Arctangent integral $\int_0^1\frac {\arctan x}{\sqrt{x(1-x)}}dx=\pi\arctan\sqrt{\frac {\sqrt2-1}2}$

Question

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?

Answer 1

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 $$\int_{0}^{1}\frac{\arctan x}{\sqrt{x(1-x)}}\,dx = \pi\arctan\left(2^{1/4}\sin\tfrac{\pi}{8}\right).$$

Answer 2

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.

Answer 3

\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}

Answer 4

Let $I(a) =\int_0^1 \frac {\tan^{-1}(x\sinh a)}{\sqrt{x(1-x)}}dx$. Then \begin{align} I’(a)=&\int_0^{1}\frac{x\cosh a}{\sqrt{x(1-x)} (1+ x^2 \sinh^2a)}\overset{x=\cos^2 t}{dx}\\ = &\int_0^{\pi/2}\frac{2\cosh a \sec^2t}{(1+\tan^2t)^2+\sinh^2a}dt=\frac\pi2\text{sech}\frac a2 \end{align} With $p=\sinh^{-1}1$ \begin{align} \int_0^1 \frac {\tan^{-1} x}{\sqrt{x(1-x)}} &dx=I(p)=\int_0^{p}I’(a)da = \frac\pi2 \int_0^{p}\text{sech}\frac a2\ da\\=& \ \pi \tan^{-1}\left( \sinh\frac a2\right)\bigg|_0^{p} = \pi\tan^{-1}\sqrt{\frac {\sqrt2-1}2} \end{align}

Answer 5

We have: $$I=\int_{0}^{1}\frac{\arctan(x)}{\sqrt{x(1-x)}}dx=\overset{x=\frac{1}{t+1}}{=}\int_{0}^{\infty}\frac{t^{\frac{-1}{2}}\arctan\left(\frac{1}{t+1}\right)}{t+1}dt$$ $$ =\int_{0}^{\infty}\frac{t^{\frac{-1}{2}}}{t+1}dt\ \Im\int_{0}^{i}\frac{1}{y+t+1}dy =\Im\int_{0}^{i}\frac{dy}{y}\int_{0}^{\infty}\left(\frac{t^{\frac{-1}{2}}}{t+1}-\frac{t^{\frac{-1}{2}}}{t+y+1}\right)dt\tag{1}$$ We will use this formula: $$\int_{0}^{\infty}\frac{t^{s-1}}{(t+a)^{n+1}}dt\overset{t=au}{=}a^{s-n-1}\int_{0}^{a^{-1}\infty}\frac{u^{s-1}}{(u+1)^{n+1}}du$$$$\overset{(*)}{=}a^{s-n-1}\int_{0}^{\infty}\frac{u^{s-1}}{(u+1)^{n+1}}du=a^{s-n-1}B(s,n+1-s)$$$$=a^{s-n-1}\frac{\Gamma(s)\Gamma(n+1-s)}{\Gamma(n+1)}=\frac{(n-s)_n}{n!}\frac{\pi}{\sin(\pi s)}a^{s-n-1} \tag{2}$$ See how to prove $(*)$. Apply $(2)$ on $(1)$: $$I=\pi\Im\int_{0}^{i}\frac{1-(y+1)^{-\frac{1}{2}}}{y}dy=\pi\Im2\log(1+\sqrt{1+y})\Big\vert_{y=0}^{y=i}$$ $$=2\pi \arg(1+\sqrt{1+i})=\pi\arctan\sqrt{\frac{\sqrt{2}-1}{2}}$$ Note that: $(1+\sqrt{1+i})^2=\left(2+\sqrt{2(\sqrt{2}+1})\right)\left(1+i\sqrt{\frac{\sqrt{2}-1}{2}}\right)$

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One advantage of this approach is that we can find solution from scratch with the power of partial fractions and Beta function after we substitute $x=\frac{1}{t+1}$ and "rationalize" $\arctan$.
Here is a similar integral: $$\int_{0}^{1}\frac{\arctan(x)}{\sqrt[3]{x^2(1-x)}}dx=$$ $$\frac{\pi}{2}\log\left(\frac{2-\sqrt[3]{4}+\sqrt[3]{16}+\sqrt[6]{432}}{2+\sqrt[3]{16}+\sqrt[3]{32}}\right)+\pi\sqrt{3}\arctan\left(\frac{-1+\sqrt{3}+\sqrt[3]{4}}{1+\sqrt{3}+\sqrt[3]{16}+\sqrt[6]{432}}\right)$$