Hard integral fraction on exponent and fraction multiplying

Question

How does one integrate this, just came across this in a project and couldn't do it

$$\int^1_{-1}\frac{a}{x-a}e^{-b/x}dx$$

I was thinking first substitution but it didn't work, then my next idea was a contour but was exactly sure how to approach it.

Answer 1

For the antiderivative, considering $$I=\int\frac{a}{x-a}e^{-\frac b x}\,dx$$first, let $x=\frac b t$ and then partial fraction decomposition to make $$I=\int \frac{a b e^{-t}}{t (a t-b)}\,dt=a\int \frac{e^{-t}}{ t-\frac ba}\,dt-a\int \frac{e^{-t}}{t}\,dt$$ For the first, make now $t=y+\frac ba$to make $$\int \frac{e^{-t}}{ t-\frac ba}\,dt=e^{-\frac{b}{a}}\int \frac{e^{-y}}{y}\,dy$$ Now $$\int \frac{e^{-z}}{z}\,dz=\text{Ei}(-z)$$ where appears the exponential integral function.

Back to $x$, we then have $$I=a \left(e^{-\frac{b}{a}} \text{Ei}\left(\frac{b}{a}-\frac{b}{x}\right)-\text{Ei}\left(-\frac{b}{x}\right) \right)$$

and then, for the integral

$$\int_{-1}^1\frac{a}{x-a}e^{-\frac b x}\,dx=a \left(e^{-\frac{b}{a}} \left(\text{Ei}\left(\left(\frac{1-a}{a}\right) b\right)-\text{Ei}\left(\left(\frac{1+a}{a}\right) b\right)\right)+\text{Ei}(b)-\text{Ei}(-b)\right)$$