Gamma Type Integral I was hoping someone could help me with a question I came across recently: essentially it's a gamma type integral that your asked to evaluate/reduce: 
P=$\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-\frac{x^{2}}{2}-\frac{cx^{4}}{4}}dx$ 
where c is a constant. The way your asked to evaluate it is to reduce the integrand using a taylor expansion to order 1 for the exponential function and then use the fact that 
$\int_{-\infty}^{\infty}e^{-\frac{x^{2}}{2}}dx=\sqrt{2\pi}$
I can't come to any plausible solutions to this problem. I mean you could say that 
$e^{-\frac{x^{2}}{2}-\frac{cx^{4}}{4}}\approx1-\frac{x^{2}}{2}-\frac{cx^{4}}{4}+\dots$
but to order 1 this would just result in the integrand becoming 1 and this doesn't make sense? If the question said using the exponential to order 2 then the integral would evaluate to $\sqrt{2\pi}$ and thus P itself would be 1 but my thoughts are that the reduced P is wanted in terms of c? Can someone please provide some guidance or a possible way to reduce P. Thanks very much.
 A: Note that this integral converges when $\text{Re}(c)\geq0$ .
When $c=0$ , the question is provided.
I only able to solve the case of positive real number $c$ :
$\dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{x^2}{2}-\frac{cx^4}{4}}~dx$
$=\dfrac{1}{\sqrt{2\pi}}\left(\int_{-\infty}^0e^{-\frac{x^2}{2}-\frac{cx^4}{4}}~dx+\int_0^\infty e^{-\frac{x^2}{2}-\frac{cx^4}{4}}~dx\right)$
$=\dfrac{1}{\sqrt{2\pi}}\left(\int_\infty^0e^{-\frac{(-x)^2}{2}-\frac{c(-x)^4}{4}}~d(-x)+\int_0^\infty e^{-\frac{x^2}{2}-\frac{cx^4}{4}}~dx\right)$
$=\dfrac{1}{\sqrt{2\pi}}\left(\int_0^\infty e^{-\frac{x^2}{2}-\frac{cx^4}{4}}~dx+\int_0^\infty e^{-\frac{x^2}{2}-\frac{cx^4}{4}}~dx\right)$
$=\dfrac{\sqrt2}{\sqrt\pi}\int_0^\infty e^{-\frac{x^2}{2}-\frac{cx^4}{4}}~dx$
$=\dfrac{\sqrt2}{\sqrt\pi}\int_0^\infty e^{-\frac{x^2(2+cx^2)}{4}}~dx$
$=\dfrac{\sqrt2}{\sqrt\pi}\int_0^\infty e^{-\frac{\bigl(\frac{\sqrt2\sinh x}{\sqrt c}\bigr)^2\bigl(2+c\bigl(\frac{\sqrt2\sinh x}{\sqrt c}\bigr)^2\bigr)}{4}}~d\biggl(\dfrac{\sqrt2\sinh x}{\sqrt c}\biggr)$
$=\dfrac{2}{\sqrt{c\pi}}\int_0^\infty e^{-\frac{\sinh^2x(2+2\sinh^2x)}{2c}}\cosh x~dx$
$=\dfrac{2}{\sqrt{c\pi}}\int_0^\infty e^{-\frac{\sinh^2x\cosh^2x}{c}}\cosh x~dx$
$=\dfrac{2}{\sqrt{c\pi}}\int_0^\infty e^{-\frac{\sinh^22x}{4c}}\cosh x~dx$
$=\dfrac{2}{\sqrt{c\pi}}\int_0^\infty e^{-\frac{\cosh4x-1}{8c}}\cosh x~dx$
$=\dfrac{e^\frac{1}{8c}}{2\sqrt{c\pi}}\int_0^\infty e^{-\frac{\cosh4x}{8c}}\cosh x~d(4x)$
$=\dfrac{e^\frac{1}{8c}}{2\sqrt{c\pi}}\int_0^\infty e^{-\frac{\cosh x}{8c}}\cosh\dfrac{x}{4}dx$
$=\dfrac{e^\frac{1}{8c}}{2\sqrt{c\pi}}K_\frac{1}{4}\left(\dfrac{1}{8c}\right)$
In fact this is a special case of the result of $\int_0^\infty e^{-(ax^2+b)(cx^2+d)}~dx$ provided in http://pi.physik.uni-bonn.de/~dieckman/IntegralsDefinite/DefInt.html.
Please feel free to edit my answer if you are able to handing the case of complex number $c$ .
A: This is not an elementary integral and I do not know how it can be on a math quiz! Here is a solution

$${\frac {{{\rm e}^{1/(8c)}}{{\rm K}_{1/4}\left(1/(8c)\right)}}{2\sqrt {c\,
\pi }}},$$

where $K_a(x)$ is the modified Bessel function of the second kind.
