Closed form of $\int_0^{\pi/2} \frac{\arctan^2 (\sin^2 \theta)}{\sin^2 \theta}\,d\theta$

I'm trying to evaluate the closed form of:

$$I =\int_0^{\pi/2} \frac{\arctan^2 (\sin^2 \theta)}{\sin^2 \theta}\,d\theta$$

So far I've tried introducing a parameters,

$\displaystyle I(a,b) = \int_0^{\pi/2} \frac{\arctan (a\sin^2 \theta)\arctan (b\sin^2 \theta)}{\sin^2 \theta}\,d\theta$

but that doesn't lead to an integral I can manage. Expanding the series for $\arctan^2 x$ leads to the sum:

$$I = \frac{\pi}{2}\sum\limits_{n=0}^{\infty}\frac{(-1)^n}{(n+1)4^{2n+1}}\binom{4n+2}{2n+1}\left(\sum\limits_{k=0}^{n}\frac{1}{2k+1}\right)$$ and using $\displaystyle \int_0^1 x^{n-\frac{1}{2}}\log (1-x)\,dx = \frac{-2\log 2 + 2\sum\limits_{k=0}^{n}\dfrac{1}{2k+1}}{n+\frac{1}{2}}$ leads to an even uglier integral:

$\displaystyle \Im \int_{0}^{1} \frac{1}{1+\sqrt{1-i\sqrt{x}}}\frac{\log(1-x)}{x}\,dx$ among others.

I got the non-square version, which seems to have a nice closed form $\displaystyle \int_0^{\pi/2} \frac{\arctan (\sin^2 \theta)}{\sin^2 \theta}\,d\theta = \frac{\pi}{\sqrt{2}\sqrt{1+\sqrt{2}}}$ but the squared version seems difficult. Any help is appreciated.

(P.S. - I'm not familiar with Hypergeometric identities, so it would be very helpful if a proof or a reference to a proof was provided, should the need arise to use them.)

• I was able to reduce it to $$I=\pi\sqrt{2}\int_0^1\frac{\sqrt{1+\sqrt{2}}-\sqrt{1+\sqrt{1+\alpha^2}}}{1- \alpha ^2 }d\alpha$$ I think that this can be evaluated in terms of elementary functions but the calculations are too tedious to do by hand. – Shobhit Bhatnagar Apr 14 '16 at 3:24

1 Answer

Let $I$ denote the integral. Then, using integration by parts we can write $$I=\int_{0}^{\pi\over 2}\frac{\left(\tan^{-1}(\sin^2 x) \right)^2}{\sin^2 x}dx = 4\int_0^{\frac{\pi}{2}}\frac{\cos^2 x\tan^{-1}(\sin^2 x)}{1+\sin^4 x}dx$$

The main idea of this evaluation is to use differentiation under the integral sign. Let us introduce the parameter $\alpha$: $$f(\alpha)=\int_0^{\frac{\pi}{2}}\frac{\cos^2 x\tan^{-1}(\alpha \sin^2 x)}{1+\sin^4 x}dx$$ Taking derivative inside the integral, \begin{align*} f'(\alpha) &= \int_0^{\pi\over 2}\frac{\cos^2 x}{1+\sin^4 x}\cdot\frac{\sin^2 x}{1+\alpha^2 \sin^4 x}dx \\ &= \frac{1}{1-\alpha^2}\int_0^{\pi\over 2}\frac{\cos^2 x\sin^2 x}{1+\sin^4 x}dx-\frac{\alpha^2}{1-\alpha^2}\int_0^{\pi\over 2}\frac{\cos^2 x\sin^2 x}{1+\alpha^2 \sin^4 x}dx \quad (\text{Partial Fractions}) \end{align*}

Let $g(\alpha)=\int_0^{\pi\over 2}\frac{\cos^2 x\sin^2 x}{1+\alpha^2 \sin^4 x}dx$. Then,

$$\frac{I}{4}=f(1)=\int_0^1\frac{g(1)-\alpha^2 g(\alpha)}{1-\alpha^2}d\alpha \tag{1}$$

Evaluation of $g(\alpha)$

\begin{align*} g(\alpha) &= \int_0^{\frac{\pi}{2}}\frac{\cos^2 x\sin^2 x}{1+\alpha^2 \sin^4 x}dx \\ &= \int_0^\infty \frac{t^2}{\left(t^4(1+\alpha^2)+2t^2+1 \right)(t^2+1)}dt \quad (t=\tan x)\\ &= -\frac{1}{\alpha^2}\int_0^\infty\frac{1}{1+t^2}dt+\frac{1}{\alpha^2}\int_0^\infty\frac{1+(1+\alpha^2)t^2}{(1+\alpha^2)t^4+2t^2+1}dt \\ &= -\frac{\pi}{2\alpha^2}+\frac{\pi \sqrt{1+\sqrt{1+\alpha^2}}}{2\sqrt{2}\alpha^2}\tag{2} \end{align*} That last integral was evaluated using an application of the residue theorem. Substitute this into equation (1) to get $$I=\pi\sqrt{2}\int_0^1\frac{\sqrt{1+\sqrt{2}}-\sqrt{1+\sqrt{1+\alpha^2}}}{1- \alpha ^2 }d\alpha \tag{3}$$

Evaluation of integral (3)

Luckily, integral (3) has a nice elementary anti-derivative. \begin{align*} &\; \int \frac{\sqrt{1+\sqrt{2}}-\sqrt{1+\sqrt{1+\alpha^2}}}{1-\alpha^2}d\alpha \\ &= \sqrt{1+\sqrt{2}}\tanh^{-1}(\alpha) -\int \frac{\sqrt{1+\sqrt{1+\alpha^2}}}{1-\alpha^2} d\alpha \\ &= \sqrt{1+\sqrt{2}}\tanh^{-1}(\alpha)-\int \frac{t\sqrt{1+t}}{(2-t^2)\sqrt{t^2-1}}dt\quad (t=\sqrt{1+\alpha^2}) \\ &= \sqrt{1+\sqrt{2}}\tanh^{-1}(\alpha)-\int \frac{t}{(2-t^2)\sqrt{t-1}}dt \\ &=\sqrt{1+\sqrt{2}}\tanh^{-1}(\alpha)- 2\int \frac{u^2+1}{2-(u^2+1)^2}du\quad (u=\sqrt{t-1}) \\ &=\sqrt{1+\sqrt{2}}\tanh^{-1}(\alpha)- 2\int \frac{u^2+1}{(\sqrt{2}-1-u^2)(\sqrt{2}+1+u^2)}du \\ &= \sqrt{1+\sqrt{2}}\tanh^{-1}(\alpha)-\int\frac{du}{\sqrt{2}-1-u^2}+\int\frac{du}{\sqrt{2}+1+u^2} \\ &=\sqrt{1+\sqrt{2}}\tanh^{-1}(\alpha)-\sqrt{\sqrt{2}+1}\tanh^{-1}\left( u \sqrt{\sqrt{2}+1}\right)+\sqrt{\sqrt{2}-1}\tan^{-1}\left(u\sqrt{\sqrt{2}-1} \right) +C\\ &= \sqrt{1+\sqrt{2}}\tanh^{-1}(\alpha)-\sqrt{\sqrt{2}+1}\tanh^{-1}\left( \sqrt{\sqrt{1+\alpha^2}-1} \sqrt{\sqrt{2}+1}\right)\\ &\quad +\sqrt{\sqrt{2} -1}\tan^{-1}\left(\sqrt{\sqrt{1+\alpha^2}-1}\sqrt{\sqrt{2}-1} \right) +C \end{align*}

Therefore, the integral is \begin{align*} I &= \pi \sqrt{2} \lim_{\alpha\to 1}\Bigg\{ \sqrt{1+\sqrt{2}}\tanh^{-1}(\alpha)-\sqrt{\sqrt{2}+1}\tanh^{-1}\left( \sqrt{\sqrt{1+\alpha^2}-1} \sqrt{\sqrt{2}+1}\right)\\ &\quad +\sqrt{\sqrt{2} -1}\tan^{-1}\left(\sqrt{\sqrt{1+\alpha^2}-1}\sqrt{\sqrt{2}-1} \right) \Bigg\} \\ &= \pi\sqrt{2}\left\{\frac{\sqrt{\sqrt{2}+1}}{2}\log\left(\frac{\sqrt{2}+2}{4} \right) +\sqrt{\sqrt{2}-1}\tan^{-1}\left(\sqrt{2}-1 \right)\right\} \\ &= \pi\sqrt{2}\left\{\frac{\sqrt{\sqrt{2}+1}}{2}\log\left(\frac{\sqrt{2}+2}{4} \right) +\frac{\pi}{8}\sqrt{\sqrt{2}-1}\right\} \\ &= \color{maroon}{\pi \log\left( \frac{2+\sqrt{2}}{4}\right) \sqrt{\frac{\sqrt{2}+1}{2}}+\frac{\pi^2}{4}\sqrt{\frac{\sqrt{2}-1}{2}}}\approx 0.576335 \end{align*}

• Fantastic answer S! :-) Thanks!! – r9m Apr 15 '16 at 15:33
• Nice solution+1. – xpaul Sep 12 '16 at 14:41