TITLE: Evaluating this integral using the Gamma function
QUESTION [12 upvotes]: I was wondering if the following integral is able to be evaluated using the Gamma Function.
$$\int_0^{\infty}t^{-\frac{1}{2}}\mathrm{exp}\left[-a\left(t+t^{-1}\right)\right]\,dt$$
I already have a tedious solution that doesn't exceed the scope of the first few semesters of calculus, but I want to tackle this with the Gamma Function. I just don't know how or if it's even possible.
If anyone can give a hint, I'd really like to finish it on my own.
EDIT:
You are allowed to use the fact that 
$$
\int_{-\infty}^{\infty}\exp(-x^2)\,dx = \sqrt{\pi}
$$

REPLY [5 votes]: Alternatively, denote the integral as $I$ and use substitution $t=x^2$, the integral becomes
$$I=2\int_0^\infty e^{-\Large a\left(x^2+\frac{1}{x^2}\right)}\,dx=2e^{-2\large a}\int_0^\infty e^{-\Large a\left(x-\frac{1}{x}\right)^2}\,dx$$
Since the integrand is an even function, then it can be rewritten as
$$\begin{align}I&=e^{-2\large a}\int_{-\infty}^\infty e^{-\Large a\left(x-\frac{1}{x}\right)^2}\,dx\\&=e^{-2\large a}\int_{-\infty}^0 e^{-\Large a\left(x-\frac{1}{x}\right)^2}\,dx+e^{-2\large a}\int_{0}^\infty e^{-\Large a\left(x-\frac{1}{x}\right)^2}\,dx\\&=I_1+I_2\end{align}$$
Making substitution $x=-e^{-\theta}$ for $I_1$ and $x=e^{\theta}$ for $I_2$ then using $\sinh\theta=\dfrac{e^{\theta}-e^{-\theta}}{2}$ and $\cosh\theta=\dfrac{e^{\theta}+e^{-\theta}}{2}$, we have
$$\begin{align}I&=e^{-2\large a}\int_{-\infty}^\infty e^{-2a\sinh^2\theta}e^{-\theta}\,d\theta+e^{-2\large a}\int_{-\infty}^\infty e^{-2a\sinh^2\theta}e^{\theta}\,d\theta\\&=2e^{-2a\large a}\int_{-\infty}^\infty e^{-2a\sinh^2\theta}\cosh\theta\,d\theta\\&=2e^{-2\large a}\int_{-\infty}^\infty e^{-2a\sinh^2\theta}\,d(\sinh\theta)\\&=e^{-2\large a}\int_{0}^\infty e^{-2az^2}\,dz\end{align}$$
We can get a gamma function by using substitution $y=2az^2$.