Trying to $\int_{0}^{\pi/2}\cos(2x)\ln^2(\tan(x/2)) dx$

Question

I would like to evaluate this integral

$$\large \int_{0}^{\pi/2}\cos(2x)\ln^2(\tan(x/2))\mathrm dx$$

Choosing

$u=\frac{x}{2}$

$\mathrm dx=2\mathrm du$

$$2\int_{0}^{\pi/4}\cos(4u)\ln^2(\tan u)\mathrm du$$

Choosing $\cos(4u)=\sin^4(u)-6\cos^2(u)\sin^2(u)+\cos^4(4)$

transforming into

$$2\int_{0}^{\pi/4}\sec^2(u)\cdot \frac{[\tan^2(u)-2\tan(u)-1][\tan^2(u)+2\tan(u)-1]\ln^2(\tan(u))}{(1+\tan^2(u))^3}du$$

Choosing

$v=\tan(u)$

$\mathrm du=\frac{1}{\sec^2(u)}\mathrm dv$

$$2\int_{0}^{\infty}\frac{(v^2-2v-1)(v^2+2v-1)\ln^2(v)}{(1+v^2)^3}\mathrm dv$$

$$2\int_{0}^{\infty}\frac{\ln^2(v)}{1+v^2}dv-4\int_{0}^{\infty}\frac{\ln^2(v)}{(1+v^2)^2}dv+4\int_{0}^{\infty}\frac{\ln^2(v)}{(1+v^2)^3}dv$$

I was thinking of applying binomial series, but it can't be don't

how would I continue this?

Answer 1

Considering $$I=\int\cos (2 x) \log ^2\left(\tan \left(\frac{x}{2}\right)\right)\,dx$$ use integration by parts $$u=\log ^2\left(\tan \left(\frac{x}{2}\right)\right)\implies du=2 \csc (x) \log \left(\tan \left(\frac{x}{2}\right)\right)\,dx$$ $$dv=\cos(2x)\,dx \implies v=\frac{1}{2} \sin (2 x)$$ $$I=\frac{1}{2} \sin (2 x) \log ^2\left(\tan \left(\frac{x}{2}\right)\right)-2\int \cos (x) \log \left(\tan \left(\frac{x}{2}\right)\right)\,dx$$ One more integration by parts to get the antiderivative and a beautiful result for the integral.

Answer 2

Although an antiderivative of the integrand is easy to find viz. Claude Leibovici's method, the limits given present a challenge. The integral is $$[\frac{1}{2}\sin 2x\ln^2\tan\frac{x}{2}-2\sin x\ln\tan\frac{x}{2}+2x]^{\pi/2}_0.$$For $n>0$, $\lim_{x\to\infty}x^n e^{-x}=0$ so $\lim_{y\to 0^+}y\ln^n y=0$. Trigonometric small-angle approximations then reduce the integral to $\pi+\lim_{x\to 0}(-x\ln^2\frac{x}{2}+2x\ln\frac{x}{2})=\pi$.