Can we simplify the integral $\int_z^\infty t^s \exp ( -\frac{1}{2t^{2s}}) \exp(-\frac{t^2}{2})dt$ using gamma functions or other elementary methods? I have an upper bound for a guassian type integral in terms of the integral in this post and I feel like I have seen something like this as an application of a change of variables and using the definition of the Gamma function.
Let $s \in (0,1)$ and $z$ be any large positive real number, that is $ z \gg 1$

Does the integral $\int_z^\infty t^{s} \exp ( -\frac{1}{2t^{2s}}) \exp(-\frac{t^2}{2})\;dt$ have a known simplification in terms of elementary functions or gamma functions or does it appear in a table of integrals?

If it helps any the upper bound we are trying to derive (and I don't know if there is an estimate of this form applicable to the inegral above) is a bound in terms of $\mathrm{polynomial} (z) \cdot e^{-z^{2-\frac{2s}{1+s}}}$.
 A: When $z$ is large, integration by parts will produce the needed approximation:
$$ \begin{eqnarray}
  &&  \int_z^\infty t^s \exp\left(-\frac{t^2}{2} - \frac{1}{2 t^{2s}} \right) \mathrm{d} t =
    \int_z^\infty t^{s-1} \exp\left( - \frac{1}{2 t^{2s}} \right) \mathrm{d} \left(-\mathrm{e}^{-\frac{t^2}{2}} \right)  \\
   && = z^{s-1} \exp\left( -\frac{z^2}{2} - \frac{1}{2 z^{2s}}\right) + \int_z^\infty \left(
  (s-1) t^{s-2} - \frac{s}{t^{s+2}} \right) \exp\left(-\frac{t^2}{2} - \frac{1}{2 t^{2s}} \right)
\end{eqnarray}
$$
Continuing this way will produce successive terms.

Added Another possibility is to write series for $\exp(-\frac{1}{2 t^{2s}})$ and integrate term-wise:
$$
   \int_z^{\infty } \frac{t^s (-1)^k \exp \left(-\frac{t^2}{2}\right)}{2^k t^{s k}} \,
   \mathrm{d}t = (-1)^k 2^{\frac{1}{2} (-k (s+2)+s-1)} \Gamma \left(\frac{1}{2} (-k
   s+s+1),\frac{z^2}{2}\right)
$$
The incomplete gamma function has known series expansion at infinity. Keeping the main term only gives the following  as the main term for your integral:
$$
  \sum_{k=0}^\infty \left(-\frac{1}{2}\right)^k z^{-1-s(k-1)} \mathrm{e}^{-\frac{z^2}{2}} = 
    \frac{z^{2 s-1}}{z^s + \frac{1}{2}} \mathrm{e}^{-\frac{z^2}{2}}
$$
A: According to the comment, I'm assuming the integrand is $t^2 \exp(-t^2/2 - t^{-2s}/2)$.  For $z=0$ and $s$ rational, Maple will find an expression for the integral in terms of hypergeometric functions.  For example,
$$\begin{split}
\int _{0}^{\infty }\!\sqrt {t}{{\rm e}^{-1/2\,{t}^{2}}}{{\rm e}^{-1/2
\,{t}^{-1}}}{dt}=1/4\,\sqrt {2} \left( 4/3\,\pi \,
{{}_0F_2(\ ;\,5/4,7/4;\,-1/32)} -\right. \\ {{\pi }^{3/2}\sqrt [4]{2}
{{}_0F_2(\ ;\,3/4,3/2;\,-1/32)}/\Gamma\left( 3/4 \right)}+\\ \left.2\,\sqrt {
\pi }\Gamma  \left( 3/4 \right) \sqrt [4]{2}
{{}_0F_2(\ ;\,1/4,1/2;\,-1/32)} \right) {\frac {1}{\sqrt {\pi }}}
\end{split}$$
