Let $b$ and $a_i, i = 1,2,3,..., n$ be positive real numbers such that
$$ \int_{0}^{\infty}\int_{0}^{\infty}\ldots \int_{0}^{\infty} e^{-b x_1^{a_1}x_2^{a_2}\ldots x_1^{a_2}} dx_1 dx_2 \ldots dx_n = 1. $$
I am looking for an expression of $b$ in terms of $a_i$.
I have obtained the solution for $i=1$ which is as follows. If
$$ \int_{0}^{\infty} e^{-b x^a} dx = 1 $$ then $$ b = \Gamma \Big(1+\frac{1}{a}\Big)^a. $$
Thus for all $a \ge 0$, $$ \int_{0}^{\infty}e^{-\Gamma(1+\frac{1}{a})^a x^a} = 1. $$
I'm not sure under what conditions this integral even converges. Let's write the integral as follows:
$$\int_0^{\infty} dx_1 \, \int_0^{\infty} dx_2 \,\cdots \int_0^{\infty} dx_n \,e^{-b x_1^{a_1} x_2^{a_2}\cdots x_n^{a_n}}$$
Note that, in evaluating the integral over $x_n$, we may treat everything else as a constant, so you end up with an integral much like the one you already evaluated. That is, consider the result
$$\int_0^{\infty} dx \, e^{-b x^a} = b^{-1/a} \Gamma\left(1+\frac{1}{a}\right)$$
Then the above multiple integral may be written as
$$b^{-1/a_n} \Gamma\left(1+\frac{1}{a_n}\right) \int_0^{\infty} dx_1 \,x_1^{-a_1/a_n}\, \int_0^{\infty} dx_2 \,x_2^{-a_2/a_n}\,\cdots \int_0^{\infty} dx_{n-1} \, x_{n-1}^{-a_{n-1}/a_n}$$
It is unclear to me how any of the above integrals will converge.