calculus - Evaluating this integral using the Gamma function

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

$$\int_{-\infty}^{\infty}\exp(-x^2)\,dx = \sqrt{\pi}$$

Answer

Let $t=u^2$, and the integral becomes

$$2 \int_0^{\infty} du \, e^{-a \left (u^2 + \frac1{u^2} \right)} = 2 e^{2 a} \int_0^{\infty} du \, e^{-a \left ( u+\frac1{u} \right )^2}$$

Let $v=u+1/u$, then

$$u = \frac12 \left (v \pm \sqrt{v^2-4} \right )$$

$$du = \frac12 \left (1 \pm \frac{v}{\sqrt{v^2-4}} \right ) dv$$

Now, it should be understood that as $u$ traverses from $0$ to $\infty$, $v$ traverses from $\infty$ down to a min of $2$ (corresponding to $u \in [0,1]$), then from $2$ back to $\infty$ (corresponding to $u \in [1,\infty)$). Therefore the integral is

$$e^{2 a} \int_{\infty}^{2} dv \left (1 - \frac{v}{\sqrt{v^2-4}}\right ) e^{-a v^2} + e^{2 a} \int_{2}^{\infty} du \left (1 + \frac{v}{\sqrt{v^2-4}}\right ) e^{-a v^2}$$

which is

$$\begin{align}2 e^{2 a}\int_2^{\infty} dv \frac{v}{\sqrt{v^2-4}} e^{-a v^2} &= e^{2 a} \int_4^{\infty} \frac{dy}{\sqrt{y-4}} e^{-a y}\\ &= e^{-2 a} \int_0^{\infty} dq \, q^{-1/2} \, e^{-a q} \end{align}$$

I guess the gamma function comes from this integral, but I find it easier to refer to gaussian integrals.