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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.

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Well, if $p(x)$ is a polynomial, then so is $p(x-a)$ for any $a$, so $\int_0^\infty exp(p(x-a))dx=\int_a^\infty exp(p(x))dx$ is asking for the INDEFINITE integral of $\exp(p(x))$. – GEdgar Feb 18 '12 at 13:32

1 Answer 1

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.. :)

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Thanks Jose. I suppose you could have $\exp(x-x^2)$. This polynomial is negative everywhere but $[0,1]$. Would the integral diverge with such a polynomial? – user001 Feb 18 '12 at 11:07
@Jose: since when $\int\!dx\,f_1(x) f_2(x) = [\int\!dx\,f_1(x)] [\int\!dx\,f_2(x)]$??? – Fabian Feb 18 '12 at 13:12
@Jose: and your first statement is also wrong... (see the comment by user001) – Fabian Feb 18 '12 at 13:14
oh yes it is wrong :( i thought he was trying a multivariable integral :( sorry :'( i apologize , in case you a ve a Polynomial there is NO method to evaluate it.. unless you use stationary phaes method. – Jose Garcia Feb 18 '12 at 15:18
You could delete the a matter of housekeeping? – daniel Jul 6 '13 at 14:41

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