Please compute $\int_0^{\infty}e^{-(x+ai)^2}dx$, where, $i^2=-1$.

Question

I'm trying to do the following Fourier (Hankel?) transform for a cylindrincally symmetric function: $$ \int_0^{\infty} \!\int_0^{\infty} \! \left( Are^{-(r^2+z^2)/\delta^2}\right)J_1(k_r r) e^{-ik_z z} r \, \mathrm{d}r\, \mathrm{d}z $$ Trying to compute the integral in $z$ I got $$\int_0^{\infty}e^{-\left(\zeta+i k_z \delta/2\right)^2}\,\mathrm{d}\zeta.$$ Any ideas on how to attempt this?

Also, any hints that might be helpful in computing the rest of the integral in $r$ would be really helpful. I have little experience with Bessel functions.

Answer 1

Consider a closed contour $\gamma$ in the complex plane consisting of four straight line segments:

1. From $z = 0$ to $z = R \in \mathbb{R}$;
2. From $z = R$ to $z = R + i a$;
3. From $z = R + i a$ to $z = i a$;
4. From $z = ia$ to $z = 0$.

Let $f(z) = e^{-z^2}$. Since this function has no poles in the complex plane, its integral around $\gamma$ is zero: $$ \oint f(z) \, dz = \int_{\gamma_1} f(z) \, dz + \int_{\gamma_2} f(z) \, dz +\int_{\gamma_3} f(z) \, dz +\int_{\gamma_4} f(z) \, dz = 0. $$ Taking each of these four integrals in turn, and looking at their limit as $R \to \infty$:

1. We have $$ \int_{\gamma_1} f(z) \, dz = \int_0^R e^{-x^2} \, dx, $$ and in the limit $R \to \infty$, this approaches $\sqrt{\pi}/2$.

2. We have $$ \int_{\gamma_2} f(z) \, dz = i \int_0^a e^{-(R + iy)^2}\, dy = i e^{-R^2} \int_0^a e^{2 i R y} e^{y^2} \, dy. $$ The absolute value of the integrand in the last step is bounded above by $e^{a^2}$, and so the integrand will always be smaller in magnitude than $a e^{a^2}$ by the estimation lemma. Thus, as $R \to \infty$, the integrand over $\gamma_2$ will vanish.

3. We have $$ \int_{\gamma_3} f(z) \, dz = \int_R^0 e^{-(x + ia) ^2} \, dx = - \int_0^R e^{-(x + ia) ^2} \, dx, $$ and so as $R \to \infty$ this approaches the negative of the integral we're looking for.

4. We have $$ \int_{\gamma_4} f(z) \, dz = i \int_a^0 e^{-(iy)^2}\, dy = - i \int_0^a e^{y^2} \, dy. $$

Thus, putting all four integrals together and taking the $R \to \infty$ limit, we conclude that $$ \frac{\sqrt{\pi}}{2} - \int_0^\infty e^{-(x + ia) ^2} \, dx - i \int_0^a e^{y^2} \, dy = 0 \\ \boxed{ \int_0^\infty e^{-(x + ia) ^2} = \frac{\sqrt{\pi}}{2} - i \int_0^a e^{y^2} dy.} $$

The remaining integral above is related to Dawson's integral. It can also be related to the imaginary error function obtained by defining the real error function $\mathrm{erf}(x)$ in the customary way and then extending it analytically to the complex plane $\mathrm{erf}(z)$. In this case, we have $$ \int_0^a e^{y^2} dy = - \frac{i \sqrt{\pi}}{2} \mathrm{erf}(i a). $$ Like the "real" error function, its values cannot be calculated exactly for arbitrary values of its argument. The NIST Digital Library of Mathematical Functions lists several interesting identities obeyed by these functions, as well as tables of values in various sources and software packages that can calculate it. Most reasonably sophisticated mathematical software (Mathematica, Maple, Octave, MATLAB, etc.) should also be able to calculate the values of the imaginary error function.

Answer 2

Let u= x+ ai. Then du= dx. When x= 0, u= ai, and as x goes to infinity so does u. That integral becomes $\int_{ai}^\infty e^{-u^2}du$ which can be written in terms of the "error function".