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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.

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up vote 7 down vote accepted

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.)

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Fantastic, thank you. – Potato May 25 '12 at 22:31
@anon: I believe you want to have $a<0$? – Fabian Jan 7 '13 at 18:06
@Fabian Oops, yes. – anon Jan 7 '13 at 18:14

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