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Given that:

$$I=\int_{0}^{\pi/2}\arctan\left(\sqrt{\sin(2x)\over \sin^2(x)}\right)\mathrm dx$$

and

$$J=\int_{0}^{\pi/2}\arctan\left(\sqrt{\sin(2x)\over \cos^2(x)}\right)\mathrm dx$$

Q1: How do we evaluate the closed for $I$?

Q2: Show that $I=J$.

Recall $$\arctan(x)+\arctan(y)=\arctan\left({x+y\over 1-xy}\right)$$

$$I+J=\int_{0}^{\pi/2}\arctan\left(\sqrt{2\cot x}+\sqrt{2\tan x}\right)dx$$

I am not sure what to do next...

Later on we notice that this integral has the same closed form given by @Jack D'aurizio

$$\int_{0}^{\pi/2}\arctan(2\tan^2x)\mathrm dx=\pi\arctan\left({1\over 2}\right)$$

2 Answers2

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$$\begin{eqnarray*}I=\int_{0}^{\pi/2}\arctan\sqrt{2\cot x}\,dx &\stackrel{x\mapsto\frac{\pi}{2}-x}{=}&\int_{0}^{\pi/2}\arctan\sqrt{2\tan x}\,dx\tag{1}\\&\stackrel{x\mapsto\arctan u}{=}&\int_{0}^{+\infty}\frac{\arctan\sqrt{2u}}{1+u^2}\,du\\&\stackrel{u\mapsto v^2/2}{=}&\int_{0}^{+\infty}\frac{v\arctan v}{1+\frac{1}{4}v^4}\,dv\end{eqnarray*}$$ and this can be easily solved through differentiation under the integral sign, by computing $$\int_{0}^{+\infty}\frac{v^2\,dv}{(1+a^2 v^2)\left(1+\frac{1}{4}v^4\right)} =\frac{\left(2^{1/4}+a \left(-2+2^{3/4} a\right)\right) \pi }{2+4 a^4}\tag{2}$$ then applying $\int_{0}^{1}\ldots\,da$ to such expression. The final outcome is: $$ I = \color{red}{\frac{\pi}{2}\arctan\frac{4}{3}}=\pi\arctan\frac{1}{2}.\tag{3} $$ $(1)$ also proves $I=J$.

Jack D'Aurizio
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  • What made you think to use DUIS here? Mere intuition and practice, or is there some general trick? – Brevan Ellefsen May 30 '17 at 16:30
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    @BrevanEllefsen: when dealing with the integral of a product between a logarithm/arctangent and a rational function, that is always the case to think to Feynman's trick. – Jack D'Aurizio May 30 '17 at 16:32
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See this formula for $J(a,b,c)$ derived by contour integration. Jack's $v$-integral is precisely

$$I = \int_0^\infty \frac{v \arctan v}{\frac14 v^4 + 1} \, dv = J \left(\frac14, 0, 1\right) = \boxed{\frac\pi2\arctan\frac43}$$

To verify, let's first clean up the formula and reduce it to the a special case below (using the principal square root).

$$\begin{align*} J\left(a,0,1\right) &= \frac{i\pi}{2\sqrt{-4a}} \arctan \frac{\frac{\sqrt{-\sqrt{-4a}}+\sqrt{\sqrt{-4a}}}{\sqrt2\left(\sqrt a-1\right)} - \frac{\sqrt{4a}}{2a}}{1 + \frac{\sqrt{-\sqrt{-4a}}+\sqrt{\sqrt{-4a}}}{\sqrt2\left(\sqrt a-1\right)}\cdot\frac{\sqrt{4a}}{2a}} \\ &= \frac{\pi}{4\sqrt{a}} \arctan \frac{\frac{\sqrt2\,a^{1/4}}{\sqrt a-1} - \frac1{\sqrt a}}{1 + \frac{\sqrt2\,a^{-1/4}}{\sqrt a-1}} \end{align*}$$

We get the expected result upon plugging in $a=\dfrac14$.

user170231
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