TITLE: Computing the integrals of the form $\exp(P(x))$, $P(x)$ a polynomial
QUESTION [12 upvotes]: Wikipedia notes that 

Exponentials of other even polynomials can easily be solved using series.  For example the solution to the integral of the exponential of a quartic polynomial is:
\begin{align}
& \int_{-\infty}^{\infty} e^{a x^4+b x^3+c x^2+d x+f}\,dx \\
& {} \quad =
    \frac12
    e^f \!\!\!\!\!\!\!\!
    \sum_{\begin{smallmatrix}n,m,p=0 \\ n+p=0 \mod 2\end{smallmatrix}}^{\infty} \!\!\!\!
    \frac{b^n}{n!}
    \frac{c^m}{m!}
    \frac{d^p}{p!}
    \frac{\Gamma(\frac{3n+2m+p+1}4)}{(-a)^{\frac{3n+2m+p+1}4}}.
\end{align}

However, Wikipedia does not provide a citation. Could someone give a reference where I could find out more about  the evaluating of such integrals and the series methods mentioned in the article? Thanks.

REPLY [11 votes]: Hint: Assume $a<0$. We have
$$\int_{-\infty}^\infty e^{\large ax^4+bx^3+cx^2+dx+f}dx=e^f\int_{-\infty}^\infty\left(\sum_{n=0}^\infty \frac{(bx^3)^n}{n!}\right)\left(\sum_{m=0}^\infty\frac{(cx^2)^m}{m!}\right)\left(\sum_{p=0}^\infty \frac{(dx)^p}{p!}\right)e^{ax^4}dx $$
$$=\int_{-\infty}^\infty \sum_{n,m,p\ge0}^{n+m\equiv0(2)}\frac{b^nc^md^p}{n!m!p!}x^{3n+2m+p}e^{ax^4}dx $$
because the powers with $n+p\equiv1\bmod2$ contribute odd functions to the integrand, which vanish when integrated over the real line. From here, interchange summation and integration, use the substitution $u=x^4$, then dilate by $-a$ appropriately...
(So basically, the techniques in play here are: series expansion, self-cancellation of odd functions, interchange of limiting operations, and power/linear substitutions.)