Exponential of a Quartic I know there is a related post regarding this, but does anyone know a 'closed' form solution for the integral
\begin{equation}I=\int_{-\infty}^{\infty} dx\,e^{ax-bx^{4}}\end{equation}
I know you can do a series expansion of the quartic term, but I'd like to find a way that avoids that. The integral should converge as $e^{-bx^{4}}$ dominates at large $x$ and so the integrand quickly goes to zero (for $b>0$ obviously). Any help/ references would be greatly appreciated. 
 A: Integrating by parts leads to $$I=\left.x\cdot e^{ax-bx^4}\right\vert_{-\infty}^\infty-\int_{-\infty}^\infty x(a-4bx^3)e^{ax-bx^4}\,dx=-a\frac{\partial I}{\partial a}-4b\frac{\partial I}{\partial b},$$but $\,\partial I/\partial b=-\int_{-\infty}^\infty x^4e^{ax-bx^4}\,dx\;$ also equals $-\partial^4I/\partial a^4$, so $I(a)+aI'(a)=(aI(a))'=4bI^{(4)}(a)$, or $\;aI(a)=C+4bI'''(a)$. Since $I(a)$ is even (and entire, of course), $I'(0)$ and $I'''(0)=0$ and so $C=0$; and using the integral formula for the gamma function, $$I(0)=2\int_0^\infty e^{-bx^4}\,dx=2\int_0^\infty e^{-y}b^{-1/4}\frac{dy}{4y^{3/4}}=2^{-1}b^{-1/4}\Gamma(1/4)$$ and $$I''(0)=2\int_0^\infty x^2e^{-bx^4}\,dx=2\int_0^\infty b^{-1/2}y^{1/2}e^{-y}\frac{b^{-1/4}dy}{4y^{3/4}}=2^{-1}b^{-3/4}\Gamma(3/4)$$I plugged all of this into Wolfram|Alpha and got the same result as @GEdgar (with $x,a$ replacing, respectively, $a,b$).
A: Maple has this...in terms of hypergeometric ${}_0F_2$ functions:
$$
\int _{-\infty }^{\infty }\!{{\rm e}^{-b{x}^{4}+ax}}{dx} =
\frac{\displaystyle  
2\,\sqrt {2}\pi \,
{\mbox{$_0$F$_2$}\left(\ ;\,1/2,3/4;\,{\frac {1}{256}}\,{\frac {{a}^{4}}{b}}\right)}
\sqrt {b}+  \Gamma  \left( 3/4 \right)   ^{2}{a}^{2}
{\mbox{$_0$F$_2$}\left(\ ;\,5/4,3/2;\,{\frac {1}{256}}\,{\frac {{a}^{4}}{b}}\right)}
 }{4 {b}^{3/4}  \Gamma  \left( 3/4 \right)  }
$$
A: This is called Gaussian, it is not integrable by elementary means. First you need to make the substitution $a \rightarrow ax$ so that the integral is easier. Now the integral is $$\int _{-\infty} ^{\infty} e^{ax^{2} -b x^{4}} dx$$
Now this is an easier integral if you make the substitution $x \rightarrow \sqrt{x}$ because the integral is now 
$$\int _{-\infty} ^{\infty} e^{ax^ - b x^{2}} dx$$
Ok, now we are really on track, because you can complete the square. Rewrite the integral as 
$$\int _{-\infty} ^{\infty} e^{-b(x-\frac{a}{2b})^{2}}e^{-(ax)^2} dx$$
Now we are almost done, just pull out the last factor to get 
$$e^{-(ax)^2} \int _{-\infty} ^{\infty} e^{-b(x-\frac{a}{2b})^{2}} dx = e^{-(ax)^2} \sqrt{\pi}$$ 
Now we have to evaluate somewhere because this is a function of x whereas the original integral is not. I picked 0, but any other value should work. So, the answer is $\sqrt{\pi}$.
