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.