Don't ask why - 17

For general integer $n \ge 1$ , note that the integral over the simplex $\big\{(x_1,x_2,\ldots,x_n)\,\big|\, x_1,x_2,\ldots,x_n > 0\,,\, x_1 + x_2 + \ldots + x_n = x\big\}\;,$ namely $g_n(x) \; = \; \int\int \cdots \int_{{x_1,x_2,\ldots,x_n > 0}\atop{x_1 + x_2 + \cdots + x_n = x}} \left(\prod_{j=1}^n \cos x_j\right)\,dx_1\,dx_2\,\ldots\,dx_{n-1}$ is equal to the $n$ -fold convolution $g_n \,=\, g^{\circ,n}$ of the function $g(x) \; =\; \cos x \;, \qquad \qquad x \ge 0$ with itself, where the convolution $f \circ g$ of two functions on $[0,\infty)$ is $(f \circ g)(x) \; = \; \int_0^x f(u) g(x-u)\,du \qquad \qquad x > 0 \;.$ If we consider the function $I_n(p) \; = \; \int_0^\infty \int_0^\infty \cdots \int_0^\infty \frac{\prod_{j=1}^n \cos x_j}{x_1 + x_2 + \cdots + x_n}\,e^{-p(x_1+x_2+\cdots + x_n)}\,dx_1\,dx_2\,\ldots\,dx_n$ for $p > 0$ , then it is clear that $I_n(p) \; = \; \int_0^\infty \frac{g_n(x)}{x}e^{-px}\,dx \qquad \qquad p > 0$ so that ( $\mathcal{L}$ denotes the Laplace transform) $I_n'(p) \; = \; -\int_0^\infty g_n(x)e^{-px}\,dx \; = \; -\big(\mathcal{L}g_n\big)(p) \; = \; -\big[(\mathcal{L}g)(p)\big]^n \; = \; -\frac{p^n}{(p^2+1)^n}$ for all $p > 0$ . Since it is clear that $I_n(p) \,\to\,0$ as $p \to \infty$ , we deduce that $I_n(p) \; = \; \int_p^\infty \frac{t^n}{(t^2+1)^n}\,dt \; = \; \frac{1}{(n-1)p^{n-1}}{}_2F_1\big(\tfrac{n-1}{2},n;\tfrac{n+1}{2};-p^{-2}\big) \;,$ and hence that $\lim_{p \to 0+} I_n(p) \; = \; \int_0^\infty \frac{t^n}{(t^2+1)^n}\,dt \; = \; \frac{\sqrt{\pi}\Gamma\big(\frac12(n-1)\big)}{2^n\Gamma\big(\frac12n\big)} \;.$ In the case $n=17$ , we obtain that $\lim_{p\to0+}I_{17}(p) \; = \; \frac{\sqrt{\pi}\Gamma(8)}{2^{17}\Gamma(\frac{17}{2})} \; = \; \frac{1}{205920}$ and hence the answer is $\boxed{205920}$ .