How do I do this integral? How do I integrate $$\int\limits_{-c}^c \frac{e^{-\gamma\, x^2}}{1-\beta \, x}\, dx $$, under the constraints (if necessary) that $\beta < 1\, /\, c$ and $\beta \ll 1$ ?
 A: You'll have a bear of a time looking for exact solutions, since the indefinite integral will be in terms of the exponential integral function (Ei(t) = -$\int_{-t}^\infty \frac{e^x}{x} dx$), which is not an elementary function. 
You can plug your integral into something like Mathematica to get $$- \frac{e^{ -\frac{\gamma}{\beta} } \text{Ei} \left( \left( \frac{1}{\beta} - x \right) \gamma \right) }{\beta}$$ and then, of course, it's plug and chug from there.
However, the fact that you were given the constraint $\beta << 1$ suggests that you may be interested in a numerical solution instead. In that case, you can approximate your function by $e^{-\gamma x}$ instead, which is close to your function in the range $(-\infty, \frac{1}{\beta})$ when $\beta << 1$ and is easy to integrate as $$\frac{e^{-\gamma x} }{ \gamma}.$$
A: HINT:


*

*Using the exponential integral:
$$\text{Ei}(x)=\int_{1}^{\infty}\frac{e^{-tx}}{t}\space\text{d}t$$



$$\int\frac{e^{-\gamma x}}{1-\beta x}\space\text{d}x=-\frac{e^{-\frac{\gamma}{\beta}}\text{Ei}\left(\gamma\left(\frac{1}{\beta}-x\right)\right)}{\beta}+\text{C}$$
So with your boundaries we get:
$$\int_{-c}^{c}\frac{e^{-\gamma x}}{1-\beta x}\space\text{d}x=\left[-\frac{e^{-\frac{\gamma}{\beta}}\text{Ei}\left(\gamma\left(\frac{1}{\beta}-x\right)\right)}{\beta}\right]_{-c}^{c}=\frac{e^{-\frac{\gamma}{\beta}}\left(\text{Ei}\left(\frac{\gamma(\beta c+1)}{\beta}\right)-\text{Ei}\left(-\frac{\gamma(\beta c+1)}{\beta}\right)\right)}{\beta}$$
