Editing Gaussian integral (section)

===Higher-order polynomials===

Applying a linear change of basis shows that the integral of the exponential of a homogeneous polynomial in ''n'' variables may depend only on [[SL(n)|SL(''n'')]]-invariants of the polynomial. One such invariant is the [[discriminant]],
zeros of which mark the singularities of the integral. However, the integral may also depend on other invariants.<ref name="morozov2009">{{cite journal | last1 = Morozov | first1 = A. | last2 = Shakirove | first2= Sh. | journal = Journal of High Energy Physics | pages = 002 | title = Introduction to integral discriminants | doi = 10.1088/1126-6708/2009/12/002 | volume = 2009 | year = 2009 | issue = 12 | arxiv = 0903.2595 | bibcode = 2009JHEP...12..002M }}</ref>

Exponentials of other even polynomials can numerically be solved using series. These may be interpreted as [[formal calculation]]s when there is no convergence. For example, the solution to the integral of the exponential of a quartic polynomial is{{citation needed|date=August 2015}}

<math display="block">\int_{-\infty}^{\infty} e^{a x^4+b x^3+c x^2+d x+f}\,dx
= \frac{1}{2} e^f \sum_{\begin{smallmatrix}n,m,p=0 \\ n+p=0 \bmod 2\end{smallmatrix}}^{\infty} \frac{b^n}{n!} \frac{c^m}{m!} \frac{d^p}{p!} \frac{\Gamma{\left (\frac{3n+2m+p+1}{4} \right)}}{{\left(-a\right)}^{\frac{3n+2m+p+1}4}}.</math>

The {{math|1=''n'' + ''p'' = 0}} mod 2 requirement is because the integral from −∞ to 0 contributes a factor of {{math|(−1)<sup>''n''+''p''</sup>/2}} to each term, while the integral from 0 to +∞ contributes a factor of 1/2 to each term. These integrals turn up in subjects such as [[quantum field theory]].