# $I_1=\int_{0}^{\frac{\pi}{2}}\frac{(\ln(\tan x))^2}{1-\sin 2x}dx$ & $I_2=\int_{0}^{\frac{\pi}{2}}(\ln(1-\sin x))(\cot x)dx$ then$|\frac{I_1}{I_2}|=?$

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

• DOes it help if you intgegrate I_1 by parts? Commented May 6 at 6:48
• The limits $0$ and $1$ are posing problems Commented May 6 at 7:14

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
$$\int_0^1 \frac{\log^2 t}{(1-t)^2} dt = -2\int_0^1 \frac{\log t}{1-t} dt \tag 2$$
Using the series $$(*)$$ for $$|t|<1$$, $$\frac{1}{1-t}=\sum_{n=0}^\infty t^n ,$$ we have \begin{aligned} I_2 & =\int_0^1 \frac{\ln t}{1-t} d t \\ & =\sum_{n=0}^{\infty} \int_0^1 t^n \ln t d t \\ & =-\sum_{n=1}^{\infty} \frac{1}{n^2} \end{aligned} For the integral $$I_1$$, we differentiate the series $$(*)$$ once and get $$\frac{1}{(1-t)^2}=\sum_{n=1}^{\infty} n t^{n-1}$$ and use integration by parts, $$I_1=2 \sum_{=1}^{\infty} n \int_0^1 t^{n-1} \ln ^2 t d t=2 \sum_{n=1}^{\infty} \frac{2}{n^2}=-4I_2$$ Hence $$\boxed{\left|\frac{I_1}{I_2} \right|=4}$$