How $\int_{0}^{\infty} \frac{\arctan(x)}{1+x}\frac{dx}{\sqrt[4]{x}}=\frac{\pi}{\sqrt2}\big(\pi/2+\ln{\beta}\big)$ $$
\int_{0}^{\infty}\frac{\arctan\left(x\right)}{1 + x}
\,\frac{\mathrm{d}x}{\sqrt[{\large 4}]{x}} =
\frac{\pi}{\,\sqrt{\,{2}\,}\,}
\left[{\pi \over 2} + \ln\left(\,{\beta}\,\right)\right]
$$
$$
\mbox{Find the value of}\quad
\beta^{4} - 28\beta^{3} + 70\beta^{2} - 28\beta.
$$
How to do this question ?. I tried conventional approaches such as substituting $x$ with $1/t^{2}$ but none of them is yielding an answer.
 A: Write
\begin{align*}
I
:= \int_{0}^{\infty}\frac{\arctan x}{(x+1)x^{1/4}} \, \mathrm{d}x
= \int_{0}^{1}\int_{0}^{\infty}\frac{x^{3/4}}{(x+1)(t^2x^2+1)} \, \mathrm{d}x\mathrm{d}t.
\end{align*}
The inner integral can be computed via a standard technique involving the contour integration along the keyhole contour. Indeed, denoting by $\operatorname{Log}$ the complex logarithm with the branch cut $[0,\infty)$ and writing
$$ f(z) = \frac{\exp\left(\frac{3}{4}\operatorname{Log}(z)\right)}{(z+1)(t^2z^2+1)}, $$
we have
\begin{align*}
\int_{0}^{\infty}\frac{x^{3/4}}{(x+1)(t^2x^2+1)} \, \mathrm{d}x
&= \frac{2\pi i}{1 - e^{3\pi i/2}} \left(
\underset{z=-1}{\mathrm{Res}} \, f(z)
+ \underset{z=i/t}{\mathrm{Res}} \, f(z)
+ \underset{z=-i/t}{\mathrm{Res}} \, f(z) \right) \\
&= -\frac{\pi}{\sin(3\pi/4)} \left( \frac{1}{t^2+1} + \frac{e^{-3i\pi/8}}{2(it-1)t^{3/4}} + \frac{e^{3i\pi/8}}{2(-it-1)t^{3/4}} \right).
\end{align*}
Now by noting that
$$ \int_{0}^{1} \frac{\omega \, \mathrm{d}t}{(\omega^4 t - 1)t^{3/4}}
= \int_{0}^{1} \frac{\omega \, \mathrm{d}u}{(\omega u)^4 - 1}
= -2\arctan(\omega) + \log\left(\frac{1-\omega}{1+\omega}\right) $$
holds for any complex $\omega$ avoiding the branch cuts $\cup_{k=0}^{3} i^k [1,\infty)$, the original integral reduces to
\begin{align*}
I
&= - \pi \sqrt{2} \biggl( \frac{\pi}{4} - \arctan(e^{3i\pi/8}) - \arctan(e^{-3i\pi/8}) \\
&\hspace{5em} + \frac{1}{2}\log\left(\frac{1-e^{3i\pi/8}}{1+e^{3i\pi/8}}\right) + \frac{1}{2}\log\left(\frac{1-e^{-3i\pi/8}}{1+e^{-3i\pi/8}}\right) \biggr) \\
&= - \pi \sqrt{2} \left( \frac{\pi}{4} - \frac{\pi}{2} + \log\tan\left(\frac{3\pi}{16}\right) \right) \\
&= \frac{\pi}{\sqrt{2}}\left( \frac{\pi}{2} - 2\log\tan\left(\frac{3\pi}{16}\right) \right).
\end{align*}
Here, the second line follows from the identities $\arctan z+\arctan(1/z) = \frac{\pi}{2}$ for $\operatorname{Re}(z) > 0$ and $\frac{1-e^{i\theta}}{1+e^{i\theta}}=-i\tan(\frac{\theta}{2})$. This shows that
$$ \beta = \cot^2\left(\frac{3\pi}{16}\right). $$
Finally, using the observation that
$$ X = \beta + \beta^{-1} = \frac{4}{\sin^2(3\pi/8)} -2 = 14 - 8\sqrt{2} $$
is a zero of the equation $X^2 - 28X + 68 = 0$, we get
$$ \beta^4 - 28\beta^3 + 70\beta^2 - 28\beta + 1 = 0, $$
and therefore the answer is $-1$.
A: I think this question is proposed by sir Srinivasa Raghava in Romanian maths magazine and in Brilliant.org too.
here is the link:
https://brilliant.org/problems/an-arctan-integral-via-a-quartic-equation/?ref_id=1585875
