Is there a closed form for the integral $\int_{0}^{\pi/4}x^{k}\ln\tan x\, dx$, where $k$ is a natural number?

Question

To start, I am aware that our integral $I(k)=\int_{0}^{\pi/4}x^{k}\ln\tan x\, dx$ is equal to $$I(k)=\int_{0}^{\pi/4}x^{k}\ln\sin x\, dx-\int_{0}^{\pi/4}x^{k}\ln\cos x\, dx$$, but I cannot seem to find a closed form for either integral above.

I believe, however, that I was able to find a method to achieve a reduction formula for the first integral, though rather hideous. I assume that the same can be done for the second integral. This is much less than what I'm looking for, which is a closed form, but I will share my progress below nonetheless.

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Employing the well known Fourier series $\ln\sin x=-\ln2-\sum\limits_{j=1}^{\infty}\frac{\cos(2jx)}{j}$, we have $$J(k)=\int\limits_{0}^{\pi/4}x^{k}\ln\sin x\, dx=-\ln 2\frac{(\pi/4)^{k+1}}{k+1}-\sum\limits_{j=1}^{\infty}\frac{1}{j}\underbrace{\int\limits_{0}^{\pi/4}x^{k}\cos(2jx)dx}_{L(k)}$$

Now consider the function $$f(t)=\int\limits_{0}^{\pi/4}e^{tx}\cos(2jx)dx=\Re\left(\int\limits_{0}^{\pi/4}e^{tx+2ijx}dx\right)$$ By differentiating under the integral sign, we quickly also arrive at the fact that $$L(k)=f^{(k)}(0)$$ and so $$f(t)=\sum\limits_{k=0}^{\infty}\frac{L(k)}{k!}t^{k}$$

Here, we split into cases.

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Case 1: $j$ is even

$$f(t)=\frac{-t}{4j^{2}+t^{2}}\left(1-(-1)^{j/2}e^{\pi t/4}\right)$$ Note that since $f(0)=\sum\limits_{k=0}^{\infty}\frac{L(k)}{k!}=0$, we also have $L(0)=0$ for even $j$.

Now, $$\left(4j^{2}+t^{2}\right)f(t)=-t-(-1)^{j/2}\sum\limits_{k=0}^{\infty}\frac{\pi^{k}}{4^{k}k!}t^{k+1}$$ but simultaneously $$\left(4j^{2}+t^{2}\right)f(t)=\sum\limits_{k=0}^{\infty}\frac{L(k)}{k!}\left(4j^{2}t^{k}+t^{k+2}\right)$$ And thus by comparing the coefficients of $t^{k}$ is the above two expressions, we obtain $$(-1)^{j/2+1}\frac{\pi^{k-1}}{4^{k-1}(k-1)!}=\frac{L(k)4j^{2}}{k!}+\frac{L(k-2)}{(k-2)!}$$ $$L(k)=\frac{k}{j^{2}}\left(\frac{k-1}{4}L(k-2)-(-1)^{j/2}\frac{\pi^{k-1}}{4^{k}}\right)$$

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Case 2: $j$ is odd

Following similar steps, we have $$f(t)=\frac{1}{4j^{2}+t^{2}}\left(t+2j(-1)^{(j-1)/2}e^{\pi t/4}\right)$$ Now, $$\left(4j^{2}+t^{2}\right)f(t)=t+2j(-1)^{(j-1)/2}\sum\limits_{k=0}^{\infty}\frac{\pi^{k}}{4^{k}k!}t^{k}$$ but simultaneously $$\left(4j^{2}+t^{2}\right)f(t)=\sum\limits_{k=0}^{\infty}\frac{L(k)}{k!}\left(4j^{2}t^{k}+t^{k+2}\right)$$ And thus by comparing the coefficients of $t^{k}$ is the above two expressions, we obtain $$2j(-1)^{(j-1)/2}\frac{\pi^{k}}{4^{k}k!}=\frac{4j^{2}}{k!}L(k)+\frac{1}{(k-2)!}L(k-2)$$ $$L(k)=\frac{1}{2j}(-1)^{(j-1)/2}\left(\frac{\pi}{4}\right)^{k}-\frac{1}{4j^{2}}k(k-1)L(k-2)$$ However, I struggle to find an expression for $L(0)$ in this scenario. All I know is $f(0)=\frac{1}{2j}(-1)^{(j-1)/2}$ which does not seem to tell me anything.

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And so $$\int\limits_{0}^{\pi/4}x^{k}\ln\sin x\, dx=-\ln 2\frac{(\pi/4)^{k+1}}{k+1}-\sum\limits_{j=1}^{\infty}\frac{1}{j}L(k)$$ where $L(k)$ is defined by the recurrence relation $$L(k)=\left\{\begin{matrix}\frac{k}{j^{3}}\left(\frac{k-1}{4}L(k-2)-(-1)^{j/2}\frac{\pi^{k-1}}{4^{k}}\right)\quad\text{, }j\text{ even}\\\frac{1}{2j^{2}}(-1)^{(j-1)/2}\left(\frac{\pi}{4}\right)^{k}-\frac{1}{4j^{3}}k(k-1)L(k-2)\quad\text{, }j\text{ odd}\end{matrix}\right.$$ with the starting condition $L(0)$ being $0$ for even $j$ but mysterious for odd $j$.

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Admittedly, this is insanely ugly. It is also missing many important pieces. I don't even know if I am on the right track to finding a closed form for the original $I(k)$ integral. If anyone has any insights, please share them below.

Answer 1

Integrate by parts \begin{align} &\int_{0}^{\pi/4}x^{k}\ln\tan x\, dx\\ =& \ \frac1{k+1} \int_{0}^{\pi/4}\ln\tan x\, d(x^{k+1}) =-\frac2{k+1} \int_{0}^{\pi/4}x^{k+1}\csc2x\, \overset{2x\to x}{dx}\\ = & \ -\frac1{2^{k+1}(k+1)} \int_{0}^{\pi/2}x^{k+1}\csc x\ dx= -\frac{I_{k+1}}{2^{k+1}(k+1)} \end{align} where $I_{n}= \int_{0}^{\pi/2}x^{n}\csc x\ dx $ is evaluated here.

Answer 2

Your integral is discussed in Theorem 2.5 in this paper, it has a closed form in terms of the Dirichlet Lambda function and the Dirichlet Beta function:

Theorem For $q\in \mathbb{N}\cup \left\{0\right\}$, $$\int_{0}^{\frac{\pi}{4}}\theta^q\ln(\tan\theta)d\theta = (-1)^{\frac{q(q-1)}{2}}(1-(-1)^q) \frac{q!}{2^{q+1}}\lambda(q+2)+\sum_{r=0}^{\left[\frac{q}{2}\right]} \frac{(-1)^{r+1}q!}{4^r(q-2r)!}\left(\frac{\pi}{4}\right)^{q-2r}\beta(2r+2) $$ where $[x]$ is the integer part of x.

Example:

$$ \int_{0}^{\frac{\pi}{4}} \theta^5\ln(\tan \theta) d\theta = \frac{1905}{512}\zeta(7)-\frac{1}{1024}\pi^5G+\frac{5\pi^3}{64}\beta(4) - \frac{15\pi}{8}\beta(6)$$