Integral $\int_{-\infty}^\infty dx e^{-nx^2/2}(z-ix)^n$

Question

$$ I\equiv\mathcal{F}_n(z)=\int_{-\infty}^\infty dx e^{-nx^2/2}(z-ix)^n. $$ Evaluate I for $n \to \infty$ and z real. We can consider $z\geq 0$ due to the symmetry of $\mathcal{F}$ given by $$ \mathcal{F}_n(-z)=(-1)^n\mathcal{F}_n(z). $$ I am looking for a solution, thanks.

Answer 1

Here is an approach. Recalling the binomial theorem, we have

$$ I = \sum_{k=0}^{n} {n\choose k}i^kz^{n-k} \int_{-\infty}^{\infty} x^ke^{-nx^2/2} dx, $$

which can be evaluated to

$$ = \sum_{k=0}^{n}( ( -1 ) ^{k}+1 ) {n\choose k}i^k n^{-\frac{k}{2}-\frac{1}{2}}\,{2}^{\frac{k}{2}-\frac{1}{2}} z^{n-k}\,\Gamma\left( \frac{k}{2}+\frac{1}{2} \right),\quad i=\sqrt{-1} $$

Note: To evaluate the integral

$$F = \int_{-\infty}^{\infty} x^ke^{-nx^2/2} dx, $$

we write $F$ as

$$ F = \int_{-\infty}^{\infty} x^ke^{-nx^2/2} dx = \int_{-\infty}^{0} x^ke^{-nx^2/2} dx + \int_{0}^{\infty} x^ke^{-nx^2/2} dx. $$

$$\implies I = (-1)^k\,\int_{0}^{\infty} t^k e^{-nt^2/2} dt + \int_{0}^{\infty} x^ke^{-nx^2/2} dx. $$

The last two integrals can be evaluated using the gamma function and the change of variables $u=nx^2$.