Integral $\int_{0}^{\pi/2} \arctan \left(2\tan^{2}\left(x\right)\right) \mathrm{d}x$

Question

The following integral may seem easy to evaluate ...

$$ \int_{0}^{\pi/2}\arctan \left(2\tan^{2}\left(x\right)\right) \mathrm{d}x = \pi\arctan\left(\frac{1}{2}\right) $$ Could you prove it ?.

Answer 1

My answer is different from that you gave. Let $$ I(a)=\int_0^{\frac{\pi}{2}}\arctan(a\tan^2x)dx. $$ Than $I(0)=0$ and \begin{eqnarray} I'(a)&=&\int_0^{\frac{\pi}{2}}\frac{\tan^2x}{1+a^2\tan^4x}dx\\ &=&\int_0^\infty\frac{u^2}{(1+u^2)(1+a^2u^4)}du\\ &=&\frac{1}{1+a^2}\int_0^\infty\left(-\frac{1}{1+u^2}+\frac{1+a^2u^2}{1+a^2u^4}\right)du\\ &=&\frac{1}{1+a^2}\left(-\int_0^\infty\frac{1}{1+u^2}du+\int_0^\infty\frac{1}{1+a^2u^4}du+\int_0^\infty\frac{a^2u^2}{1+a^2u^4}\right)du\\ &=&\frac{1}{1+a^2}\left(-\frac{\pi}{2}+\frac{\pi}{2\sqrt{2}\sqrt{a}}+\frac{\sqrt{a}\pi}{2\sqrt{2}}\right) \end{eqnarray} and hence $$ I=\int_0^2I'(a)da=\pi\arctan(1+\sqrt{2a})\Big]_0^2=-\frac{\pi^2}{4}+\pi\arctan 3=\frac{\pi}{2}\arctan\frac{4}{3}.$$

Answer 2

Here is a result avoiding differentiation with respect to a parameter.

Set$$ I(\alpha):= \int_{0}^{\Large\frac{\pi}{2}} \arctan \left(\frac{2\alpha \:\sin^2 x}{\alpha^2-1+\cos^2 x}\right)\: \mathrm{d}x, \quad \alpha>0. $$ Then $$ I(\alpha)= \pi \arctan \left(\frac{1}{2\alpha}\right) \quad ({\star}) $$

With $ \alpha:=1$, we get $$ \int_{0}^{\Large\frac{\pi}{2}} \arctan \left(2 \tan^2 x\right) \mathrm{d}x = \pi \arctan \left( \frac{1}{2} \right). $$ To obtain $({\star})$ use the standard evaluation extended to complex numbers

$$ \int_{0}^{\Large\frac{\pi}{2}} \log \left(1+ t \sin^2 x\right) \mathrm{d}x = \pi \log \left( \frac{1+\sqrt{1+t}}{2} \right) $$

and observe that $$ \arctan (z) = \frac{i}{2} \left(\log (1-i z)-\log (1+i z)\right), \quad\Re z \neq 0. $$

Answer 3

$$I = \int_0^\tfrac\pi2 \arctan\left(2 \tan^2x\right) \, dx \stackrel{2\tan^2x\to x}= \frac1{\sqrt2} \int_0^\infty \frac{\arctan x}{\sqrt x (2+x)}$$

This can be evaluated with the residue theorem using the keyhole contour discussed here. The integral of

$$f(z) = \frac{\arctan z}{\sqrt z (2+z)} = -\frac i2 \frac{\log\left\lvert\frac{i-z}{i+z}\right\rvert + i \arg\left(\frac{i-z}{i+z}\right)}{\sqrt{\lvert z\rvert}\,e^{\tfrac i2 \arg z} (2+z)}$$

along the overall contour $C$ breaks down and converges to

$$\begin{align*} \oint_C f(z) \, dz &= \left(1-e^{-i\pi}\right) \int_0^\infty f(x) \, dx \\ &\qquad + i\pi \int_1^\infty \left(\frac{e^{i\pi/4}}{\sqrt x (2-ix)} - \frac{e^{-i\pi/4}}{\sqrt x (2+ix)}\right) \, dx \\ &= 2 \int_0^\infty f(x) \, dx - \sqrt2\,\pi\int_1^\infty \frac{2+x}{\sqrt x\left(4+x^2\right)} \, dx \\ &= 2\sqrt2\,I - 2\sqrt2\,\pi \int_0^1 \frac{1+2x^2}{1+4x^4} \, dx & x\to\frac1{x^2} \end{align*}$$

The remaining integral is elementary. $f$ has a simple pole at $z=-2$ with residue $\dfrac i{\sqrt2} \arctan2$, so by the residue theorem and careful application of $\arctan$ identities we conclude

$$\begin{align*} I &= \pi \underbrace{\int_0^1 \frac{1+2x^2}{1+4x^4} \, dx}_{\tfrac\pi8 + \tfrac12 \arctan 3} - \frac\pi2 \arctan2 \\ &= \frac\pi2 \left(\frac\pi4 + \arctan\frac17\right) \\ &= \frac\pi2 \arctan \frac43 \\ &= \boxed{\pi \arctan\frac12} \end{align*}$$

Answer 4

\begin{align} &\int_{0}^{\pi/2}\tan^{-1}(2\tan^2x)dx\\ =&\ \frac12\int_{-\pi/2}^{\pi/2}[\tan^{-1}\left(\cot x+1\right)- \tan^{-1}\left(\cot x-1\right)]dx \\ =&\ \frac12\int_{-\pi/2}^{\pi/2}\int_{-1}^1 \frac{1}{1+(\cot x+ t)^2} dt\ dx = \int_{-1}^1 \frac{\pi}{t^2+4} dt= \pi\cot^{-1}2 \\ \end{align}