Evaluating the definite integral $\int_0^\infty \mathrm{e}^{\sum_ia_ix^{n_i}}\mathrm{d}x $ As a follow-up to a question on evaluating the definite integral $\int_0^\infty \mathrm{e}^{-x^n}\,\mathrm{d}x$, I wish to know if there is a general analytic solution to the related integral where $-x^n$ is replaced by a polynomial of arbitrary degree, namely $$ \int_0^\infty \mathrm{e}^{\sum_ia_ix^{n_i}}\mathrm{d}x $$ for $n_i\in\mathbb{Z}$ and where the individual coefficients $a_i\in\mathbb{R}$ can be positive or negative, but in a manner such that the argument of the exponent (i.e,. the polynomial) is negative.
ALL the $ a_{i} $ are negative aren't they ??.. otherwise the integral would be DIVERGENT

from the property of the exponential $ f(a+b)=f(a)f(b) $ and the fact that

$ \int_{0}^{\infty}dxexp(-x^{n}) = \int_{0}^{\infty}duu^{1/n-1}exp(-u)= \frac{1}{n}\Gamma(1/n)$

so your itnegral will be equal to the product $ \prod _{i}\frac{1}{a_{i}^{1/n_{i}}}\Gamma(1/n_{i}) $ or similar.. :)
ALL the $ a_{i} $ are negative aren't they ??.. otherwise the integral would be DIVERGENT

from the property of the exponential $ f(a+b)=f(a)f(b) $ and the fact that

$ \int_{0}^{\infty}dxexp(-x^{n}) = \int_{0}^{\infty}duu^{1/n-1}exp(-u)= \frac{1}{n}\Gamma(1/n)$

so your itnegral will be equal to the product $ \prod _{i}\frac{1}{a_{i}^{1/n_{i}}}\Gamma(1/n_{i}) $ or similar.. :)