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 

$$
I=\int_{-\infty}^\infty \exp{\big(\alpha x^4+\beta x^3+\gamma x^2 +\delta x+\epsilon}\big)dx, \ \alpha &lt;0.
$$
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 
$$
I=e^{\epsilon} \sum_{n,m,p=0} \frac{\beta^{4n}}{(4n)!}\frac{\gamma^{2m}}{(2m)!}\frac{\delta^{4p}}{(4p)!}\frac{\Gamma(3n+m+p+\frac{1}{4})}{\alpha^{3n+m+p+\frac{1}{4}}}
$$
OP mentions in a comment that the integral 

$$ \tag{1} I~:=~\int_{-\infty}^{\infty} \! dx~\exp \left(\alpha x^4+\beta x^3+\gamma x^2 +\delta x+\epsilon \right), \qquad \alpha ~&lt;~0,$$

appears on the Superstring theory Wikipedia page.
The $\epsilon$ dependence is trivial, so let's put $\epsilon=0$ in what follows.

Define 
$$ a~:=~\sqrt[4]{-\alpha}&gt;0, \qquad b~:=~\frac{\beta}{a^3}, \qquad c~:=~\frac{\gamma}{a^2}, \qquad d~:=~\frac{\gamma}{a}, $$ 
$$ \tag{2} y~:=~ax, \qquad t~:=~y^4. $$

Then 

$$\begin{align} I &amp;= \frac{1}{a}\int_{-\infty}^{\infty} \! dy~\exp \left(- y^4+by^3+cy^2 +dy\right)\\
&amp;=\frac{1}{a}\sum_{n,m,p\in \mathbb{N}_0}  \frac{b^n}{n!}\frac{c^m}{m!}\frac{d^p}{p!}\int_{-\infty}^{\infty} \! dy~y^{3n+2m+p}e^{- y^4}\\  
&amp;\stackrel{(4)}{=} \frac{1}{a}\sum_{k,n,m,p\in \mathbb{N}_0} \frac{b^n}{n!}\frac{c^m}{m!}\frac{d^p}{p!} ~\delta_{2k,3n+2m+p} ~\frac{1}{2} \Gamma\left(\frac{2k+1}{4}\right)\\
&amp;\tag{3}~=~ \frac{1}{2}  \sum^{n\equiv p \pmod 2}_{n,m,p\in \mathbb{N}_0}  \frac{\beta^n}{n!}\frac{\gamma^m}{m!}\frac{\delta^p}{p!} \frac{\Gamma\left(\frac{3n+2m+p+1}{4}\right)}{a^{3n+2m+p+1}}\\
\end{align}$$

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.
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 &lt;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 !