calculus - Integration between infinity and -infinity

Given that
$$\int_{-\infty}^{\infty}(ce^{\frac{-\alpha r}{2a}})^2dr = 1$$

express c in terms of a.

I assume this is meant to be done by solving the integral, but how do you evaluate

$$\frac{2ac}{\alpha}e^{\frac{-\alpha r}{2a}}$$

between $-\infty$ and $\infty$? The answer isn't finite, is it? For $\infty$, the $e$ term will tend to zero, but what about the other one? Or is evaluating the integral the wrong approach?

To give the context of this question, it's related to quantum mechanics. The expression to be integrated is the wavefunction. I am supposed to use the 'normalisation condition' to solve the problem, which I presume is the same normalisation condition which states that the integral between infinity and -infinity of the magnitude of the wavefunction is equal to 1.

Answer

New answer:

Given that $r>0$, we can solve the integral, as the limits now become $0$ and $\infty$:

$$\int_{0}^{\infty}c^2 e^{\frac{-\alpha r}{a}}=-c^2 \frac{a}{\alpha}\int_{0}^{\infty} \frac{d}{dr} e^{\frac{-\alpha r}{a}}=-c^2 \frac{a}{\alpha}\left[0-1 \right]=c^2 \frac{a}{\alpha}$$
so
$$c^2 \frac{a}{\alpha}=1 \leftrightarrow c=\sqrt{\frac{\alpha}{a}}$$ is your normalization constant.

Previous answer:

As it is now, then no, your integral does not converge. Assuming that it was meant to be a Gaussian (which I think was the intention), it can be evaluated:

$$\int_{-\infty}^{\infty}ce^{\frac{-\alpha r^2}{2a}}=c\sqrt{\frac{2r\pi}{\alpha}}$$
which can be shown using a very cute trick of squaring the integral.

In this case $c$ is easily found.