log^{k}(u) }{u^2+1} du $$ Then $$I(x)=\sum_{k=0}^\infty I_k\,\frac{x^k}{k!} = \frac{1}{2}\int_0^\infty \frac{ u^x + u^{-x} }{u^2+1} du $$ where an interchange of $\sum$ and $\int$ has been made. The integral can be solved in closed form, $I(x) = \pi/2 \cdot \sec{(\pi x/2)}.$ Expanding the sec in a power series will give the last answer that Stafan Lafon gave, in terms of Euler numbers.

Solution 3:

Slightly different way - use well known results

$$\int_0^{\frac{\pi}{2}}\tan^ay\;dy=\frac{\pi}{2}\frac{1}{\sin \frac{\pi}2(a+1) } $$

( this integral is considered in this site probably many times.) $$\frac{1}{\sin x}=\frac{1}{x}+\sum _{n=1}^\infty (-1)^n\left ( \frac{1}{x-n\pi}+ \frac{1}{x+n\pi}\right )$$

and differentiate with respect to $a$ as many times as necessary.