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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

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

2 Answers 2

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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

$$\int_0^1 \frac{\log^2 t}{(1-t)^2} dt = -2\int_0^1 \frac{\log t}{1-t} dt \tag 2$$

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  • $\begingroup$ The limits can be easily evaluated by using standard limits, but still the answer deserves an upvote. +1 $\endgroup$
    – Paramanand Singh
    Commented May 6 at 10:07
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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 $$\boxed{\left|\frac{I_1}{I_2} \right|=4}$$

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