Integral, Definite Integral $ \int_{-\infty}^\infty \exp{\big(\alpha x^4+\beta x^3+\gamma x^2 +\delta x+\epsilon}\big)dx, \ \alpha <0. $ Calculate  the integral 
$$
I=\int_{-\infty}^\infty \exp{\big(\alpha x^4+\beta x^3+\gamma x^2 +\delta x+\epsilon}\big)dx, \ \alpha <0.
$$
The answer can be expressed analytically in terms of a series which will have a $\Gamma$ function in it.  I am not sure how to approach it since I do not know how to work with the argument of the exponential.  The other constants we can assume to be real.  The answer can be expressed as 
$$
I=e^{\epsilon} \sum_{n,m,p=0} \frac{\beta^{4n}}{(4n)!}\frac{\gamma^{2m}}{(2m)!}\frac{\delta^{4p}}{(4p)!}\frac{\Gamma(3n+m+p+\frac{1}{4})}{\alpha^{3n+m+p+\frac{1}{4}}}
$$
 A: OP mentions in a comment that the integral 
$$ \tag{1} I~:=~\int_{-\infty}^{\infty} \! dx~\exp \left(\alpha x^4+\beta x^3+\gamma x^2 +\delta x+\epsilon \right), \qquad \alpha ~<~0,$$
appears on the Superstring theory Wikipedia page.
The $\epsilon$ dependence is trivial, so let's put $\epsilon=0$ in what follows.
Define 
$$ a~:=~\sqrt[4]{-\alpha}>0, \qquad b~:=~\frac{\beta}{a^3}, \qquad c~:=~\frac{\gamma}{a^2}, \qquad d~:=~\frac{\gamma}{a}, $$ 
$$ \tag{2} y~:=~ax, \qquad t~:=~y^4. $$
Then 
$$\begin{align} I &= \frac{1}{a}\int_{-\infty}^{\infty} \! dy~\exp \left(- y^4+by^3+cy^2 +dy\right)\\
&=\frac{1}{a}\sum_{n,m,p\in \mathbb{N}_0}  \frac{b^n}{n!}\frac{c^m}{m!}\frac{d^p}{p!}\int_{-\infty}^{\infty} \! dy~y^{3n+2m+p}e^{- y^4}\\  
&\stackrel{(4)}{=} \frac{1}{a}\sum_{k,n,m,p\in \mathbb{N}_0} \frac{b^n}{n!}\frac{c^m}{m!}\frac{d^p}{p!} ~\delta_{2k,3n+2m+p} ~\frac{1}{2} \Gamma\left(\frac{2k+1}{4}\right)\\
&\tag{3}~=~ \frac{1}{2}  \sum^{n\equiv p \pmod 2}_{n,m,p\in \mathbb{N}_0}  \frac{\beta^n}{n!}\frac{\gamma^m}{m!}\frac{\delta^p}{p!} \frac{\Gamma\left(\frac{3n+2m+p+1}{4}\right)}{a^{3n+2m+p+1}}\\
\end{align}$$
where we used that 
$$\tag{4}  2\int_{0}^\infty \! dy~y^{2k} e^{- y^4}
~=~\frac{1}{2}\int_{0}^{\infty}  \! dt~t^{\frac{2k-3}{4}}e^{-t}~=~\frac{1}{2}\Gamma\left(\frac{2k+1}{4}\right).$$
The formula (3) is not the formula from the Superstring theory Wikipedia page. Without having checked the physics behind the integral $I$, it seems that there might be an implicit procedure assumed, where one e.g. should somehow replace $a\to \omega a$, where $\omega =\sqrt[4]{1}\in\{1,i,-1,-i\}$ is a fourth root of unity, and then somehow average over these four roots. Or something along these lines. We do not plan to pursuit or check such loose speculations here.
A: why not use the definition of $\Gamma$ function ?
you can use the  $\Gamma$ function at first :
$
I=e^{\epsilon} \sum_{n,m,p=0} \frac{\beta^{4n}}{(4n)!}\frac{\gamma^{2m}}{(2m)!}\frac{\delta^{4p}}{(4p)!}\frac{\Gamma(3n+m+p+\frac{1}{4})}{\alpha^{3n+m+p+\frac{1}{4}}}
$$=e^{\epsilon} \sum_{n,m,p=0} \frac{\beta^{4n}}{(4n)!}\frac{\gamma^{2m}}{(2m)!}\frac{\delta^{4p}}{(4p)!}\frac{\int{e^{-t}t^{{3n+m+p+\frac{1}{4}}-1}dt}}{\alpha^{3n+m+p+\frac{1}{4}}}
$
then, you can expand the follwoing two series :


*

*$ \exp{\big(\alpha x^4+\beta x^3+\gamma x^2 +\delta x+\epsilon}\big), \ \alpha <0$

*${e^{-t}t^{{3n+m+p+\frac{1}{4}}-1}}$
and compare their value nextly, a point you should be careful is that the integral doman of the two formulas above are different !  
the hint is to assume :
$y=\alpha^{1/4}x$
then, $ \exp{\big(\alpha x^4+\beta x^{3}+\gamma x^2 +\delta x}\big)$$ =\exp{\big(y^4+\frac{\beta y^3}{\alpha^{3/4}}+\frac{\gamma y^2}{\alpha^{1/2}} +\frac{\delta y}{\alpha^{1/4}}}\big)$
therefore, we can let $n=1/4$$,$$m=1/2$$,$$p=1/4$
which lead that :
$ =\exp{\big(y^4+\frac{\beta y^3}{\alpha^{3/4}}+\frac{\gamma y^2}{\alpha^{1/2}} +\frac{\delta y}{\alpha^{1/4}}}\big)=$
$ =\exp{\big(y^4+\frac{(\beta^{1/4}y^{3/4})^{4}}{\alpha^{3/4}}+\frac{({\gamma^{1/2} y})^2}{\alpha^{1/2}} +\frac{（\delta^{1/4} y^{1/4})^{4}}{\alpha^{1/4}}}\big)$
consequently, we can put the following terms out :
$\sum_{n,m,p=0} \frac{\beta^{4n}}{(4n)!}\frac{\gamma^{2m}}{(2m)!}\frac{\delta^{4p}}{(4p)!}\frac{1}{\alpha^{3n+m+p+\frac{1}{4}}}$
then we can make our assumption :
$z=(ay)^{4}$
so the only thing you should compare is the relation below :
${e^{-t}t^{{3n+m+p+\frac{1}{4}}-1}}$$=e^{-t}{t^{3/4}}=e^{zy+z^{3/4}+z^{1/2}+z^{1/4}}$$=exp$$(z^{3/4}(z^{1/4}+1+z^{-1/4}+z^{-1/2}))$$\Longrightarrow$$(z^{3/4}(z^{1/4}+1++z^{-1/4}+z^{-1/2}))$$=e^{-t}{t^{3/4}}-1$
$\Longrightarrow$$\infty=\infty$
the condition is :
$t,z\longrightarrow\epsilon$$\Longrightarrow$$t^{3/4}=z^{3/4}$$\Longrightarrow$$t^{-3/4}=z^{-1/4-1/2}$
then your solution is right , thank you !
