Let $0\le r \le 1$. By generalising A definite integral involving a logarithm and trigonometric functions. we consider a following integral: \begin{equation} \mu_d(r):=\int\limits_{||\vec{x}||=1} \log\left[1+r \cdot \sum\limits_{j=1}^d x_j \right] d S \end{equation} here $dS$ is a surface element of the $d$-dimensional sphere. This quantity can be viewed as an expectation of the log of wealth of a fund placing a constant bet $r$ on a cumulative random walk process with iid uniformly distributed returns subject to sum of the squared returns being equal to unity. Note that since the function $\log(1+\cdot)$ is not symmetric with respect to the origin the quantity above will always be negative. In other words such a fund is supposed to be loosing money on average. Now by using spherical coordinates and the expansion of the logarithm we found the following result. \begin{equation} \mu_d(r) = -\pi^{d/2} \sum\limits_{n=1} \frac{(\frac{d}{4} r^2)^n}{n \cdot n!} \cdot \frac{(2 n)!}{(n+\frac{d-2}{2})!} \end{equation} It turns out that the series above have neat closed form expressions. In fact we have: \begin{eqnarray} \mu_2(r) &=& 2 \pi \log \left(\frac{1}{2} \left(\sqrt{1-2 r^2}+1\right)\right) \\ \mu_3(r) &=& \frac{2 \pi \left(3 r \log \left(1-3 r^2\right)-6 r+2 \sqrt{3} \tanh ^{-1}\left(\sqrt{3} r\right)\right)}{3 r} \\ \mu_4(r) &=& \frac{\pi ^2 \left(-\sqrt{1-4 r^2}+2 r^2 \left(\log \left(r^2\right)+2 \tanh ^{-1}\left(\sqrt{1-4 r^2}\right)-1\right)+1\right)}{2 r^2} \end{eqnarray} Now, my question is twofold. Firstly, can we find closed form expressions for the quantity in question for arbitrary values of $d$. Secondly how do we compute the following quantity: \begin{equation} \Theta_d := \int\limits_0^\infty \mu_d(r) r^{d-1} \frac{e^{-\frac{r^2}{2}}}{(2 \pi)^{d/2}} dr \end{equation} meaning the expected logarithm of wealth over all possible Gaussian distributed random walks.
Surface integral over a $d$-dimensional unit sphere
1 Answers
Here we address the first question namely we derive closed form expressions for the quantity $\mu_d(r)$. We need to distinguish two cases of $d$ being either even or odd. We have: \begin{equation} \mu_d(r) = \left\{ \begin{array}{rr} -(\sqrt{\pi})^{d-1} \cdot \sum\limits_{n=1}^\infty \frac{(-d r^2)^n}{n^{(\tilde{d})}} \cdot \binom{-\frac{1}{2}}{n} & \mbox{if $d=2 \tilde{d}$}\\ -(\sqrt{\pi})^{d-1} \cdot \sum\limits_{n=1}^\infty \frac{(d r^2)^n}{(n+\frac{1}{2})^{(\tilde{d})}} \cdot \frac{1}{n} & \mbox{if $d=2\tilde{d}+1$} \end{array} \right. \end{equation} Now we use the integral representation of the beta function to eliminate the inverses of the Pochammer symbols. This leads to the following results: \begin{eqnarray} &&\mu_{2 \tilde{d}}(r) = -\frac{2 \pi^{\tilde{d}}}{(\tilde{d}-1)!} \cdot \\ && \left( -\log\left[\frac{1}{2}(1+\sqrt{1-d r^2}\right] + \sum\limits_{p=1}^{\tilde{d}-1} \binom{\tilde{d}-1}{p} (\frac{-2}{\tilde{d} r^2})^p \cdot (\beta(p+1,p) - \beta_{\frac{1}{2}(1+\sqrt{1-d r^2}}(p+1,p))) \right)\\ &&\mu_{2 \tilde{d}+1}(r) = \frac{\pi^{d-1}}{(\tilde{d}-1)!} \cdot \\ &&\sum\limits_{p=0}^{\tilde{d}-1} \binom{\tilde{d}-1}{p} \frac{(-1)^p}{(d r^2)^{p+1/2}} \cdot \left(\frac{(d r^2)^{p+1/2}}{p+1/2} \log(1-d r^2) + \beta_{d r^2} \left(p+\frac{3}{2},0\right)\right) \end{eqnarray} where $\beta_x(a,b)$ is the incomplete beta function.
After some simplifications we write the result as follows: \begin{eqnarray} &&\mu_{2 \tilde{d}}(r) = -\frac{2 \pi^{\tilde{d}}}{(\tilde{d}-1)!} \cdot \\ &&\left( -\log(X) + \sum\limits_{p=1}^{\tilde{d}-1} \binom{\tilde{d}-1}{p} (-Y)^p \frac{p!(p-1)!}{(2 p)!} - \sum\limits_{P=1}^{2 \tilde{d}-2} \sum\limits_{p=P/2}^{P \vee (\tilde{d}-1)} \binom{\tilde{d}-1}{p} \binom{p}{P-p} Y^p (-X)^P \frac{1}{P} \right) \end{eqnarray} where $X := \left(1+\sqrt{1-2 \tilde{d} r^2}\right)/2$ and $Y:=2/(\tilde{d} r^2)$. Likewise: \begin{eqnarray} &&\mu_{2 \tilde{d}+1}(r) = \frac{\pi^{\tilde{d}}}{(\tilde{d}-1)!}\cdot \\ &&\left(\log\left(1-X\right) \frac{\sqrt{\pi} (-1+\tilde{d})!}{(-1/2+\tilde{d})!} + 2 \log(\frac{1+\sqrt{X}}{1-\sqrt{X}}) \sum\limits_{p=0}^{\tilde{d}-1} \binom{\tilde{d}-1}{p} \frac{(-1)^p}{(2 p+1) X^{p+1/2}}\right. \\ && \left. -4 \sum\limits_{p=0}^{\tilde{d}-1} \sum\limits_{j=1}^{\tilde{d}-p} \binom{\tilde{d}-1}{p+j-1} \frac{(-1)^{p+j-1}}{(2(p+j-1)+1)(2 j-1)} \frac{1}{X^p} \right) \end{eqnarray} where $X := d r^2$.