I would like help approximating the surface integral $$\int_{S^{n-1}_{\ge 0}}\frac{1}{\hat n\cdot p}dS$$ where $\hat n$ is the unit normal to the sphere at the given point, $p\in\mathbb{R}^n$ is a fixed vector with all positive components, and $S^{n-1}_{\ge 0}$ is the portion of the sphere in $\mathbb{R}^n$ with all positive components.
The problem has arisen in the following manner. Suppose that one has a budget dependent on time $v(t)$ in a market of $n$ items each with a positive-price varying over time given by the vector $p(t)=\langle p_1,\dots,p_n\rangle$. One way to consider a "purchasing power" would be the number of ways to purchase different amounts of these items, that is, $$\text{PP}(t)=\#\{x\in\mathbb{N}^n \; |\; x\cdot p(t)\le v(t)\}$$ To approximate this, we want to "undescretize" it. Assuming both to be evenly distributed, the number of such integer-component $x$ ought to be proportional to the volume of the set when extended to real valued components, that is, $$\text{PP}(t)\sim |\{x\in\mathbb{R}_{\ge 0}^n\; |\; x\cdot p(t)\le v(t)\}|$$ Suppose now that some $x_0$ satisfies $x_0\cdot p(t)\le v(t)$, then considering all vectors in the direction of $x_0$ that also satisfy the inequality we get a line of length $v(t)/x_0\cdot p(t)$. Thus this volume can be well approximated by integrating the lengths of these lines over all possible directions resulting in $$\text{PP}(t)\sim v(t)\int_{S^{n-1}_{\ge 0}}\frac{1}{\hat n\cdot p}dS$$ The best I could do to approximate this integral was to consider the following, $$\int_{S^{n-1}_{\ge 0}}\frac{1}{\hat n\cdot p}dS=\frac{1}{\|p\|}\int_{S^{n-1}_{\ge 0}}\frac{1}{\cos\theta}dS$$ where $\theta$ is the angle between $p$ and the current normal $\hat n$. To approximate this I decided to integrate with respect to $\theta$ sweeping outwards on the sphere by the angle from $p$ starting at $p$ where $\theta = 0$. If we let $\theta_0=\min\{\angle(e_1,p),\dots, \angle(e_n,p)\}$ clearly we can sweep outwards to $\theta_0$ without leaving the bounds of $S^{n-1}_{\ge 0}$, however moving past this angle from $p$ will leave these bounds. However, I figure that (possibly incorrectly) that the contribution at a given angle will be roughly proportional to the fraction of the axes within that angle to $p$. That is $$\frac{1}{\|p\|}\int_{S^{n-1}_{\ge 0}}\frac{1}{\cos\theta}dS\sim \int_{0}^{\pi/2}\frac{\#\{e_i \; |\; \angle(e_i, p)\le \theta\}}{n}\cdot \frac{1}{\cos \theta}\cdot \frac{2 \pi^{n/2}}{\Gamma(n/2)}\theta^{n-1}d\theta$$ where the upper bound of $\pi/2$ was relatively arbitrary simply because $\#\{e_i \; |\; \angle(e_i, p)\le \theta\}$ is guaranteed to be zero for $\theta$ any larger. Since we know that $\angle(e_i,p)=\arccos(p_i/\|p\|)$ we can simplify this integral to $$\begin{aligned} &\int_{0}^{\pi/2}\frac{\#\{e_i \; |\; \angle(e_i, p)\le \theta\}}{n}\cdot \frac{1}{\cos \theta}\cdot \frac{2 \pi^{n/2}}{\Gamma(n/2)}\theta^{n-1}d\theta \\ =&\frac{2\pi^{n/2}}{n\Gamma(n/2)}\sum_{i=1}^n\int_{0}^{\arccos(p_i/\|p\|)}\theta^{n-1}\sec\theta \;d\theta \end{aligned}$$ All in all I thus feel as if (although I lack a fully formal proof) that $$\text{PP}(t)\sim \frac{v(t)}{\|p(t)\|}\cdot\frac{2\pi^{n/2}}{n\Gamma(n/2)}\sum_{i=1}^n\int_{0}^{\arccos(p_i/\|p\|)}\theta^{n-1}\sec\theta \;d\theta$$ but this is the best I can do. I can't seems to simplify or approximate that final integral either. I could possibly use a series expansion of $\sec(x)$ and term-wise integrate.
You want the volume of the set $\{x\in\mathbb{R}_{\ge 0}^n\; |\; x\cdot p\le v\}$. This is the volume of $\mathbb{R}_{\ge 0}^n$ defined by the coordinate planes and the plane $x\cdot p=v$. This is a simplex with coordinate axis intercepts $x_i=v/p_i$. Thus it has the volume $1/n!$ of the unit simplex scaled by $\prod_iv/p_i$.