How to solve this semi gaussian definite complex integral?

Question

How can I solve this definite complex integral :

$L=\int_{-\infty}^{\infty} e^{-{\lambda r^2}-jr \zeta_1+j\zeta_2}dr$
where : $\lambda , \zeta_1, \zeta_2$ are real and constant values and $\,j =\sqrt{-1} \,$.

Answer 1

You need to complete the square in the $r$. That is, write $$ -\lambda r^2 - ir\zeta_1 + i\zeta_2 = -\lambda\Bigl(r+\frac{i\zeta_1}{2\lambda}\Bigr)^2+i\zeta_2-\frac{\zeta_1^2}{4\lambda} $$ We can now write your integral as $$ \exp\Bigl[i\zeta_2-\frac{\zeta_1^2}{4\lambda}\Bigr]\int_{-\infty}^{\infty}\exp\biggl[-\lambda\Bigl(r+\frac{i\zeta_1}{2\lambda}\Bigr)^2\biggr]dr $$ Let $u = r+\frac{i\zeta_1}{2\lambda}$. Then $dr = du$ so we have $$ \exp\Bigl[i\zeta_2-\frac{\zeta_1^2}{4\lambda}\Bigr]\int_{-\infty}^{\infty}e^{-\lambda u^2}du $$ For $\lambda > 0$, the integral is $\sqrt{\frac{\pi}{\lambda}}$. Note the following integral $$ \int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}\tag{1} $$ The goal with integrals like yours is to manipulate into the form of equation $(1)$.

Answer 2

We use the noation $j=i$.
Use that $\int_{-\infty}^{\infty}e^{-ax^2+bx+c}dx=\sqrt{\pi\over a}e^{{b^2\over 4a}+c}$
to get the answer $\sqrt{\pi\over \lambda}e^{{-{\xi_1}^2\over 4\lambda}+i\xi_2}$

Answer 3

Write

$$L = \int_{-\infty}^{\infty} e^{-\lambda r^2 - jr\zeta_1}e^{j\zeta_2}\,dr = e^{j\zeta_2}\int_{-\infty}^\infty e^{-\lambda\left(r + \frac{j\zeta}{2\lambda}\right)^2+\lambda\left(\frac{j\zeta_1}{2\lambda}\right)^2}\, dr = e^{j\zeta_2}e^{-\frac{\zeta_1^2}{4\lambda}}\int_{-\infty}^\infty e^{\lambda\left(r - \frac{j\zeta_1}{2\lambda}\right)^2}\, dr.$$

By contour shifting, the integral on the right is $\int_{-\infty}^\infty e^{-\lambda r^2}\, dr$. To justify this, consider the contour integral

$$\int_{\Gamma(R)} e^{-\lambda z^2}\, dz$$

where $\Gamma(R)$ is a rectangle parallel to the axes, with counterclockwise orientation, and with vertices at $-R$, $R$, $-R + j\zeta_1/(2\lambda)$, and $R + j\zeta_1/(2\lambda)$,. The integrals along the vertical edges are $O(e^{-\lambda R^2})$, which is negligible as $R \to \infty$. Further, since $e^{-\lambda z^2}$ is entire, $\int_{\Gamma(R)} e^{-\lambda z^2}\, dz = 0$ by Cauchy's integral theorem. Hence, the contour shifting is valid and

$$\int_{-\infty}^\infty e^{-\lambda\left(r - \frac{j\zeta_1}{2\lambda}\right)^2}\, dr = \int_{-\infty}^\infty e^{-\lambda r^2}\, dr = \sqrt{\frac{\pi}{\lambda}}.$$ So

$$L = e^{j\zeta_2 - \frac{\zeta_1^2}{4\lambda}} \sqrt{\frac{\pi}{\lambda}}.$$