Evaluate the integral $\int_0^{\infty} e^{\frac{-t(s-1)^2}{2}} \left( \frac{t(s-1)^3}{3} \right) ds$ I am attempting to evaluate the integral (where $t \rightarrow \infty$)
$$I(t) = \int_0^{\infty} e^{\frac{-t(s-1)^2}{2}} \left( \frac{t(s-1)^3}{3} \right) ds$$
which occurs in the calculation of the second term in the asymptotic form of the gamma function. I believe the answer should be $\frac{1}{12t}\sqrt{\frac{2\pi}{t}}$.
Edit: Apparently this integral evaluates to zero. As mentioned I was attempting to calculate the asymptotics of the Gamma function using 
$$\Gamma(t+1) \sim t^{t+1}e^{-t}\int_0^{\infty}e^{-t(s-1)^2/2}\left(1 + \frac{t(s-1)^3}{3} + \left( \frac{t(s-1)^3}{3} \right)^2 + \dots \right)^2 $$
Hmm...

Given the integral
$$I(t) = \int_0^{\infty} e^{\frac{-t(s-1)^2}{2}} \left( \frac{t(s-1)^3}{3} \right) ds$$
then by integration by parts
$$I = \left[ -\frac{(s-1)^2}{3}e^{-t(s-1)^2/2}\right]_0^{\infty} + \frac{2}{3}\int_0^{\infty}(s-1)e^{-t(s-1)^2/2} ds.$$
The remaining integral is $t^{-1}$ times the derivative of the exponential and leads to
\begin{align}
I &= \left[ -\frac{(s-1)^2}{3}e^{-t(s-1)^2/2}\right]_0^{\infty} + \frac{2}{3}\int_0^{\infty}(s-1)e^{-t(s-1)^2/2} ds \\
&= \frac{1}{3} \, e^{-t/2} - \frac{2}{3 \, t} \, \int_{0}^{\infty} \frac{d}{ds} \left( e^{- \frac{t (s-1)^{2}}{2}} \right) \, ds \\
&= \frac{1}{3} \, e^{- t/2} + \frac{2}{3 \, t} \, e^{-t/2}
\end{align}
From this
\begin{align}
\int_0^{\infty} e^{\frac{-t(s-1)^2}{2}} \left( \frac{t(s-1)^3}{3} \right) ds = \frac{1}{3} \, \left( 1 + \frac{2}{t} \right) \, e^{-t/2}.
\end{align}
In the case $t \to \infty$ the value of the integral tends to zero. 
