# 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 $\mathrm{polynomial} (z) \cdot e^{-z^{2-\frac{2s}{1+s}}}$.

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Suppose $s=1/2$. It is clear that your integral diverges at $+\infty$. Do you mean to say $s \in (-1,0)$, maybe ? –  Sasha Oct 7 '11 at 19:37
Thank you there was a factor of $exp(-t^2/2)$ I left out. –  user7980 Oct 7 '11 at 19:45

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: $$\sum_{k=0}^\infty \left(-\frac{1}{2}\right)^k z^{-1-s(k-1)} \mathrm{e}^{-\frac{z^2}{2}} = \frac{z^{2 s-1}}{z^s + \frac{1}{2}} \mathrm{e}^{-\frac{z^2}{2}}$$

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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, $$\begin{split} \int _{0}^{\infty }\!\sqrt {t}{{\rm e}^{-1/2\,{t}^{2}}}{{\rm e}^{-1/2 \,{t}^{-1}}}{dt}=1/4\,\sqrt {2} \left( 4/3\,\pi \, {{}_0F_2(\ ;\,5/4,7/4;\,-1/32)} -\right. \\ {{\pi }^{3/2}\sqrt [4]{2} {{}_0F_2(\ ;\,3/4,3/2;\,-1/32)}/\Gamma\left( 3/4 \right)}+\\ \left.2\,\sqrt { \pi }\Gamma \left( 3/4 \right) \sqrt [4]{2} {{}_0F_2(\ ;\,1/4,1/2;\,-1/32)} \right) {\frac {1}{\sqrt {\pi }}} \end{split}$$

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