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Inspired by the question in the post, I started to generalise the integral $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{\ln^n (\sin x) \ln (\cos x)}{\tan x} d x =& \frac{1}{n+1} \int_0^{\frac{\pi}{2}} \ln (\cos x) d\left(\ln ^{n+1}(\sin x)\right) \\ =& \frac{1}{n+1}\left[\ln (\cos x) \ln ^{n+1}(\sin x)\right]_0^{\frac{\pi}{2}}+\frac{1}{n+1} \int_0^{\frac{\pi}{2}} \frac{\sin x \ln ^{n+1}\left(\sin x\right)}{\cos x} d x\\\stackrel{IBP}{=} &\frac{1}{2^{n+2}(n+1)} \int_0^{\frac{\pi}{2}} \frac{\ln ^{n+1}\left(\sin ^2 x\right)}{\cos ^2 x} d\left(\sin ^2 x\right)\\=& \frac{1}{2^{n+2}(n+1)} \int_0^{1} \frac{\ln ^{n+1} y}{1-y} d y \quad \textrm{ where }y=\sin^2 x\end{aligned} $$ We then deal the last integral using an infinite series on the integral $$ \begin{aligned} \int_0^1 \frac{y^a}{1-y} d y=\sum_{k=0}^{\infty} \int_0^1 y ^a\cdot y^k d y =\sum_{k=1}^{\infty} \frac{1}{a+k} \end{aligned} $$ Differentiating both sides w.r.t. $a$ by $n$ times yields $$ \int_0^1 \frac{\ln^{n+1} y}{1-y} d y=\left.\frac{\partial^{n+1}}{\partial a^{n+1}} \int_0^1 \frac{y^a}{1-y} d t\right|_{a=0} =(-1)^{n+1}(n+1) ! \sum_{k=1}^{\infty} \frac{1}{k^{n+2}}= (-1)^{n+1}(n+1) !\zeta(n+2) $$

Then we can conclude that $$\boxed{\int_0^{\frac{\pi}{2}} \frac{\ln^n (\sin x) \ln (\cos x)}{\tan x} d x = \frac{(-1)^{n+1} n !}{2^{n+2}} \zeta(n+2)} $$

For examples, $$ \begin{aligned} &I_1=\frac{1}{8} \zeta(3);\quad I_2=-\frac{1}{16} \cdot 2 \cdot \zeta(4)=-\frac{\pi^4}{720} ;\quad I_{12}=-\frac{12 !}{2^{14} }\zeta(12)=-\frac{691 \pi^{12}}{21840} \end{aligned} $$

Similarly, I want to generalise the integral further as

$$I(n,m)=\int_0^{\frac{\pi}{2}} \frac{\ln ^n(\sin x) \ln ^m(\cos x)}{\tan x} d x =\frac{m}{2^{n+m}(n+1)} \int_0^1 \frac{\ln ^{n+1} y \ln ^{m-1}(1-y)}{1-y} d y $$

Noticing that

$$ \begin{aligned} \int_0^1 y^a(1-y)^bd y =& \int_0^1 y^a(1-y)^{b} d y =B(a+1, b+1) \end{aligned} $$ Therefore

$$ \boxed{I(n, m)= \frac{1}{2^{n+2}(n+1)} \left.\frac{\partial^{n+1}}{\partial a^{n+1}} \frac{\partial^{m-1}}{\partial b^{m-1}}\left(\frac{\Gamma(a+1) \Gamma(b+1)}{\Gamma(a+b+2)}\right)\right|_{a=0,b=-1}} $$ My question: Is there a closed form for the last high derivative?

Your comments and suggestions are highly appreciated.

Lai
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  • You may be able to develop a general form for the partial derivative starting with e.g. https://functions.wolfram.com/GammaBetaErf/Beta/20/ShowAll.html then once you've found that, take the limit as $a\to 0$ and $b\to 0$. – pshmath0 Sep 20 '22 at 08:54
  • Thank you, but is rather complicated. – Lai Sep 20 '22 at 09:35

1 Answers1

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After submitting the post, I suddenly recognise that $I(n,m) $ is actually a high derivative of a Beta function as $$ \begin{aligned} \int_0^{\frac{\pi}{2}} \frac{\sin ^a x \cos ^b x}{\tan x} d x &=\int_0^{\frac{\pi}{2}} \sin ^{a-1} x \cos ^{b+1} x d x =\frac{1}{2} B\left(\frac{a}{2} , \frac{b}{2} +1\right) \end{aligned} $$ Then we can write down $$ \begin{aligned}\int_0^{\frac{\pi}{2}} \frac{\ln ^n(\sin x) \ln ^m(\cos x)}{\tan x} d x & =\left.\frac{1}{2} \frac{\partial^{n+m}}{\partial a^n \partial b ^m}\left(B\left(\frac{a}{2} , \frac{b}{2} +1\right)\right)\right|_{a,b \rightarrow 0}\\&=\frac{1}{2^{n+m+1}} \lim _{a \rightarrow 0} \lim _{b \rightarrow 1}\left[\frac{\partial^{n+m}}{\partial a^n\partial b^m} B(a, b)\right]\end{aligned} $$

Lai
  • 20,421