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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.

hululu
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Tolaso
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    The integral is indeed $0$ as verified by Wolfram alpha – Samrat Mukhopadhyay Feb 23 '15 at 06:37
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    Do you have any ideas on how to attack this monster? – Tolaso Feb 23 '15 at 13:07
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    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. – Samrat Mukhopadhyay Feb 23 '15 at 14:21
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    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. – Tolaso Feb 24 '15 at 08:08
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    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. – Calum Gilhooley Feb 25 '15 at 16:50
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    Wolfram Alpha also suggests that the two values which cancel out to $0$ may be $\pm \pi^2/96$. I meant to ask earlier, where did you come across this problem? – Calum Gilhooley Feb 25 '15 at 21:37
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    Well I peaked it somewhere , long time ago.. that I don't remember. What two values cancel out to $0$. – Tolaso Feb 25 '15 at 22:17
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    This may belong to a family of integrals related to something called "Ahmed's integral", about which a considerable literature seems to have developed since 2002. The "two values" I referred to are the integrals from $0$ to $\pi/6$ and $\pi/6$ to $\pi/4$. – Calum Gilhooley Feb 25 '15 at 22:52
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    MathWorld mentions the similar-looking $C(a) = \int_0^1\frac{\tan^{-1}(\sqrt{x^2 + a^2})}{\sqrt{x^2 + a^2}(x^2 + 1)},dx$, with a reference to a book by Bailey et al., Experimental Mathematics in Action (2007). Also likely to be relevant is a book by P. J. Nahin, Inside Interesting Integrals (2004). – Calum Gilhooley Feb 25 '15 at 23:26
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    Also relevant is H. S. M. Coxeter, A challenging definite integral, Am. Math. Mon. 95, No. 4 (April 1988), p.330, with the associated literature on "Coxeter integrals"; and this MathSE question, from 6 April 2014, which leads to a downloadable PDF by Sangchul Lee, Some properties on generalized Ahmed’s integral (1 December 2013). But I still haven't seen an actual answer to the question anywhere - possibly because my eyes are glazing over. – Calum Gilhooley Feb 25 '15 at 23:52
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    I have seen about Ahmed integral and a generalization of it.. by sos440.. which I really like.. but i do not think you can relate that. But a Coxeter would be a nice category to put it in. – Tolaso Feb 25 '15 at 23:54
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    User sos440 says (in a comment on the question I just referred to) that Ahmed's and Coxeter's integrals are related. – Calum Gilhooley Feb 26 '15 at 00:03
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    Sorry, I did not know that... – Tolaso Feb 26 '15 at 08:09
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    By the way, I was thinking of a double integral here... to kill that $\arctan $ using a double integral. – Tolaso Feb 26 '15 at 08:10
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    Bounty's going to end soon; have you tried contacting the integration masters @Tolaso ? – Panglossian Oporopolist Feb 26 '15 at 11:14
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    It looked to me that the tricky part is just to apply integration by parts, then write the resulting simplified arctangent as an integral, then switch the integration order. Voilà, les jeux sont faits. :D – Jack D'Aurizio Feb 26 '15 at 14:41
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    @JackD'Aurizio What does "Voila, les jeux sont faits" mean? I know it is French and I know what voila means since I use it my self but I cannot understand the other.. – Tolaso Feb 26 '15 at 16:33
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    @Tolaso: it is the phrase that the croupier says when the roulette bets are closed. More or less, it means "game is over". – Jack D'Aurizio Feb 26 '15 at 16:36
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    @JackD'Aurizio Thank you. – Tolaso Feb 26 '15 at 16:45

2 Answers2

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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: $$ \int \frac{du}{(1+u^2)\sqrt{2+u^2}}=\arctan\frac{u}{\sqrt{2+u^2}}.$$

Jack D'Aurizio
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    Perfect... You are amazing... ! Thank you! – Tolaso Feb 26 '15 at 14:53
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    Damn that's awesome! – Panglossian Oporopolist Feb 26 '15 at 15:19
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    Could I give the bounty instead of you @Tolaso? Jack deserves it... – Panglossian Oporopolist Feb 26 '15 at 15:26
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    I just awarded the bounty to Jack.... He really does after this marvellous, astonishing answer. – Tolaso Feb 26 '15 at 16:32
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    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 $\int_0^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{1 - \tan^2\phi}{2}},d\phi = \frac{\pi^2}{24}.$ – Calum Gilhooley Feb 27 '15 at 02:14
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    @CalumGilhooley: right, hence we just need to prove the last identity through $(4)$ or similar arguments. – Jack D'Aurizio Feb 27 '15 at 10:58
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    In a new question, I've expanded some details of Jack's argument (these would be obvious to an expert in such calculations, but I had to struggle with them), and treated the parallel case where the upper limit of integration is $t = \frac{1}{\sqrt{3}}$, which leads to a connection with Ahmed's integral, and proves that the value of the present integral is a sum of values $\pm\frac{\pi^2}{96}$, which are the integrals from $0$ to $\frac{\pi}{6}$ and $\frac{\pi}{6}$ to $\frac{\pi}{4}$ respectively. – Calum Gilhooley Feb 28 '15 at 19:36
  • but what's $u$? – WuKong Apr 23 '19 at 15:22
2

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

$$ I = \int_{0}^{1}\frac{1}{1+t^2}\,\arctan\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)\,dt =\frac{\pi^2}{8}-\int_{0}^{1}\frac{3\sqrt{2}\, t \arctan(t)}{(3-t^2)\sqrt{1-t^2}}\,dt.$$

Evaluation of $\int_{0}^{1}\frac{3\sqrt{2}\, t \arctan(t)}{(3-t^2)\sqrt{1-t^2}}\,dt.$

$\begin{aligned} & \Omega \stackrel{t=\tan x}{\leftrightharpoons} 3 \sqrt{2} \int_0^1 \int_0^{\infty} \frac{1}{\left(1+2 t^2\right)\left(2+y^2+2 t^2\right)} d t d y=3 \sqrt{2} \int_0^1 \frac{1}{1+y^2} \int_0^{\infty}\left(\frac{1}{1+2 t^2}-\frac{1}{2+y^2+2 t^2}\right) d t d y \\ & \Omega=3 \sqrt{2} \int_0^1 \frac{1}{1+y^2}\left[\frac{1}{\sqrt{2}} \tan ^{-1}(t \sqrt{2})-\frac{1}{\sqrt{2} \sqrt{2+y^2}} \tan ^{-1}\left(\frac{t \sqrt{2}}{\sqrt{2+y^2}}\right)\right]_0^{\infty} d y \\ & \Omega=\frac{3 \pi}{2} \int_0^1 \frac{d y}{1+y^2}-\frac{3 \pi}{2} \int_0^1 \frac{d y}{\left(1+y^2\right) \sqrt{2+y^2}} \stackrel{y=\sqrt{2} \tan \theta}{\leftrightharpoons} \frac{3 \pi^2}{8}-\frac{3 \pi}{2} \int_0^{\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)} \frac{\cos \theta}{1+\sin ^2 \theta} d \theta \\ & \Omega=\frac{3 \pi^2}{8}-\frac{3 \pi}{2}\left[\tan ^{-1}(\sin \theta)\right]_0^{\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right)}=\frac{3 \pi^2}{8}-\frac{3 \pi}{2} \tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{3 \pi^2}{8}-\frac{\pi^2}{4} \\ & \int_0^1 \frac{3 \sqrt{2} t \arctan t}{\left(3-t^2\right) \sqrt{1-t^2}} d t=\frac{\pi^2}{8} \quad \\ & \end{aligned}$

And we are done

Martin.s
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