Using contour methods, evaluate the following integral:
$$\int_{-\infty}^{\infty} e^{i(k+iV)x^2}\,\mathrm{d}x$$
as $V$ approaches positive zero.
As far as I'm concerned I don't see any singularity so I thought that I should use a semi circle contour and set the contour integral to zero. That way the integral on the real line is the negative of the integral of the semi circle. However I feel that that approach is wrong because I get an integral with an exponential to an exponential. Also as $r$ goes to infinity it looks like it goes to zero which cannot be right.
Any help will be appreciated.
If you really meant $\int_{-\infty}^\infty e^{(ik-v^2) x^2}dx$, then let $$F(z) = \int_{-\infty}^\infty e^{-z x^2}dx$$ For $z > 0$ you can do the change of variable $y = z^{1/2}x$ to get $$F(z) = \frac{1}{z^{1/2}} \int_{-\infty}^\infty e^{- y^2}dy = \frac{\pi^{1/2}}{z^{1/2}}$$ Then for $z \in \mathbb{C}, Re(z) > 0$ note that $F(z) = \int_{-\infty}^\infty e^{-z x^2}dx$ and $\frac{\pi^{1/2}}{z^{1/2}}$ both are analytic in $z$,
hence by the identity theorem for analytic functions $F(z) = \frac{\pi^{1/2}}{z^{1/2}}$ stays true.
Overall : $$\int_{-\infty}^\infty e^{(ik-v^2) x^2}dx = \frac{\pi^{1/2}}{(v^2-ik )^{1/2}}$$ (where ${}^{1/2}$ is the branch of the square root analytic on $Re(z) > 0$ and such that $z^{1/2} > 0$ for $z > 0$)