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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

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

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    $\begingroup$ Use $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ $\endgroup$ Commented May 30, 2017 at 11:29
  • $\begingroup$ It seems it can be computed only numerically. $\endgroup$
    – Jon
    Commented May 30, 2017 at 12:21
  • $\begingroup$ The substitution $u = \pi/2-x$ shows that $I=J. \qquad$ $\endgroup$ Commented May 30, 2017 at 13:03
  • $\begingroup$ @Jon: that's not true, $\pi\arctan\frac{1}{2}$ is a nice closed form. $\endgroup$ Commented May 30, 2017 at 13:38
  • $\begingroup$ @JackD'Aurizio I was cheated by cas tools. Of course, I upvoted your fine solution. $\endgroup$
    – Jon
    Commented May 30, 2017 at 14:21

3 Answers 3

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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 $I=J$.

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  • $\begingroup$ What made you think to use DUIS here? Mere intuition and practice, or is there some general trick? $\endgroup$ Commented May 30, 2017 at 16:30
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    $\begingroup$ @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. $\endgroup$ Commented May 30, 2017 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 $a=\dfrac14$.

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An alternative solution using residues:

$$\begin{align*} &\int_0^\tfrac\pi2 \arctan \sqrt{\frac{\sin(2x)}{\sin^2x}} \, dx \\ &= \left\{\int_{-\tfrac\pi4}^0+\int_0^\tfrac\pi4\right\} \arctan \sqrt{\frac{2\cos(2x)}{(\cos x+\sin x)^2}} \, dx & \left\{\begin{matrix}x\to x+\frac\pi4\\\text{split at }x=0\end{matrix}\right. \\ &= \int_0^\tfrac\pi4 \left(\arctan \sqrt{\frac{2\cos(2x)}{(\cos x-\sin x)^2}} + \arctan \sqrt{\frac{2\cos(2x)}{(\cos x+\sin x)^2}}\right) \, dx & x\to-x \\ &= \int_0^\tfrac\pi4 \left(\pi - \arctan\left[\sqrt{2 \frac{1+\tan x}{1-\tan x}} + \sqrt{2 \frac{1-\tan x}{1+\tan x}}\right]\right) \, dx & (*)\\ &= \frac{\pi^2}4 - \int_0^\tfrac\pi4 \arctan\left[\sqrt{2\cot x}+\sqrt{2\tan x}\right] \, dx & x\to\frac\pi4-x \\ &= \frac{\pi^2}4 - \int_0^1 \frac{2y}{1+y^4} \arctan\left[\sqrt2\left(\frac1y+y\right)\right] \, dy & x=\arctan y^2 \\ &= \frac{\pi^2}4 - 4 \int_{2\sqrt2}^\infty \frac{\arctan z}{\left(z^2-4\right) \sqrt{z^2-8}} \, dz & z=\sqrt2\left(\frac1y+y\right) \end{align*}$$

Simplification of $(*)$ follows from a special case of the $\arctan$ identity in the OP.

A flavor of the remaining integral is evaluated here.

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