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Evaluate the integral:

$$\displaystyle \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{\sqrt{2}\cos3 \phi}{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi$$

I have no clue on how to attack it. The only thing I noticed is that there exists a symmetry around $\pi/8$, meaning that from $\pi/8$ to $\pi/4$ is the negative of zero to $\pi/4$. But, there exists a root of the integrand at $\pi/6$ and the limit of the integrand at $\pi/4$ is $-\infty$.

Conjecture: The integral is $0$ for the reason of symmetry I mentioned above.

However I cannot prove that. I would appreciate your help.

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    $\begingroup$ The integral is indeed $0$ as verified by Wolfram alpha $\endgroup$ Feb 23, 2015 at 6:37
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    $\begingroup$ Do you have any ideas on how to attack this monster? $\endgroup$
    – Tolaso
    Feb 23, 2015 at 13:07
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    $\begingroup$ If you look at the Wolfram alpha result, there you'll find the graph of the function and from the looks of it I believe that the integral in the range $\pi/8$ to $\pi/4$ is just the same but of opposite sign of the one in $0$ to $\pi/8$. So if we can somehow show this, we are done. $\endgroup$ Feb 23, 2015 at 14:21
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    $\begingroup$ Well, if you look above to what I've written well this holds but we have some trouble because there exists a root and the limit at $\pi/4$ is $-\infty$. So, it's not that easy. $\endgroup$
    – Tolaso
    Feb 24, 2015 at 8:08
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    $\begingroup$ Changing the variable of integration to $y = 1 - 2\cos2\phi$ brings out the apparent symmetry more clearly, as in this Wolfram Alpha plot. But the symmetry (at least with this choice of variable) is only approximate. Weird! Basically I haven't a clue, either. $\endgroup$ Feb 25, 2015 at 16:50

1 Answer 1

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By replacing $\phi$ with $\arctan(t)$, then using integration by parts, we have:

$$ I = \int_{0}^{1}\frac{1}{1+t^2}\,\arctan\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)\,dt =\frac{\pi^2}{8}-\int_{0}^{1}\frac{3\sqrt{2}\, t \arctan(t)}{(3-t^2)\sqrt{1-t^2}}\,dt.$$ Now comes the magic. Since: $$\int \frac{3\sqrt{2}\,t}{(3-t^2)\sqrt{1-t^2}}\,dt = -3\arctan\sqrt{\frac{1-t^2}{2}}\tag{1}$$ integrating by parts once again we get:

$$ I = \frac{\pi^2}{8}-3\int_{0}^{1}\frac{1}{1+t^2}\arctan\sqrt{\frac{1-t^2}{2}}\,dt \tag{2}$$ hence we just need to prove that: $$ \int_{0}^{1}\frac{dt}{1+t^2}\,\arctan\sqrt{\frac{1-t^2}{2}}=\int_{0}^{\frac{1}{\sqrt{2}}}\frac{\arctan\sqrt{1-2t^2}}{1+t^2}\,dt=\color{red}{\frac{\pi^2}{24}}\tag{3}$$ and this is not difficult since both $$\int_{0}^{1}\frac{dt}{1+t^2}(1-t^2)^{\frac{2m+1}{2}},\qquad \int_{0}^{\frac{1}{\sqrt{2}}}\frac{(1-2t^2)^{\frac{2m+1}{2}}}{1+t^2}\,dt $$ can be computed through the residue theorem or other techniques. For instance: $$\int_{0}^{1}\frac{(1-t)^{\frac{2m+1}{2}}}{t^{\frac{1}{2}}(1+t)}\,dt = \sum_{n\geq 0}(-1)^n \int_{0}^{1}(1-t)^{\frac{2m+1}{2}} t^{n-\frac{1}{2}}\,dt=\sum_{n\geq 0}(-1)^n\frac{\Gamma\left(m+\frac{3}{2}\right)\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(m+n+2)}$$ or just: $$\int_{0}^{1}\frac{\sqrt{\frac{1-t^2}{2}}}{(1+t^2)\left(1+\frac{1-t^2}{2}u^2\right)}\,dt = \frac{\pi}{2(1+u^2)}\left(1-\frac{1}{\sqrt{2+u^2}}\right)\tag{4}$$ from which: $$\int_{0}^{1}\frac{dt}{1+t^2}\,\arctan\sqrt{\frac{1-t^2}{2}}=\frac{\pi}{2}\int_{0}^{1}\frac{du}{1+u^2}\left(1-\frac{1}{\sqrt{2+u^2}}\right) =\color{red}{\frac{\pi^2}{24}} $$ as wanted, since: $$ \int \frac{du}{(1+u^2)\sqrt{2+u^2}}=\arctan\frac{u}{\sqrt{2+u^2}}.$$

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    $\begingroup$ Perfect... You are amazing... ! Thank you! $\endgroup$
    – Tolaso
    Feb 26, 2015 at 14:53
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    $\begingroup$ Damn that's awesome! $\endgroup$ Feb 26, 2015 at 15:19
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    $\begingroup$ Could I give the bounty instead of you @Tolaso? Jack deserves it... $\endgroup$ Feb 26, 2015 at 15:26
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    $\begingroup$ I just awarded the bounty to Jack.... He really does after this marvellous, astonishing answer. $\endgroup$
    – Tolaso
    Feb 26, 2015 at 16:32
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    $\begingroup$ Rather than integrating by parts twice (in opposite directions, so to speak), at the start of this proof, one could show by basic trigonometry that $\tan^{-1}\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right) = 3\tan^{-1}\sqrt{\frac{1-t^2}{2}} - \frac{\pi}{2}.$ Similarly rewriting the original equation to be proved transforms it directly to $\int_0^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{1 - \tan^2\phi}{2}}\,d\phi = \frac{\pi^2}{24}.$ $\endgroup$ Feb 27, 2015 at 2:14

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