Evaluate $\int_{0<x_1,\cdots,x_n<1,\ 0<(x_1\cdots x_n)^{\frac{1}{n}}<a}dx_1\cdots dx_n$

$$1$$. How to prove that for $$n\in\mathbb{N}, a\in(0,1)$$ one have $$f(a,0):=\int_{0 This identity arises from probability theory, but I wonder if it's solvable using calculus alone.

$$2$$. Moreover, for $$p\in \mathbb{R}$$, can we give a closed form to the generalized $$f(a,p):=\int_{0 This one is rather open. Thanks in advance!

1. Let $$f_n(a)=\idotsint\limits_{\substack{0 for $$a>0$$ (for $$a<1$$ it is your $$f(a^{1/n},0)$$; for $$a\geqslant 1$$ it is $$1$$). Then $$x_n=x\implies(x_1\cdots x_n thus $$f_n(a)=\int_0^1 f_{n-1}(a/x)\,dx=a+\int_a^1 f_{n-1}(a/x)\,dx$$ for $$0. (Now induction works.)
2. This time, for $$g_n(a)=\idotsint\limits_{\substack{0 (and $$p>0$$), we get $$g_1(a)=\min\{a^{1/p},1\}$$ and $$g_n(a)=\int_0^{a}g_{n-1}(a-x^p)\,dx,$$ which doesn't look explicitly solvable ($$n=2$$ already looks clumsy).