TITLE: Evaluate $\frac{1}{2}\int_0^1\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2}}\sqrt{(x-h)^2+y^2}dy dx dh$
QUESTION [5 upvotes]: What I'm trying to get is the value of
$$\delta=\frac{1}{2}\int_0^1\int_{-1}^1\frac{1}{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2}}\sqrt{(x-h)^2+y^2}dy dx dh$$

I know that
$$\frac{1}{\sqrt{1-x^2}}\int_0^{\sqrt{1-x^2}}\sqrt{(x-h)^2+y^2}dy=\frac{2 \sqrt{1-x^2} \sqrt{h^2-2 h x+1}-(h-x)^2 \left(\log \left((h-x)^2\right)-2 \log \left(\sqrt{h^2-2 h x+1}+\sqrt{1-x^2}\right)\right)}{4 \sqrt{1-x^2}}$$
but I certainly don't know how to follow from there.

REPLY [4 votes]: Let the value of $\delta$ be given by the iterated integral
$$\delta:=\frac12\int_{0}^{1}\mathrm{d}h\int_{-1}^{1}\frac{\mathrm{d}x}{\sqrt{1-x^{2}}}\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\,\sqrt{\left(x-h\right)^{2}+y^{2}}.\tag{1}$$
We will show that

$$\delta=\frac{1}{12}+\frac56C,$$
  where $C$ denotes the Catalan constant.


Making use of a few simple substitutions and symmetries, we can rewrite the iterated integral for $\delta$ so that we have a triple integral over the unit cube:
$$\begin{align}
\delta
&=\frac12\int_{0}^{1}\mathrm{d}h\int_{-1}^{1}\frac{\mathrm{d}x}{\sqrt{1-x^{2}}}\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\,\sqrt{\left(x-h\right)^{2}+y^{2}}\\
&=\frac12\int_{0}^{1}\mathrm{d}h\int_{-1}^{1}\mathrm{d}x\int_{0}^{\sqrt{1-x^{2}}}\mathrm{d}y\,\frac{\sqrt{\left(x-h\right)^{2}+y^{2}}}{\sqrt{1-x^{2}}}\\
&=\frac12\int_{0}^{1}\mathrm{d}h\int_{-1}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(x-h\right)^{2}+\left(1-x^{2}\right)v^{2}};~~~\small{\left[\frac{y}{\sqrt{1-x^{2}}}=v\right]}\\
&=\int_{0}^{1}\mathrm{d}h\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(2u-1+h\right)^{2}+4u\left(1-u\right)v^{2}};~~~\small{\left[\frac{1-x}{2}=u\right]}\\
&=\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(t-2u\right)^{2}+4u\left(1-u\right)v^{2}}\\
&=4\int_{0}^{\frac12}\mathrm{d}w\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(w-u\right)^{2}+u\left(1-u\right)v^{2}};~~~\small{\left[t=2w\right]}\\
&=2\int_{0}^{\frac12}\mathrm{d}w\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(w-u\right)^{2}+u\left(1-u\right)v^{2}}\\
&~~~~~+2\int_{0}^{\frac12}\mathrm{d}w\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(w-u\right)^{2}+u\left(1-u\right)v^{2}}\\
&=2\int_{0}^{\frac12}\mathrm{d}x\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(x-u\right)^{2}+u\left(1-u\right)v^{2}};~~~\small{\left[w=x\right]}\\
&~~~~~+2\int_{\frac12}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(1-x-u\right)^{2}+u\left(1-u\right)v^{2}};~~~\small{\left[w=1-x\right]}\\
&=2\int_{0}^{\frac12}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(x-y\right)^{2}+y\left(1-y\right)v^{2}};~~~\small{\left[u=y\right]}\\
&~~~~~+2\int_{\frac12}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(x-y\right)^{2}+y\left(1-y\right)v^{2}};~~~\small{\left[u=1-y\right]}\\
&=2\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(x-y\right)^{2}+y\left(1-y\right)v^{2}}.\tag{2}\\
\end{align}$$
Wolfram Alpha yields the following numerical approximation for the triple integral $(2)$:
$$\delta\approx0.846638.\tag{3}$$
The approximation $(3)$ will be helpful verifying the final result after the fact because the derivation that follows is admittedly convoluted. In particular, the evaluation will make use of the Gauss hypergeometric function, which may be defined via Euler's integral representation: given complex parameters and argument $\left(\alpha,\beta,\gamma,z\right)\in\mathbb{C}^{4}$ such that
$$\Re{\left(\gamma\right)}>\Re{\left(\beta\right)}>0\land\left(\left[z\neq1\land\left|\arg{\left(1-z\right)}\right|<\pi\right]\lor\left[z=1\land\Re{\left(\gamma-\alpha-\beta\right)}>0\right]\right),$$
Euler's integral representation for the Gauss hypergeometric function states
$$\int_{0}^{1}\frac{t^{\beta-1}\left(1-t\right)^{\gamma-\beta-1}}{\left(1-zt\right)^{\alpha}}\,\mathrm{d}t:=\operatorname{B}{\left(\beta,\gamma-\beta\right)}\,{_2F_1}{\left(\alpha,\beta;\gamma;z\right)},\tag{4a}$$
where $\operatorname{B}{\left(\alpha,\beta\right)}$ denotes the beta function, defined for $\left(\alpha,\beta\right)\in\mathbb{C}^{2}\land\Re{\left(\alpha\right)}>0\land\Re{\left(\beta\right)}>0$ via the integral representation
$$\operatorname{B}{\left(\alpha,\beta\right)}:=\int_{0}^{1}t^{\alpha-1}\left(1-t\right)^{\beta-1}\,\mathrm{d}t.$$
Setting $\alpha=-\frac12\land\beta=\frac12\land\gamma=\frac32$ and assuming real argument $z\in\left[-1,\infty\right)$, Euler's integral formula $(4a)$ implies
$$\int_{0}^{1}\frac{\sqrt{1+zt}}{2\sqrt{t}}\,\mathrm{d}t={_2F_1}{\left(-\frac12,\frac12;\frac32;-z\right)};~~~\small{z\ge-1}.\tag{4b}$$

Starting from the last line of $(2)$ above, we obtain the following result for $\delta$:
$$\begin{align}
\delta
&=2\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(t-u\right)^{2}+u\left(1-u\right)v^{2}}\\
&=2\int_{0}^{1}\mathrm{d}t\,\bigg{[}\int_{0}^{t}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(t-u\right)^{2}+u\left(1-u\right)v^{2}}\\
&~~~~~+\int_{t}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(t-u\right)^{2}+u\left(1-u\right)v^{2}}\bigg{]}\\
&=2\int_{0}^{1}\mathrm{d}t\,\bigg{[}\int_{0}^{t}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(t-u\right)^{2}}\sqrt{1+\frac{u\left(1-u\right)v^{2}}{\left(t-u\right)^{2}}}\\
&~~~~~+\int_{t}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\sqrt{\left(t-u\right)^{2}}\sqrt{1+\frac{u\left(1-u\right)v^{2}}{\left(t-u\right)^{2}}}\bigg{]}\\
&=2\int_{0}^{1}\mathrm{d}t\,\bigg{[}\int_{0}^{t}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\left(t-u\right)\sqrt{1+\frac{u\left(1-u\right)v^{2}}{\left(t-u\right)^{2}}}\\
&~~~~~+\int_{t}^{1}\mathrm{d}u\int_{0}^{1}\mathrm{d}v\,\left(u-t\right)\sqrt{1+\frac{u\left(1-u\right)v^{2}}{\left(u-t\right)^{2}}}\bigg{]}\\
&=2\int_{0}^{1}\mathrm{d}t\,\bigg{[}\int_{0}^{t}\mathrm{d}u\,\left(t-u\right)\int_{0}^{1}\mathrm{d}v\,\sqrt{1+\frac{u\left(1-u\right)v^{2}}{\left(t-u\right)^{2}}}\\
&~~~~~+\int_{t}^{1}\mathrm{d}u\,\left(u-t\right)\int_{0}^{1}\mathrm{d}v\,\sqrt{1+\frac{u\left(1-u\right)v^{2}}{\left(u-t\right)^{2}}}\bigg{]}\\
&=2\int_{0}^{1}\mathrm{d}t\,\bigg{[}\int_{0}^{t}\mathrm{d}u\,\left(t-u\right)\int_{0}^{1}\mathrm{d}w\,\frac{\sqrt{1+\frac{u\left(1-u\right)w}{\left(t-u\right)^{2}}}}{2\sqrt{w}}\\
&~~~~~+\int_{t}^{1}\mathrm{d}u\,\left(u-t\right)\int_{0}^{1}\mathrm{d}w\,\frac{\sqrt{1+\frac{u\left(1-u\right)w}{\left(u-t\right)^{2}}}}{2\sqrt{w}}\bigg{]};~~~\small{\left[v^{2}=w\right]}\\
&=2\int_{0}^{1}\mathrm{d}t\,\bigg{[}\int_{0}^{t}\mathrm{d}u\,\left(t-u\right)\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-\frac{u\left(1-u\right)}{\left(t-u\right)^{2}}\right)}\\
&~~~~~+\int_{t}^{1}\mathrm{d}u\,\left(u-t\right)\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-\frac{u\left(1-u\right)}{\left(t-u\right)^{2}}\right)}\bigg{]},\tag{5}\\
\end{align}$$
where in the last line above we have invoked identity $(4b)$.
Next, using the quadratic transformation
$$\frac{u\left(1-u\right)}{\left(t-u\right)^{2}}=v\implies u=\frac{\left(2tv+1\right)\pm\sqrt{1+4t\left(1-t\right)v}}{2\left(v+1\right)},$$
we find
$$\begin{align}
\delta
&=2\int_{0}^{1}\mathrm{d}t\,\bigg{[}\int_{0}^{t}\mathrm{d}u\,\left(t-u\right)\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-\frac{u\left(1-u\right)}{\left(t-u\right)^{2}}\right)}\\
&~~~~~+\int_{t}^{1}\mathrm{d}u\,\left(u-t\right)\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-\frac{u\left(1-u\right)}{\left(t-u\right)^{2}}\right)}\bigg{]}\\
&=\small{2\int_{0}^{1}\mathrm{d}t\,\bigg{[}\int_{0}^{+\infty}\mathrm{d}v\,\frac{\left(1-2t-\sqrt{1+4t\left(1-t\right)v}\right)^{2}}{4\left(v+1\right)^{2}\sqrt{1+4t\left(1-t\right)v}}\cdot\frac{2t-1+\sqrt{1+4t\left(1-t\right)v}}{2\left(v+1\right)}}\\
&~~~~~\small{\times\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-v\right)}};~~~\small{\left[u=\frac{\left(2tv+1\right)-\sqrt{1+4t\left(1-t\right)v}}{2\left(v+1\right)}\right]}\\
&~~~~~\small{+\int_{+\infty}^{0}\mathrm{d}v\,\frac{\left(-1\right)\left(1-2t+\sqrt{1+4t\left(1-t\right)v}\right)^{2}}{4\left(v+1\right)^{2}\sqrt{1+4t\left(1-t\right)v}}\cdot\frac{1-2t+\sqrt{1+4t\left(1-t\right)v}}{2\left(v+1\right)}}\\
&~~~~~\small{\times\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-v\right)}\bigg{]}};~~~\small{\left[u=\frac{\left(2tv+1\right)+\sqrt{1+4t\left(1-t\right)v}}{2\left(v+1\right)}\right]}\\
&=\frac14\int_{0}^{1}\mathrm{d}t\int_{0}^{\infty}\mathrm{d}v\,\bigg{[}\frac{\left(-1\right)\left(1-2t-\sqrt{1+4t\left(1-t\right)v}\right)^{3}}{\left(v+1\right)^{3}\sqrt{1+4t\left(1-t\right)v}}\\
&~~~~~+\frac{\left(1-2t+\sqrt{1+4t\left(1-t\right)v}\right)^{3}}{\left(v+1\right)^{3}\sqrt{1+4t\left(1-t\right)v}}\bigg{]}\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-v\right)}\\
&=\frac14\int_{0}^{1}\mathrm{d}t\int_{0}^{\infty}\mathrm{d}v\,\bigg{[}\left(1-2t+\sqrt{1+4t\left(1-t\right)v}\right)^{3}\\
&~~~~~-\left(1-2t-\sqrt{1+4t\left(1-t\right)v}\right)^{3}\bigg{]}\\
&~~~~~\times\frac{1}{\left(v+1\right)^{3}\sqrt{1+4t\left(1-t\right)v}}\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-v\right)}\\
&=\frac14\int_{0}^{1}\mathrm{d}t\int_{0}^{\infty}\mathrm{d}v\,\\
&~~~~~\times\frac{2\left[3-12t+12t^{2}+\left(1+4t\left(1-t\right)v\right)\right]\sqrt{1+4t\left(1-t\right)v}}{\left(v+1\right)^{3}\sqrt{1+4t\left(1-t\right)v}}\\
&~~~~~\times\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-v\right)}\\
&=2\int_{0}^{1}\mathrm{d}t\int_{0}^{\infty}\mathrm{d}v\,\frac{1+t\left(1-t\right)\left(v-3\right)}{\left(v+1\right)^{3}}\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-v\right)}\\
&=2\int_{0}^{\infty}\mathrm{d}v\int_{0}^{1}\mathrm{d}t\,\frac{1+t\left(1-t\right)\left(v-3\right)}{\left(v+1\right)^{3}}\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-v\right)}\\
&=\frac13\int_{0}^{\infty}\mathrm{d}v\,\frac{\left(v+3\right)}{\left(v+1\right)^{3}}\,{_2F_1}{\left(-\frac12,\frac12;\frac32;-v\right)}\\
&=\frac13\int_{0}^{1}\mathrm{d}w\,\left(3-2w\right)\,{_2F_1}{\left(-\frac12,\frac12;\frac32;\frac{w}{w-1}\right)};~~~\small{\left[\frac{v}{1+v}=w\right]}.\tag{6}\\
\end{align}$$
The hypergeometric function in the last line of $(6)$ may be written in terms of elementary functions as,
$${_2F_1}{\left(-\frac12,\frac12;\frac32;\frac{w}{w-1}\right)}=\frac{1}{2\sqrt{1-w}}+\frac{\sqrt{1-w}\,\tanh^{-1}{\left(\sqrt{w}\right)}}{2\sqrt{w}}.\tag{7}$$
A derivation of identity $(7)$ is given in Appendix 1 below.
Continuing with our evaluation of $\delta$ from the last line of $(6)$, we find
$$\begin{align}
\delta
&=\frac13\int_{0}^{1}\mathrm{d}w\,\left(3-2w\right)\,{_2F_1}{\left(-\frac12,\frac12;\frac32;\frac{w}{w-1}\right)}\\
&=\frac13\int_{0}^{1}\mathrm{d}w\,\left(3-2w\right)\left[\frac{1}{2\sqrt{1-w}}+\frac{\sqrt{1-w}\,\tanh^{-1}{\left(\sqrt{w}\right)}}{2\sqrt{w}}\right]\\
&=\frac16\int_{0}^{1}\mathrm{d}w\,\frac{\left(3-2w\right)}{\sqrt{1-w}}+\frac16\int_{0}^{1}\mathrm{d}w\,\frac{\left(3-2w\right)\sqrt{1-w}\,\tanh^{-1}{\left(\sqrt{w}\right)}}{\sqrt{w}}\\
&=\frac16\int_{0}^{1}\mathrm{d}w\,\frac{\left(3-2w\right)}{\sqrt{1-w}}\\
&~~~~~+\frac16\int_{0}^{1}\mathrm{d}w\,\left(3-2w\right)\sqrt{1-w}\,{_2F_1}{\left(\frac12,1;\frac32;w\right)}\\
&=\frac16\int_{0}^{1}\mathrm{d}x\,\frac{\left(1+2x\right)}{\sqrt{x}};~~~\small{\left[1-w=x\right]}\\
&~~~~~+\frac12\int_{0}^{1}\mathrm{d}w\,\sqrt{1-w}\,{_2F_1}{\left(\frac12,1;\frac32;w\right)}\\
&~~~~~-\frac13\int_{0}^{1}\mathrm{d}w\,w\sqrt{1-w}\,{_2F_1}{\left(\frac12,1;\frac32;w\right)}\\
&=\frac13\int_{0}^{1}\mathrm{d}y\,\left(1+2y^{2}\right);~~~\small{\left[\sqrt{x}=y\right]}\\
&~~~~~+\frac12\operatorname{B}{\left(1,\frac32\right)}\,{_3F_2}{\left(\frac12,1,1;\frac32,1+\frac32;1\right)}\\
&~~~~~-\frac13\operatorname{B}{\left(2,\frac32\right)}\,{_3F_2}{\left(\frac12,1,2;\frac32,2+\frac32;1\right)}\\
&=\frac59+\frac13\,{_3F_2}{\left(\frac12,1,1;\frac32,\frac52;1\right)}\\
&~~~~~-\frac{4}{45}\,{_3F_2}{\left(\frac12,1,2;\frac32,\frac72;1\right)}\\
&=\frac59+\frac13\left(3C-\frac32\right)-\frac{4}{45}\left(\frac{15C}{8}-\frac{5}{16}\right)\\
&=\frac{1+10C}{12}.\blacksquare\\
\end{align}$$

Appendix 1:
For $z<1$, we have
$$\begin{align}
{_2F_1}{\left(-\frac12,\frac12;\frac32;z\right)}
&=\int_{0}^{1}\frac{\sqrt{1-zt}}{2\sqrt{t}}\,\mathrm{d}t\\
&=\frac12\int_{0}^{1}\frac{1-zt}{\sqrt{t\left(1-zt\right)}}\,\mathrm{d}t\\
&=\frac12\int_{0}^{1}\frac{\left(1-z\right)^{2}}{\left(1-z+zu\right)^{3}}\cdot\frac{\left(1-z+zu\right)}{\sqrt{\left(1-z\right)u}}\,\mathrm{d}u;~~~\small{\left[\frac{\left(1-z\right)t}{1-zt}=u\right]}\\
&=\frac12\left(1-z\right)^{3/2}\int_{0}^{1}\frac{\mathrm{d}u}{\left(1-z+zu\right)^{2}\sqrt{u}}.\\
\end{align}$$
Then, setting $z=:\frac{w}{w-1}$, we have for $0<w<1$ the following:
$$\begin{align}
{_2F_1}{\left(-\frac12,\frac12;\frac32;\frac{w}{w-1}\right)}
&=\frac12\left(\frac{1}{1-w}\right)^{3/2}\int_{0}^{1}\frac{\mathrm{d}u}{\left[\left(\frac{1}{1-w}\right)+\left(\frac{w}{w-1}\right)u\right]^{2}\sqrt{u}}\\
&=\frac{\left(1-w\right)^{2}}{2\left(1-w\right)^{3/2}}\int_{0}^{1}\frac{\mathrm{d}u}{\left(1-wu\right)^{2}\sqrt{u}}\\
&=\sqrt{1-w}\int_{0}^{1}\frac{\mathrm{d}u}{2\left(1-wu\right)^{2}\sqrt{u}}\\
&=\frac{\sqrt{1-w}}{\sqrt{w}}\int_{0}^{w}\frac{\mathrm{d}x}{2\left(1-x\right)^{2}\sqrt{x}};~~~\small{\left[wu=x\right]}\\
&=\frac{\sqrt{1-w}}{\sqrt{w}}\int_{0}^{\sqrt{w}}\frac{\mathrm{d}y}{\left(1-y^{2}\right)^{2}};~~~\small{\left[\sqrt{x}=y\right]}\\
&=\small{\frac{\sqrt{1-w}}{\sqrt{w}}\int_{0}^{\sqrt{w}}\left[\frac{1}{4\left(1-y\right)^{2}}+\frac{1}{4\left(1+y\right)^{2}}+\frac{1}{2\left(1-y^{2}\right)}\right]\,\mathrm{d}y};~~~\small{P.F.D.}\\
&=\frac{\sqrt{1-w}}{2\sqrt{w}}\left[\frac{y}{1-y^{2}}+\tanh^{-1}{\left(y\right)}\right]_{y=0}^{y=\sqrt{w}}\\
&=\frac{\sqrt{1-w}}{2\sqrt{w}}\left[\frac{\sqrt{w}}{1-w}+\tanh^{-1}{\left(\sqrt{w}\right)}\right]\\
&=\frac{\sqrt{1-w}}{2}\left[\frac{1}{1-w}+\frac{\tanh^{-1}{\left(\sqrt{w}\right)}}{\sqrt{w}}\right]\\
&=\frac{1}{2\sqrt{1-w}}+\frac{\sqrt{1-w}\,\tanh^{-1}{\left(\sqrt{w}\right)}}{2\sqrt{w}}.\\
\end{align}$$