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This question was posted on I&S

Prove the following

$$\int_0^\frac{\pi}{2} \cot^{-1}{\sqrt{1+\csc{\theta}}}\,\mathrm d\theta =\frac{\pi^2}{12}$$

The numerical value of the integral seems to agree with the answer.

Maybe someone could use that

$$1+\csc \theta = \csc(\theta) (\cos(\theta/2) + \sin(\theta/2))^2$$

I am sure there is a smart substitution or some trigonometric properties that I fail to see.

Zaid Alyafeai
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3 Answers3

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The first integral is $$ I_1=\frac{\pi^2}{4}-\int_{0}^{\pi/2}\arctan\sqrt{\frac{1+\sin t}{\sin t}}\,dt=\frac{\pi^2}{4}-2\int_{0}^{\pi/4}\arctan\sqrt{\frac{1+\cos(2t)}{\cos (2t)}}\,dt$$ and $$\int_{0}^{\pi/4}\arctan\sqrt{\frac{1+\cos(2t)}{\cos (2t)}}\,dt=\int_{0}^{1}\arctan\sqrt{\frac{2}{1-u^2}}\frac{du}{1+u^2}$$ is a variant of Ahmed's integral that can be tackled through differentiation under the integral sign: it is enough to be able to integrate $\frac{\sqrt{1-u^2}}{(1+a-u^2)(1+u^2)}$.

Credit goes to Jack D'Aurizio

Jack Tiger Lam
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    $\displaystyle \int_0^1\arctan\left(\sqrt{\dfrac{2}{1-x^2}}\right)\cdot \dfrac{1}{1+x^2}dx=\int_0^{\infty}\dfrac{\arctan\left(\sqrt{2+x^2}\right)}{(1+x^2)\sqrt{2+x^2}}dx$ – FDP Apr 25 '17 at 17:14
  • @FDP, could you provide more information regarding how you derived that relationship? – a-cyclohexane-molecule Jul 21 '17 at 23:29
  • Consider the change of variable $y=\sqrt{\frac{x^2}{1-x^2}}$ – FDP Jul 29 '17 at 00:49
  • It's natural to try $\sqrt{\frac{2}{1-x^2}}=\sqrt{2+y^2}$. this leads to the change of variable mentioned above. – FDP Jul 29 '17 at 00:55
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\begin{align} &\int_0^{{\pi}/{2}} \cot^{-1}{\sqrt{1+\csc{\theta}}} \ {d\theta}\>\>\>\>\>\>\>(\theta \to \frac\pi2-\theta)\\ = &\int_0^{\pi/2} \cot^{-1}{{\sqrt{\frac2{1-\tan^2\frac {\theta}2}}}}{d\theta}\>\>\>\>\>(\tan\frac {\theta}2=\sin t)\\ =& \int_0^{\pi/2}\frac{2\cos t}{2-\cos^2 t} \tan^{-1}\frac{\cos t}{\sqrt2}\ dt\\ =& \int_0^{\pi/2}\int_0^{\pi/4}\frac{2\sqrt2\cos^2 t \sec^2 y}{(2-\cos^2 t)(2+\tan^2y\cos^2t)}dy\ dt \\ =& \ \pi \int_0^{\pi/4} \bigg( 1-\frac{\cos y}{\sqrt{2-\sin^2y}}\bigg)dy=\frac{\pi^2}{12} \end{align}

Quanto
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Expand the integrand with the identity $\operatorname{arccot}x=\dfrac\pi2-\arctan x$, then substitute $t=\sqrt{1+\csc\theta}$. The resulting integral is evaluated here using the residue theorem.

$$\begin{align*} I &= \int_0^\tfrac\pi2 \operatorname{arccot} \sqrt{1+\csc\theta} \, d\theta \\ &= \frac{\pi^2}4 - \int_0^\tfrac\pi2 \arctan \sqrt{1+\csc\theta} \, d\theta \\ &= \frac{\pi^2}4 - 2 \int_{\sqrt2}^\infty \frac{\arctan t}{\left(t^2-1\right) \sqrt{t^2-2}} \, dt \\ &= \frac{\pi^2}4 - \frac{\pi^2}{6} = \boxed{\frac{\pi^2}{12}} \end{align*}$$

user170231
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