I need help with the following integral:
$$ \int_{-\infty}^{\infty}\frac{e^{-x^2}e^{iax}}{1-x^2}dx$$
Where $a$ is real.
Obviously the integral doesn't converge due to the singularities at $|x|=1$ but I am interested in the principal value of this integral. However I am confused about whether to take my contour above or below each of the singularities.
Any help will be appreciated :)
I'm not sure you can find this PV in closed form. But you can set the odd part to 0 and use symmetry to reduce it to $$ 2 PV \int_{0}^{\infty} \dfrac{e^{-x^2}\cos ax}{1-x^2} dx $$ For a real PV, you would omit an interval of length $\epsilon$ on either side of the singularity. $$ I = 2 \left ( \lim_{\epsilon \rightarrow 0 } \int_{0}^{1-\epsilon}\dfrac{e^{-x^2}\cos ax}{1-x^2} dx + \lim_{\epsilon \rightarrow 0 } \int_{1+\epsilon}^{\infty}\dfrac{e^{-x^2}\cos ax}{1-x^2} dx \right ) $$
In the complex plane, you would join the ends of the excised intervals by a semi-circle of radius $\epsilon$ above the x-axis to get a contour that excluded the singularities.
Don't know if that helps.
