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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.

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  • $\begingroup$ @YuriyS The inner-most integral relates to the variable $y$ the middle one $x$ and the outer-most one $h$. But I thought this was implied by $dydxdh$. $\endgroup$
    – Garmekain
    Dec 15, 2016 at 18:10
  • $\begingroup$ There are different conventions on writing multiple integrals, so I couldn't be sure without clarification, thanks $\endgroup$
    – Yuriy S
    Dec 15, 2016 at 18:11
  • $\begingroup$ 2 $\delta$ appears to be around $1.693276$ according to Mathematica, but I was not able to reliably increase the precision (apparently, the integral is nasty enough, so Mathematica has some trouble computing it numerically) $\endgroup$
    – Yuriy S
    Dec 15, 2016 at 18:19

1 Answer 1

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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:

$$\begin{align} {_2F_1}{\left(-\frac12,\frac12;\frac32;\frac{w}{w-1}\right)} &=\frac12\left(\frac{1}{1-w}\right)^{3/2}\int_{0}^{1}\frac{\mathrm{d}u}{\left[\left(\frac{1}{1-w}\right)+\left(\frac{w}{w-1}\right)u\right]^{2}\sqrt{u}}\\ &=\frac{\left(1-w\right)^{2}}{2\left(1-w\right)^{3/2}}\int_{0}^{1}\frac{\mathrm{d}u}{\left(1-wu\right)^{2}\sqrt{u}}\\ &=\sqrt{1-w}\int_{0}^{1}\frac{\mathrm{d}u}{2\left(1-wu\right)^{2}\sqrt{u}}\\ &=\frac{\sqrt{1-w}}{\sqrt{w}}\int_{0}^{w}\frac{\mathrm{d}x}{2\left(1-x\right)^{2}\sqrt{x}};~~~\small{\left[wu=x\right]}\\ &=\frac{\sqrt{1-w}}{\sqrt{w}}\int_{0}^{\sqrt{w}}\frac{\mathrm{d}y}{\left(1-y^{2}\right)^{2}};~~~\small{\left[\sqrt{x}=y\right]}\\ &=\small{\frac{\sqrt{1-w}}{\sqrt{w}}\int_{0}^{\sqrt{w}}\left[\frac{1}{4\left(1-y\right)^{2}}+\frac{1}{4\left(1+y\right)^{2}}+\frac{1}{2\left(1-y^{2}\right)}\right]\,\mathrm{d}y};~~~\small{P.F.D.}\\ &=\frac{\sqrt{1-w}}{2\sqrt{w}}\left[\frac{y}{1-y^{2}}+\tanh^{-1}{\left(y\right)}\right]_{y=0}^{y=\sqrt{w}}\\ &=\frac{\sqrt{1-w}}{2\sqrt{w}}\left[\frac{\sqrt{w}}{1-w}+\tanh^{-1}{\left(\sqrt{w}\right)}\right]\\ &=\frac{\sqrt{1-w}}{2}\left[\frac{1}{1-w}+\frac{\tanh^{-1}{\left(\sqrt{w}\right)}}{\sqrt{w}}\right]\\ &=\frac{1}{2\sqrt{1-w}}+\frac{\sqrt{1-w}\,\tanh^{-1}{\left(\sqrt{w}\right)}}{2\sqrt{w}}.\\ \end{align}$$


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    $\begingroup$ This is more tham impressive ! $\endgroup$ Dec 29, 2016 at 9:45
  • $\begingroup$ Are you doing every single step by hand or do you use CAS to do some of the algebra? But nice work anyway (+1) $\endgroup$
    – tired
    Jan 3, 2017 at 12:34
  • $\begingroup$ @tired I use Wolfram Alpha whenever possible to speed things along, as well as to check for typographical errors. Without it, large-scale calculations would be all but impossible for me! ;) $\endgroup$
    – David H
    Jan 3, 2017 at 20:41

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