Hot Integral - 26

Calculus Level 3

Let $\mathcal{J}$ is the expected distance of a random point from any vertex of the unit tesseract. Define

$\large \large \mathcal{I}=\int\limits_{0}^{1} \frac{\tanh^{-1}\left(\frac{1}{\sqrt{3+x^2}}\right)}{1+x^2} dx$

We have

$\displaystyle \mathcal{I}-5\mathcal{J} = A\sqrt{B}\tan^{-1} \sqrt{C}+\frac{\pi}{D}\log(E+\sqrt{F})-\frac{G}{H}\pi\sqrt{K}-L\log M -N,$

where $G,H$ are coprime numbers, $F$ is a prime number and $K$ is square free.

Calculate $A+B+C+D+E+F+G+H+K+L+M+N$ .

The answer is 41.

1 solution

Mark Hennings
Aug 14, 2016

Someone has solved this!

In the notation of the seminal paper on the topic of box integrals, the expected distance $\mathcal{J}$ from any vertex of the unit tesseract (four-dimensional cube) is $\mathcal{J} = B_4(1)$ , and $\mathcal{J} = B_4(1) \; = \; \begin{array}{c}[t]{c}\tfrac25 + \tfrac{7}{20}\pi\sqrt{2} - \tfrac{1}{20}\pi\ln(1 + \sqrt{2})\\ + \ln3 - \tfrac75\sqrt{2}\tan^{-1}\sqrt{2} + \tfrac1{10}\mathcal{K}_0\end{array}$ where $\mathcal{K}_0 = 2\int_0^1 \frac{\tanh^{-1}\frac{1}{\sqrt{3+y^2}}}{1+y^2}\,dy \; = \; 2\mathcal{I}$ (although the factor of $2$ is dropped in later papers). Thus $\mathcal{I} - 5\mathcal{J} = 7\sqrt{2}\tan^{-1}\sqrt{2} + \tfrac14\pi\ln(1+\sqrt{2}) - \tfrac74\pi\sqrt{2} - 5\ln3 - 2$ which makes the answer $7+2+2+4+1+2+7+4+2+5+3+2 = \boxed{41}$ .

Hot Integral - 26

The answer is 41.

1 solution

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