Calculating $\int_0^1 \frac{\operatorname{arctanh}\left(\sqrt{1-\frac{u}{2}}\right)\sqrt{\frac{2 \pi \sqrt{1-u}}{u-2}+\pi } }{u\sqrt{1-u}} \, du$

What real tools would you employ for calculating the integral below? Some useful ideas I mean.

$$\int_0^1 \frac{\operatorname{arctanh}\left(\sqrt{1-\frac{u}{2}}\right)\sqrt{\frac{2 \pi \sqrt{1-u}}{u-2}+\pi } }{u\sqrt{1-u}} \, du$$

Solution 1:

For the integral \begin{align} I = \int_0^1 \frac{\operatorname{arctanh}\left(\sqrt{1-\frac{u}{2}}\right)\sqrt{\frac{2 \pi \sqrt{1-u}}{u-2}+\pi } }{u\sqrt{1-u}} \, du \end{align} let $t^{2} = 1-u$ to obtain the form \begin{align}\tag{1} I = 2 \sqrt{\pi} \, \int_{0}^{1} \tanh^{-1}\left(\sqrt{\frac{1 + t^{2}}{2}}\right) \, \frac{dt}{(1+t) \, \sqrt{1+t^{2}}} \end{align}

Method 1: Jack D'Aurizio has provided one way to continue with equation (1)

Method 2: Another way to continue with equation (1) is to make the substitution $t = \sqrt{\cos \theta}$ to obtain \begin{align}\tag{2} I = \sqrt{2 \pi} \, \int_{0}^{\pi/2} \ln\left(\cot\left(\frac{\theta}{4}\right)\right) \, \frac{\sin\theta \, d\theta}{\sqrt{\cos\theta} \, (1 + \sqrt{\cos\theta})} \end{align} or, by letting $\theta = 4 \phi$, \begin{align}\tag{2.1} I = 4 \, \sqrt{2 \pi} \, \int_{0}^{\pi/8} \frac{\ln(\cot\phi)) \, \sin(4\phi) \, d\phi}{\sqrt{\cos(4\phi)} \, (1 + \sqrt{\cos(4\phi)})} \end{align}

Utilizing the series \begin{align}\tag{3} \ln(\cot\phi) = 2 \, \sum_{k=0}^{\infty} \frac{\cos(4k+2)\phi}{2k+1} \end{align} then the integral of (2.1) becomes \begin{align} I = 8 \sqrt{2 \pi} \, \sum_{k=0}^{\infty} \frac{1}{2k+1} \, \int_{0}^{\pi/8} \frac{ \cos(4k+2)\phi \, \sin(4\phi)}{\sqrt{\cos(4\phi)} \, (1 + \sqrt{\cos(4\phi)})} \, d\phi. \end{align} Now let $t = \cos(4 \phi)$ to obtain \begin{align}\tag{4} I &= 2 \, \sqrt{2\pi} \, \sum_{k=0}^{\infty} \frac{1}{2k+1} \, \int_{0}^{1} \frac{\cos\left(\left(k + \frac{1}{2}\right) \, \cos^{-1}(t)\right)}{\sqrt{t} \, (1 + \sqrt{t})} \, dt \\ &= 2 \, \sqrt{2\pi} \, \sum_{k=0}^{\infty} \frac{1}{2k+1} \, \int_{0}^{1} \frac{T_{k+\frac{1}{2}}(t)}{\sqrt{t} \, (1 + \sqrt{t})} \, dt \end{align} where $T_{n}(x)$ is the Chebyshev polynomial of the first kind. Making the substitution $t = x^{2}$ leads to \begin{align}\tag{5} I = 4 \, \sqrt{2\pi} \, \sum_{k=0}^{\infty} \frac{1}{2k+1} \, \int_{0}^{1} T_{k+\frac{1}{2}}(x^{2}) \, \frac{dx}{1 + x} \end{align}

Let \begin{align}\tag{6} J_{k} = \int_{0}^{1} T_{k+\frac{1}{2}}(x^{2}) \, \frac{dx}{1+x} \end{align} and use $T_{n+1}(x) = 2\, x \, T_{n}(x) - T_{n-1}(x)$ to obtain \begin{align} J_{k+1} &= 2 \, \int_{0}^{1} T_{k+1/2}(x^2) \, \frac{x^{2} \, dx}{1+x} - J_{k-1} \\ &= 2 \, \int_{0}^{1} \left[ (x+1) - 2 + \frac{1}{1+x} \right] \, T_{k+1/2}(x^2) \, dx - J_{k-1} \\ &= 2 \, \int_{0}^{1} x \, T_{k+1/2}(x^2) \, dx - 2 \, \int_{0}^{1} T_{k+1/2}(x^2) \, dx + 2 \, J_{k} - J_{k-1}. \tag{7} \end{align} It can be quickly determined that \begin{align} \int_{0}^{1} x \, T_{k+\frac{1}{2}}(x^{2}) \, dx = \frac{1}{2((2k+1)^{2} - 4)} \, \left[ \sqrt{2} \, (2k+1) \, \left( \cos\left(\frac{k \pi}{2}\right) + \sin\left(\frac{k \pi}{2}\right) \right) - 4 \right] = \frac{P_{k}}{2((2k+1)^{2} - 4)} \end{align} and let \begin{align} \sigma_{k} = \int_{0}^{1} T_{k+\frac{1}{2}}(x^2) \, dx \end{align} to take (7) into the form \begin{align}\tag{8} J_{k+1} - 2 \, J_{k} + J_{k-1} = \frac{P_{k}}{(2k+1)^{2} - 4} + \sigma_{k}. \end{align} By the same pattern the recurrence relation for $\sigma_{k}$ is seen to be \begin{align}\tag{9} \sigma_{k+1} + \sigma_{k-1} = \frac{P_{k}}{(2k+1)^{2} - 4}. \end{align} The first few values of $J_{k}$ and $\sigma_{k}$ are \begin{align}\tag{10} J_{0} &= \left(1 - \frac{1}{\sqrt{2}}\right) \, \left( 1 + \sinh^{-1}(1) \right) \\ J_{1} &= \frac{\sqrt{2}}{3} \, \left( -8 + 7 \sqrt{2} + 3 (2 \sqrt{2} - 3) \, \sinh^{-1}(1) \right) - J_{0} \\ &= \frac{\sqrt{2}}{3} \, (-8 + 7 \sqrt{2}) + \frac{3 \sqrt{2} - 5}{\sqrt{2}} \, \sinh^{-1}(1) \\ \sigma_{0} &= \frac{1}{4} \, \left( 2 + \sqrt{2} \, \sinh^{-1}(1) \right) \\ \sigma_{1} &= \frac{1}{8} \, \left( 2 - 3 \sqrt{2} \, \sinh^{-1}(1) \right) \end{align} Condensing the results of equations (5-9) leads to \begin{align}\tag{11} I = 4 \sqrt{2 \pi} \, \sum_{k=0}^{\infty} \frac{J_{k}}{2k+1} \end{align} where \begin{align}\tag{12} & J_{k+1} - 2 \, J_{k} + J_{k-1} = \frac{P_{k}}{(2k+1)^{2} - 4} + \sigma_{k} \\ & \sigma_{k+1} + \sigma_{k-1} = \frac{P_{k}}{(2k+1)^{2} - 4} \\ & P_{k} = \sqrt{2} \, (2k+1) \, \left(\cos\left(\frac{k \, \pi}{2}\right) + \sin\left(\frac{k \, \pi}{2}\right) \right) - 4 \end{align} and the first few values are obtained from (10). The general form of the desired integral can be expressed as \begin{align} \int_0^1 \frac{\operatorname{arctanh}\left(\sqrt{1-\frac{u}{2}}\right)\sqrt{\frac{2 \pi \sqrt{1-u}}{u-2}+\pi } }{u\sqrt{1-u}} \, du = A \, \sinh^{-1}(1) + B \end{align} where $A$ and $B$ can be determined from the above collected set of recurrence relations.

It can be seen that \begin{align} J_{k+2} - 2 \, J_{k+1} + 2 \, J_{k} - 2 \, J_{k-1} + J_{k-2} = \frac{P_{k+1}}{(2k+3)^{2} - 4} + \frac{P_{k}}{(2k+1)^{2} - 4} + \frac{P_{k-1}}{(2k-1)^{2} - 4}. \end{align} which makes the series (11) slightly easier to calculate term by term.

Solution 2:

Continuing from Leucippus' answer, through $\text{arctanh}(z)=\frac{1}{2}\log\frac{1+z}{1-z}$, we just need to compute:

$$ I_1 = \int_{0}^{1}\frac{dt}{(1+t)\sqrt{1+t^2}}=\frac{1}{\sqrt{2}}\text{arcsinh}(1)=\frac{1}{\sqrt{2}}\log(1+\sqrt{2}),$$ $$ I_2 = \int_{0}^{1}\frac{\log(\sqrt{2}+\sqrt{1+t^2})}{(1+t)\sqrt{1+t^2}}\,dt,\qquad I_3 = \int_{0}^{1}\frac{\log(\sqrt{2}-\sqrt{1+t^2})}{(1+t)\sqrt{1+t^2}}\,dt$$ and since $$ \int\frac{dt}{(1+t)\sqrt{1+t^2}} = \frac{1}{\sqrt{2}}\,\log\left(1-t+\sqrt{2}\sqrt{1+t^2}\right) $$ integration by parts looks like a good way to find $I_2$ and $I_3$ in terms of dilogarithms.

I have the strong feeling that the Catalan-Harmonic identity is involved, or some special value of the Rogers L-function.

If we substitute $t=\tan\theta$ in Leucippus' $(1)$ we get: $$ I = 2\sqrt{\pi}\int_{0}^{\pi/4}\text{arctanh}\left(\frac{1}{\sqrt{2}\cos\theta}\right)\frac{d\theta}{\sin\theta+\cos\theta}$$ so we just need to compute: $$ J_k = \int_{0}^{\pi/4}\frac{d\theta}{\cos^{2k+1}(\theta)(\sin\theta+\cos\theta)}=\int_{0}^{1}(1+t^2)^{k+1}\frac{dt}{1+t}$$ then exploit the Taylor series of $\text{arctanh}$.