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

Question

$$I=\int_0^{\frac{\pi}{2}}\sin^2x\ln(\sin^2(\tan x))dx \hspace{15mm}(1)$$ 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?

How about http://mathworld.wolfram.com/IntegrationbyParts.html — lab bhattacharjee, Dec 28 '16 at 16:04
@labbhattacharjee: that still leaves us with the problem of computing $$J_1 = \int_{0}^{\pi/2}\arctan(x)\cot(x),dx,\qquad J_2 = \int_{0}^{\pi/2}\frac{x\cot x}{1+x^2},dx.$$ — Jack D'Aurizio, Dec 28 '16 at 16:14
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. — Jack D'Aurizio, Dec 28 '16 at 16:42
Anyway when you replace $\cot x$ with $m$ the boundaries of the integration range do not stay the same. — Jack D'Aurizio, Dec 28 '16 at 18:15

score 7 · Accepted Answer · edited Apr 13 '17 at 12:21

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

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

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

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

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

ADDENDUM

As required A proof using complex analysis for

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

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.

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

Could you please elaborate on how complex analysis helps us get that result? — Panglossian Oporopolist, Dec 29 '16 at 03:36
@Kugelblitz, no problem. Let me know if there is anything else that is not clear. — Zaid Alyafeai, Dec 29 '16 at 09:02
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). — Panglossian Oporopolist, Dec 29 '16 at 09:11

score 0 · Answer 2 · answered Mar 22 '24 at 20:49

After the substitution $\tan{x}\mapsto x$, we get \begin{align} \mathscr{R} =&\int^\infty_0\frac{x^2}{(1+x^2)^2}\ln(\sin^2{x})\ {\rm d}x\\ =&\Re\int^\infty_{-\infty}\frac{x^2\ln(1-e^{i2x})}{(1+x^2)^2}{\rm d}x-\ln{2}\underbrace{\int^\infty_{-\infty}\frac{x^2}{(1+x^2)^2}{\rm d}x}_{\frac{\pi}{2}} \end{align} Even though the function $\displaystyle f(z)=\frac{z^2\ln(1-e^{i2z})}{(1+z^2)^2}$ has infinitely many branch points at $z=n\pi$, once we close the contour along the upper half of $|R|$ and make semicircular bumps around the branch points, one may check (by letting $z=n\pi+ \epsilon e^{i\theta}$) that the contribution along the bumps vanishes. The integral along the big arc also tends to $0$. Hence \begin{align} \mathscr{R} =&2\pi i{\rm Res}(f,i)-\frac{\pi}{2}\ln{2}\\ \end{align} Using WolframAlpha to compute the residue and simplify terms, $$\mathscr{R}=\frac{\pi}{e^2-1}-\pi-\frac{\pi}{2}\ln{2}+\frac{\pi}{2}\ln(e^2-1)$$

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

2 Answers2