What is $\int \frac{e^{a x}}{1+x^2} dx $?

86 Views Asked by At

In my answer to another question (here: Upper and lower bound on different of ${\rm erf}(\frac{x+c}{b})-{\rm erf}(\frac{x-c}{b})$), I came up with this integral: $\int \frac{e^{a x}}{1+x^2} dx $. I was not able to evaluate it, so I asked Wolfram Alpha, which came up with this:

$$\int \frac{e^{a x}}{1+x^2} dx = \frac{-1}{2 i e^{-i a}} (e^{2 i a} \mbox{Ei}(a (x-i))-\mbox{Ei}(a (x+i))) + \mbox{constant} $$

where $\mbox{Ei}$ is the exponential integral.

Is there a simpler form of this?

I don't know if this will help, but the actual integral in my answer has symmetric limits around zero, so its form is

$$\int_{-b}^b \frac{e^{a x}}{1+x^2} dx $$

Simplifying this would be sufficient.