I have the following function
$$ L(T) = N \frac{\int_{-\infty}^{\infty}ye^{- v(y)/kT}\text{d}y}{\int_{-\infty}^{\infty}e^{- v(y)/kT}\text{d}y} $$
where
$$ v(y) = \frac{k_0}{2}(y-a)^2-\gamma (y-a)^3 $$
with $\gamma > 0$
I am supposed to approximate this using a Taylor series expansion to first order in $\gamma$, which I think means that I expand around $y=\gamma$. I am supposed to end up with
$$ L(T) \approx N \left[a + \frac{3\gamma kT }{ k_0^2}\right] $$
Setting $f(y)= \frac{L(T)}{N}$, I get that
$$ f(y) = \frac{\int_{-\infty}^{\infty} ye^{-\frac{k_0 }{2kT} (y-a)^2 + \frac{\gamma}{kT}(y-a)^3}\text{d}y}{\int_{-\infty}^{\infty}e^{-\frac{k_0 }{2kT} (y-a)^2 + \frac{\gamma}{kT}(y-a)^3}\text{d}y} $$
The first order Taylor series expansion will be
$$ f(y) \approx f(\gamma) + \frac{\text{d} f(\gamma)}{\text{d}\gamma}(y-\gamma) $$
Looking at the zeroth order term first, I plug in for $y=\gamma$, and expand the squared and cubed parantheses in the exponent. I get that zeroth order term becomes (the factors outside the integrals cancel)
$$ \frac{e^{-\frac{ k_0 a^2}{2kT}} \int_{-\infty}^{\infty} \gamma e^{-\frac{ k_0 \gamma^2}{2kT}} e^{ \frac{k_0 a \gamma}{kT}} e^{-\frac{ k_0 \gamma^4}{2kT}} e^{\frac{3 k_0 \gamma^3 a}{2kT}} e^{-\frac{3 k_0 \gamma^2 a}{2kT}} e^{\frac{ k_0 a^3 \gamma}{2kT}} \text{d} \gamma } {e^{-\frac{ k_0 a^2}{2kT}} \int_{-\infty}^{\infty} e^{-\frac{ k_0 \gamma^2}{2kT}} e^{\frac{ k_0 a \gamma}{kT}} e^{-\frac{ k_0 \gamma^4}{2kT}} e^{\frac{3 k_0 \gamma^3 a}{2kT}} e^{-\frac{3 k_0 \gamma^2 a}{2kT}} e^{\frac{ k_0 a^3 \gamma}{2kT}} \text{d}\gamma} $$
However, I don't know how to proceed to solve this integral. I am not used to doing Taylor expansions, so may have started off completely wrong, here.