If $$I_1=\int_{0}^{\frac{\pi}{2}}\frac{(\ln(\tan x))^2}{1-\sin 2x}dx$$ and $$I_2=\int_{0}^{\frac{\pi}{2}}(\ln(1-\sin x))(\cot x)dx$$ then evaluate $$\left|\frac{I_1}{I_2}\right|$$
My Attempt
$$I_1=\int_{0}^{\frac{\pi}{2}}\frac{(\ln(\tan x))^2}{1-\sin 2x}dx=2\int_{0}^{\frac{\pi}{4}}\frac{(\ln(\tan x))^2}{1-\sin 2x}dx$$
Taking $t=\tan x$, the integral transforms to
$$I_1=2\int_{0}^{1}\frac{(\ln t)^2}{(1-t)^2}dt$$
Putting $t=1-\sin x$ in $I_2$ we have $$I_2=\int_{0}^{1}\frac{\ln t}{1-t}dt$$
What to do after this
Best Answer
Integrating by parts: $$\int \frac{\log^2 t}{(1-t)^2} dt = \left[\frac{\log^2 t}{1-t} \right]-2\int \frac{\log t}{t(1-t)} dt $$ $$ = \frac{\log^2 t}{1-t} -2\int \frac{\log t}{t} dt -2\int \frac{\log t}{1-t} dt $$ $$ = \frac{\log^2 t}{1-t} -\log^2 t -2\int \frac{\log t}{1-t} dt $$ $$ = \frac{t \log^2 t}{1-t} -2\int \frac{\log t}{1-t} dt $$ Therefore $$\int_0^1 \frac{\log^2 t}{(1-t)^2} dt = \lim_{t \to 1-} \frac{t \log^2 t}{1-t} - \lim_{t \to 0+} \frac{t \log^2 t}{1-t} -2\int_0^1 \frac{\log t}{1-t} dt \tag 1$$
For both of the limits, set $t=e^x$, then $$\lim_{t \to 1-} \frac{t \log^2 t}{1-t}=\lim_{x \to 0-} \frac{x^2 e^x}{1-e^x}=\lim_{x \to 0-} \frac{x^2 }{e^{-x}-1}=\lim_{x \to 0-} \frac{2x}{-e^{-x}}=0$$ $$\lim_{t \to 0+} \frac{t \log^2 t}{1-t}=\lim_{x \to -\infty} \frac{x^2 e^x}{1-e^x} =\lim_{x \to -\infty} x^2 e^x \lim_{x \to -\infty} \frac{1}{1-e^x} $$ The second limit is $1$ and the first is $$\lim_{x \to -\infty} x^2 e^x =\lim_{x \to -\infty} \frac{x^2}{e^{-x}}=\lim_{x \to -\infty} \frac{2x}{-e^{-x}}=\lim_{x \to -\infty} \frac{2}{e^{-x}}=0$$ and so both limits in (1) are zero and so
$$\int_0^1 \frac{\log^2 t}{(1-t)^2} dt = -2\int_0^1 \frac{\log t}{1-t} dt \tag 2$$