TITLE: A definite integral related to Ahmed's integral
QUESTION [8 upvotes]: Further work on a recent question which has just been answered leads me to make this conjecture:
$$
\int_0^1 \frac{\tan^{-1}\left(\frac{1}{\sqrt{2 + x^2}}\right)}
{(1 + x^2)\sqrt{2 + x^2}}\,du = \frac{\pi^2}{32}.
$$
I haven't made any effort to prove it (and I wouldn't even know how to begin), but I'll explain the connection with the other problem (later, by editing this question), if anyone expresses an interest.
Hasty update:
I should have looked again at more of the references I looked at yesterday! Google Books shows pages 190 to 193 (your mileage may vary) of the book by Paul J. Nahin, Inside Interesting Integrals, in which this very integral occurs in the course of an evaluation of Ahmed's integral. (Indeed, with hindsight, I have to admit that the connection is "obvious", especially because it follows from an elementary trigonometric identity I already used yesterday.)
What is the etiquette for a situation like this?  Should I simply delete this question, or should I explain the details of how this integral is related to Ahmed's integral, and to the other MathSE question I referred to? (Please excuse a blundering old newbie!)

REPLY [7 votes]: Ahmed's integral is:
$$
A = \int_0^1 \frac{\tan^{-1}\sqrt{2 + x^2}}
{(1 + x^2)\sqrt{2 + x^2}}\,dx.
$$
If we denote the unknown integral above by $B$, then:
\begin{gather*}
A + B =
 \int_0^1 \frac{\tan^{-1}\sqrt{2 + x^2}
    + \tan^{-1}\left(\frac{1}{\sqrt{2 + x^2}}\right)}
{(1 + x^2)\sqrt{2 + x^2}}\,dx  \\
 = \frac{\pi}{2}\int_0^1 \frac{dx}{(1 + x^2)\sqrt{2 + x^2}}
 = \frac{\pi}{2}\left[
      \tan^{-1}\left(\frac{x}{\sqrt{2 + x^2}}\right)\right]_0^1
 = \frac{\pi^2}{12}.
\end{gather*}
So the conjecture $B = \pi^2/32$ is equivalent to the
known theorem $A = 5\pi^2/96$.
On p.190ff. of Paul J. Nahin's book Inside Interesting
Integrals (with an introduction to contour integration): A
Collection of Sneaky Tricks, Sly Substitutions, and Numerous Other
Stupendously Clever, Awesomely Wicked, and Devilishly Seductive
Maneuvers for Computing Nearly 200 Perplexing Definite Integrals
From Physics, Engineering, and Mathematics (Plus 60 Challenge
Problems with Complete, Detailed Solutions (2014), there is an
ingenious and well-explained derivation of these two results, which
amounts to proving, in the same way, the above equation
$A + B = \pi^2/12$, along with the simultaneous equation
$A - B = \pi^2/48$.
Part of the highly condensed argument given by MathSE user Jack D'Aurizio
in answer to a recent question
is equivalent to the case $\tau = 1$ of the following argument,
for $0 < \tau \leqslant 1$, which I've expanded considerably for
the sake of those unused to this kind of calculation (i.e. me).
For all real $a$,
$$
\tan^{-1} a = \int_0^1 \frac{a}{1 + a^2u^2}\,du.
$$
Using Wolfram Alpha to evaluate the inner integral after
changing the order of integration:
\begin{gather*}
\int_0^\tau \frac{\tan^{-1}\sqrt{\frac{1 - t^2}{2}}}{1 + t^2}\,dt =
\int_0^\tau \int_0^1 \frac{\sqrt{\frac{1 - t^2}{2}}}
   {(1 + t^2)\left(1 + \frac{1 - t^2}{2}u^2\right)}\,du\,dt \\
= \int_0^1\int_0^\tau \frac{\sqrt{\frac{1 - t^2}{2}}}
   {(1 + t^2)\left(1 + \frac{1 - t^2}{2}u^2\right)}\,dt\,du \\
= \int_0^1 \left(
\frac{\tan^{-1}\left(\frac{\sqrt{2}\tau}{\sqrt{1 - \tau^2}}\right)}
{u^2 + 1}
 - \frac{\tan^{-1}\left(\frac{\sqrt{2}\tau}{\sqrt{1 - \tau^2}\sqrt{u^2 + 2}}\right)}
   {(u^2 + 1)\sqrt{u^2 + 2}}
\right)du.
\end{gather*}
But we have already seen the values of these integrals:
$$
\int_0^1 \frac{du}{u^2 + 1}
= \frac{\pi}{4}, \ \
\int_0^1 \frac{du}{(u^2 + 1)\sqrt{u^2 + 2}}
= \frac{\pi}{6}, \ \
\int_0^1 \frac{\tan^{-1}\left(\frac{1}{\sqrt{u^2 + 2}}\right)}
{(u^2 + 1)\sqrt{u^2 + 2}}\,du
= \frac{\pi^2}{32}.
$$
Hence, taking $\tau = 1$ and $\tau = \frac{1}{\sqrt{3}}$ respectively:
$$
\int_0^1
\frac{\tan^{-1}\sqrt{\frac{1 - t^2}{2}}}{1 + t^2}\,dt
= \frac{\pi}{2}\cdot\frac{\pi}{4} - \frac{\pi}{2}\cdot\frac{\pi}{6}
= \frac{\pi^2}{24},
$$
and:
$$
\int_0^{\frac{1}{\sqrt{3}}}
\frac{\tan^{-1}\sqrt{\frac{1 - t^2}{2}}}{1 + t^2}\,dt
= \frac{\pi}{4}\cdot\frac{\pi}{4} - \frac{\pi^2}{32}
= \frac{\pi^2}{32}.
$$
The familiar trigonometrical identities:
$$
\tan\left(\theta - \frac{\pi}{2}\right) = -\frac{1}{\tan\theta}, \ \
\tan3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}
$$
combine to give:
$$
\tan^{-1}\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)
= 3\tan^{-1}\sqrt{\frac{1-t^2}{2}} - \frac{\pi}{2}.
$$
Therefore:
\begin{align*}
\int_0^1
\frac{\tan^{-1}
\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)}
{1 + t^2}\,dt
& = 3\cdot\frac{\pi^2}{24} - \frac{\pi}{2}\cdot\frac{\pi}{4}
= 0, \\
\int_0^{\frac{1}{\sqrt{3}}}
\frac{\tan^{-1}
\left(\frac{\sqrt{2}(1-3t^2)}{(5+t^2)\sqrt{1-t^2}}\right)}
{1 + t^2}\,dt
& = 3\cdot\frac{\pi^2}{32} - \frac{\pi}{2}\cdot\frac{\pi}{6}
= \frac{\pi^2}{96}.
\end{align*}
Finally, making the change of variables $t = \tan\phi$, and
subtracting these two integrals, we get:
\begin{align*}
\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
& = 0, \\
\int_{0}^{\frac{\pi}{6}}\tan^{-1}\!\!
\left(\frac{\sqrt{2}\cos3 \phi}
{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi
& = \frac{\pi^2}{96}, \\
\int_{\frac{\pi}{6}}^{\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
& = -\frac{\pi^2}{96}
\end{align*}