$$I=\int_0^{\frac{\pi}{2}}\sin^2x\ln(\sin^2(\tan x))dx \hspace{15mm}(1)$$ Now, using definite integral property of $\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$ $$I=\int_0^{\frac{\pi}{2}}\cos^2x\ln(\sin^2(\cot x))dx\hspace{15mm}(2)$$ After $\tan x=t$ substitution in $(1)$ and $\cot x=m$ in $(2)$, to get: $$I=\int_0^{\frac{\pi}{2}}\frac{t^2}{(1+t^2)^2}\ln(\sin^2 t)dt=\int_0^{\frac{\pi}{2}}\frac{m^2}{(1+m^2)^2}\ln(\sin^2 m)dm$$ After variable change and addition, I get:

$$2I=\int_0^{\frac{\pi}{2}}\frac{x^2}{(1+x^2)^2}\ln(\sin^4 x)dx\implies \frac{I}{2}=\int_0^{\frac{\pi}{2}}\frac{x^2}{(1+x^2)^2}\ln(\sin x)dx$$

How could I proceed? Any other solutions which happen to be more efficient/simple?

  • $\begingroup$ How about mathworld.wolfram.com/IntegrationbyParts.html $\endgroup$ – lab bhattacharjee Dec 28 '16 at 16:04
  • $\begingroup$ @labbhattacharjee: that still leaves us with the problem of computing $$J_1 = \int_{0}^{\pi/2}\arctan(x)\cot(x)\,dx,\qquad J_2 = \int_{0}^{\pi/2}\frac{x\cot x}{1+x^2}\,dx.$$ $\endgroup$ – Jack D'Aurizio Dec 28 '16 at 16:14
  • $\begingroup$ That is not big issue, since $$\int_{0}^{\pi/2}x^{2n+1}\cot(x)\,dx $$ is always a combination of $\log(2)$, powers of $\pi$ and values of the Riemann $\zeta$ function at positive odd integers. $\endgroup$ – Jack D'Aurizio Dec 28 '16 at 16:42
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    $\begingroup$ Anyway when you replace $\cot x$ with $m$ the boundaries of the integration range do not stay the same. $\endgroup$ – Jack D'Aurizio Dec 28 '16 at 18:15
  • $\begingroup$ Noted! @JackD'Aurizio $\endgroup$ – Kugelblitz Dec 29 '16 at 3:11

$$I=\int_0^{\frac{\pi}{2}}\cos^2x\ln(\sin^2(\tan x))dx$$

Use the substitution $t = \tan(x)$

$$I=2\int_0^{\infty}\frac{\ln|\sin t|}{(1+t^2)^2}dt $$

Then use the fourier expansion of $\log|\sin x|$

$$I=-2\sum\frac{1}{k}\int_0^{\infty}\frac{\cos(2kt)}{(1+t^2)^2}dt-2\int_0^{\infty}\frac{\log(2)}{(1+t^2)^2}dt $$

For the second integral


We can use the beta function to deduce that

$$2\int_0^{\infty}\frac{\log(2)}{(1+t^2)^2} =\frac{\pi}{2}\log(2) $$

For the second integral

$$\int_0^{\infty}\frac{\cos(2kt)}{(1+t^2)^2}dt = Re\int_0^{\infty}\frac{e^{2kit}}{(1+t^2)^2}dt$$

It can be showing by complex analysis that

$$\int_0^{\infty}\frac{\cos(2kt)}{(1+t^2)^2}dt = \frac{\pi}{4}e^{-2k}(1+2k) $$

Finally we have the integral

$$I=-\frac{\pi}{2}\sum \frac{e^{-2k}(1+2k)}{k } -\frac{\pi}{2}\log(2)$$

The sum can be computed using

$$-\log(1-x) = \sum \frac{x^k}{k}$$

Hence we deduce that

$$I=-\frac{\pi}{2}\left\{\log\left(\frac{2}{e^2-1} \right) +\frac{2e^ 2}{e^2-1}\right\}$$


As required A proof using complex analysis for


This can be done by considering

$$F(z) = \frac{e^{2ki z}}{(1+z^2)^2}$$

The poles are $z=\pm i$, then consider half a circle in the upper half plane.

enter image description here

Note there is only one pole of order 2 inside the contour. Hence by the residue theorem $$\int_C \frac{e^{2ki z}}{(1+z^2)^2}+\int_{-\infty}^0\frac{e^{2ki x}}{(1+x^2)^2} \,dx+ \int^{\infty}_0\frac{e^{2ki x}}{(1+x^2)^2}\,dx = 2\pi i \text{Res} \frac{e^{2ki z}}{(1+z^2)^2}|_{z = i} $$

The integral along the arc converges to 0

$$2\int^{\infty}_0\frac{\cos(2k x)}{(1+x^2)^2} \,dx = 2\pi i \text{Res} \frac{e^{2ki z}}{(1+z^2)^2}|_{z = i} $$

The pole of order 2 can be calculated using

$$\lim_{z \to i} \frac{d}{dz} (z-i)^2 \frac{e^{2ki z}}{(1+z^2)^2}$$

This simplifies to

$$2\int^{\infty}_0\frac{\cos(2k x)}{(1+x^2)^2} \,dx = \frac{\pi}{2}e^{-2k}(1+2k)$$

$$\int^{\infty}_0\frac{\cos(2k x)}{(1+x^2)^2} \,dx = \frac{\pi}{4}e^{-2k}(1+2k)$$

For many other approaches see Calculating the integral $\int_{0}^{\infty} \frac{\cos x}{1+x^2}\mathrm{d}x$ without using complex analysis

  • $\begingroup$ Could you please elaborate on how complex analysis helps us get that result? $\endgroup$ – Kugelblitz Dec 29 '16 at 3:36
  • $\begingroup$ @Kugelblitz, I edited my post. $\endgroup$ – Zaid Alyafeai Dec 29 '16 at 4:39
  • $\begingroup$ You are awesome! Thank you! $\endgroup$ – Kugelblitz Dec 29 '16 at 9:00
  • $\begingroup$ @Kugelblitz, no problem. Let me know if there is anything else that is not clear. $\endgroup$ – Zaid Alyafeai Dec 29 '16 at 9:02
  • $\begingroup$ No issues. I recently started complex analysis so I just wanted a proper contour integration solution if possible (which you've provided in the addendum). $\endgroup$ – Kugelblitz Dec 29 '16 at 9:11

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