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I am interested into the following integral

$I = \int_{-\infty}^{+\infty} \frac{f(z)}{z-z_0} dz$

where $f$ is typically a Gaussian function $e^{-z^2}$. I am tempted to simply use Cauchy formula to get

$I = 2\pi i f(z_0) = 2\pi i \ e^{-z_0^2}$

But I am not sure that the contribution of the half circle containing $z_0$ goes to $0$ as the radius of the circle $R$ goes to $+\infty$ and I could not find a description of the analytic properties of the complex Gaussian to justify this integration on the internet. On top of that, Mathematica gives me $0$ as a result too.. Can the integral be carried out like written or is not so simple? Thank you.

RobPratt
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J. Rouxel
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  • It is tempting to say $|e^{-z^2}|$ is small when $|z|$ is large, but it is not true. The real part of $-z^2$ is large if the imaginary component of $z$ is large and the real component is small. You can't say that the contour of the semicircle is goes to 0. – user317176 Feb 07 '20 at 23:00
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    I am not sure if the integral converges for any $z_0$. Note that the integrand is odd for $z_0=0$, but the integral still doesn’t converge – Maximilian Janisch Feb 07 '20 at 23:01
  • The inetgral converges if $z_0$ is not real. – Kavi Rama Murthy Feb 07 '20 at 23:16
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    @j.rouxel I have provided a solution that relies on Feynman's trick and validated it using Wolfram Alpha. Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Feb 09 '20 at 20:46
  • Thank you for your feedback. How do you choose between $e^{-z \times z}$ and $e^{-z \times z^*} = e^{-|z|^2}$ as the correct analytic extension of the Gaussian function? Both seems reasonable to me but $e^{-|z|^2}$ would definitely converge when $|z|$ is getting larger. – J. Rouxel Feb 10 '20 at 23:44

1 Answers1

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One can deform the real line contour in the complex plane, but it is rather dubious that such a defomation facilitates evaluation of the integral of interest via Cauchy's Integral Theorem. However, we can use Feynman's trick to evaluate the integral in terms of the Imaginary Error Function. To that end we now proceed.

We first note that

$$\begin{align} \int_{-\infty}^\infty \frac{e^{-z^2}}{z-z_0}\,dz&=2z_0 \int_0^\infty \frac{e^{-x^2}}{x^2-z_0^2}\,dx\tag1 \end{align}$$

Second, let $I(a)$ be defined as

$$\begin{align} I(a)&=2z_0 \int_0^\infty \frac{e^{-ax^2}}{x^2-z_0^2}\,dx\tag2 \end{align}$$

Observe that $I(1)$ is simply the integral of interest in $(1)$.

Third, differentiate $(2)$ to find

$$\begin{align} I'(a)&=-2z_0 \int_0^\infty \frac{x^2\,e^{-ax^2}}{x^2-z_0^2}\,dx\\\\ &=-2z_0 \int_0^\infty \frac{(x^2-z_0^2+z_0^2)\,e^{-ax^2}}{x^2-z_0^2}\,dx\\\\ &=-2z_0\int_0^\infty e^{-ax^2}\,dx-z_0^2I(a)\\\\ I'(a)+z_0^2I(a)&=-z_0\sqrt{\pi/a}\tag3 \end{align}$$

Now, solving the ODE in $(3)$ for $I(a)$ we find that

$$I(a)=I(0)e^{-az_0^2}-2\sqrt{\pi}e^{-az_0^2}\int_0^{\sqrt{a}z_0}e^{x^2}\,dx\tag4$$

Using the initial condition $I(0)=i\pi\,\text{sgn}\left(\text{Im}\left(z_0\right)\right)$ and setting $a=1$ in $(4)$ yields the coveted result

$$\begin{align} \int_{-\infty}^\infty \frac{e^{-z^2}}{z-z_0}\,dz&=i\pi\,\text{sgn}\left(\text{Im}\left(z_0\right)\right)e^{-z_0^2}-2\sqrt{\pi}e^{-z_0^2}\int_0^{z_0}e^{x^2}\,dx\\\\ &=\bbox[5px,border:2px solid #C0A000]{i\pi\,\text{sgn}\left(\text{Im}\left(z_0\right)\right)e^{-z_0^2}-\pi e^{-z_0^2}\text{erfi}(z_0)}\tag5 \end{align}$$

where $\text{sgn}(x)$ is the Sign Function and $\text{erfi}(z)$ is the Imaginary Error Function.

EXAMPLES:

As an example, if $z_0=2+i5$, the result is $i\pi e^{21-20i}-\pi e^{21-20i} \text{erfi}(2+i5)$, which is verified using Wolfram Alpha (See Here).

And if $z_0=2-i5$ the result is $-i\pi e^{21+20i}-\pi e^{21+20i} \text{erfi}(2-i5)$, which was also verified using Wolfram Alpha.

Thus, we see that the sign of the imaginary part of the answer depends on the sign of $\text{Im}(z_0)$.

Note that if $z_0$ is purely real, we can evaluate $(1)$ as a principal value. Setting $z_0$ to $x_0\in \mathbb{R}$ in $(5)$ shows that

$$\text{PV}\int_{-\infty}^\infty \frac{e^{-z^2}}{z-x_0}\,dz=-\pi e^{-x_0^2}\text{erfi}(x_0)$$

Mark Viola
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  • @j.rouxel My solution is correct. As an example, see THIS. – Mark Viola Feb 09 '20 at 20:25
  • @j.rouxel I have provided a solution that relies on Feynman's trick and validated it using Wolfram Alpha. Please let me know how I can improve my answer. I really want to give you the best answer I can. – Mark Viola Feb 09 '20 at 20:46
  • It took me some time to follow and redo your derivation + do some numerical crosschecks and yes, I get the correct answer. Thank you for showing the nice trick with the derivation. 2 small questions: 1) the ODE becomes singular for a=0. We don't need it to compute I(0) but does it means that the function I(a) is ill-defined in a = 0? 2) I understand the erfi function well as a power series. But for its integral representation, since z0 is complex, is the integral valid for any contour from 0 to z0? – J. Rouxel Feb 11 '20 at 01:35
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    Inasmuch as $e^{z^2}$ is entire, the value if the integral representation of rhe imaginary error function is independent of the path of integration from $0$ to $z_0$. – Mark Viola Feb 11 '20 at 05:06