I need some help in evaluating $I$, which is the integral:
$$I = \int_0^{\frac{\pi}{4}} \frac{\ln^3(\cos x) + \ln^2(\cos x)\tan x – \ln(\cos x)\cot^2 x}{\sin^3 x}\,dx$$
Clarifications:
- For better readability, the expression $\ln^3(\cos x)$ represents $(\ln(\cos x))^3$.
- The term $\ln^2(\cos x)\tan x$ denotes $(\ln(\cos x))^2 \times \tan x$.
- If any other interpretations are possible, please advise.
First Steps:
\begin{align*}
I &= \int_0^{\frac{\pi}{4}} \frac{\ln^3(\cos x) + \ln^2(\cos x)\tan x – \ln(\cos x)\cot^2 x}{\sin^3 x} \, dx \\
&= \int_0^{\frac{\pi}{4}} \frac{\ln^3(\cos x) + \ln^2(\cos x)\tan x – \ln(\cos x)\frac{\cos^2 x}{\sin^2 x}}{\sin^3 x} \, dx \\
&= \int_0^{\frac{\pi}{4}} \frac{\ln^3(\cos x) + \ln^2(\cos x)\tan x – \ln(\cos x)\csc^2 x}{\sin x} \, dx
\end{align*}
$$ = \int_0^{\frac{\pi}{4}} \frac{\ln^3(\cos x) + \ln^2(\cos x)\tan x – \ln(\cos x)\cot(x)\csc(x)}{\sin x} \, dx$$
Best Answer
First, look at what happens around $x=0$.
Around this value, a Taylor series of the integrand gives $$\text{integrand} =\frac{1}{2 x^3}-\frac{13 x}{240}+O\left(x^2\right)$$ Do you see the problem ?
For the antiderivatives, let $x=\cos^{-1}(t)$ to face $$-\int\frac{\log ^3(t)}{\left(1-t^2\right)^2}\,dt-\int\frac{\log ^2(t)}{t \left(1-t^2\right)^{3/2}}\,dt+\int\frac{t^2 \log (t)}{\left(1-t^2\right)^3}\,dt$$ which has an explicit solution (in particular in terms of generalized hypergeometric functions and polylogarithms).