# Arctangent integral that I'm having difficulty on

Question: Show that$$\int\limits_0^1 dx\,\frac {\arctan x}{\sqrt{x(1-x)}}=\pi\arctan\sqrt{\frac {\sqrt2-1}2}$$

I'm just having a hard time figuring out what to do. I tried to make the substitution $x=\frac {1-t}{1+t}$ but that didn't help very much because the denominator is slightly different. My next thought was to try to represent $\arctan x$ as an infinite series$$\arctan x=\sum\limits_{n\geq1}\frac {(-1)^{n-1}x^n}n\sin\left(\frac {\pi n}2\right)$$ But seeing as to how the result is in terms of $\arctan(\cdot)$, I doubt an infinite series would help much. Especially if the argument is a nested radical. Perhaps there is some sort of hidden symmetry one may exploit for this one?

• Some definite integrals of $\int \arctan$ have a tricky formula using substitution, I would check first them. – peterh Jul 2 '18 at 18:10
• A sort of beta function and using integration by parts may work! – Nosrati Jul 2 '18 at 18:26
• Mathematica confirms the result instantly. – David G. Stork Jul 2 '18 at 19:35
• @DavidG.Stork Yeah, Wolfram Alpha can evaluate it too! – Crescendo Jul 2 '18 at 19:55

We have $$\int_{0}^{1}\frac{x^{2n+1}}{\sqrt{x(1-x)}}\,dx = \frac{\pi}{4^{2n}}\binom{4n}{2n}\frac{4n+1}{4n+2}$$ hence the given integral equals $$\frac{\pi}{2} \sum_{n\geq 0}\frac{(-1)^n}{4^{2n}}\binom{4n}{2n}\frac{4n+1}{(2n+1)^2}$$ where by the generating function for Catalan numbers we have $$\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\frac{2n+1}{n+1}z^n = \frac{2}{z\sqrt{1-z}}-\frac{2}{z}\tag{A}$$ hence by integration $$\sum_{n\geq 0}\frac{1}{4^n}\binom{2n}{n}\frac{2n+1}{(n+1)^2}z^{n+1} = 4\log\left(\frac{2}{1+\sqrt{1-z}}\right)\tag{B}$$ and the given result can be proved by evaluating $(B)$ at $z=\pm i$, with some care in managing the determinations of the complex logarithm / square root. In a equivalent form $$\int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x)}}\,dx = \pi\arctan\left(2^{1/4}\sin\tfrac{\pi}{8}\right).$$

For a straightforward approach, use that

$$\int_0^1 \frac{x\arccos(x)}{1+a x^2}\text{d}x=\frac{\pi}{2a}\log((1+\sqrt{1+a})/2),\tag{1}$$ since your integral is (by the integration by parts and a variable change) $\displaystyle 4\int_0^1 \frac{x\arccos(x)}{1+x^4}\text{d}x$.

$\textbf{Q.E.D.}$

NOTE: the integral in $(1)$ is straightforward with the integration by parts and D.U.I.S.

• @Crescendo the integrals of this type are meant to be done under a minute with the proper training. Practice more, and use less Mathematica. – user 1357113 Jul 2 '18 at 22:10
• math.stackexchange.com/questions/2799950/… and math.stackexchange.com/questions/2798619/… have a similar structure but I doubt them can be done in less than a minute... Still I am fairly sure they have closed forms in terms of polylogarithms. – Jack D'Aurizio Jul 2 '18 at 22:27
• Don't you mean $1+ax^4$ in the denominator? – Isaac Browne Jul 3 '18 at 0:31
• @user23571113 Just because I said Wolfram Alpha provides the same result doesn't mean I didn't give the integral any thought. I also don't have Mathematica (I wish I did though). – Crescendo Jul 3 '18 at 3:08
• So many steps are skipped here it was hard to follow, but I think I see now. We can plug in $a=i$ and $a=-i$ after doing an imaginary partial fraction decomposition. – Isaac Browne Jul 3 '18 at 3:30

\begin{align} \int_0^1\frac{\arctan x}{\sqrt{x(1-x)}} \,dx &= \int_0^1\frac{1}{\sqrt{x(1-x)}}\left(x-\dfrac{x^3}{3}+\dfrac{x^5}{5}+\cdots\right)\,dx\\ &= \int_0^1 \left(x^\frac{1}{2}(1-x)^\frac{-1}{2}-\dfrac13x^\frac{5}{2}(1-x)^\frac{-1}{2} + \dfrac15x^\frac{9}{2}(1-x)^\frac{-1}{2}-\cdots\right)\,dx\\ &= \beta\left(\dfrac{3}{2},\dfrac{1}{2}\right) -\frac13\beta\left(\dfrac{7}{2},\dfrac{1}{2}\right)+\frac15\beta\left(\dfrac{11}{2},\dfrac{1}{2}\right)-\cdots\\ &= \pi\sum_{n=0}^{\infty}\dfrac{(-1)^n}{(2n+1)2^{4n+1}}{4n+1\choose2n+1} \end{align}