Can the integral,

$$I = \int^{-x}_{-\infty} e^{-at^2}\ \mathrm dt,\ a \in \mathbb{C},\ Re(a) > 0,$$

be written as an error function?

I tried, by substitution,

$$\int^{-x}_{-\infty} e^{-at^2}\ \mathrm dt = \frac{1}{\sqrt{a}}\int^{-x\sqrt{a}}_{-\infty\sqrt{a}} e^{-u^2}\ \mathrm du.$$ Reducing this to a complementary error function, $erfc(z) = \frac{2}{\sqrt{\pi}} \int_{z}^{\infty}e^{-u^2}du$, seems to require that the lower bound of the integral obtained after the substitution, $-\sqrt{a}\cdot \infty$, is real.

Despite this, Mathematica computes the integral as defined above to,

$\int^{-x}_{-\infty} e^{-at^2}\ dt = \frac{\sqrt{\pi}}{2\sqrt{a}}erfc(\sqrt{a}x)$,

subject to the above constraints on the parameter $a$. The limit of the error function for $x \rightarrow \infty$ depends on the real and imaginary parts of $\sqrt{a}$. [see http://functions.wolfram.com/GammaBetaErf/Erfc/ for limiting values at infinity]. For the initial integral, $I(x \rightarrow \infty) \rightarrow 0$, independent of $a$. I would like to understand where this apparent discrepancy arises.

Any thoughts would be very helpful!

  • $\begingroup$ Do you have a simple numerical example of this discrepancy? $\endgroup$ – Somos Mar 30 at 21:22
  • $\begingroup$ For $a >0, \int_{-\infty}^x e^{-at^2}dt = \int_{-\infty}^{xa^{1/2}} e^{-u^2}d(ua^{-1/2})= a^{-1/2} erfc(xa^{1/2})$. Then $\int_{-\infty}^x e^{-at^2}dt-a^{-1/2} erfc(xa^{1/2})$ is complex analytic in $a,\Re(a) > 0$ and since it vanishes on $a > 0$ it must vanish for every $a, \Re(a) > 0$. As you noticed this is equivalent to saying for $\Re(a) > 0$ then $\lim_{R \to \infty}erfc(-Ra^{1/2}) =\lim_{R \to \infty}\int_{-R}^{-R a^{1/2}} e^{-u^2}du = 0$. $\endgroup$ – reuns Mar 31 at 14:30
  • $\begingroup$ Abramowitz & Stegun suggests that this limit holds only for $Re(a^{1/2}) > Im(a^{1/2})$, ($arg(|a^{1/2}|) < \pi/4$), see 7.1.16 in people.math.sfu.ca/~cbm/aands/page_298.htm. I think that if this condition is not met, $\int_{-\infty}^{x} e^{-at^2}dt \neq \int_{-\infty}^{xa^{1/2}} e^{-u^2}d(ua^{-1/2}) = a^{-1/2}erfc(xa^{1/2})$, since the integral is along a path from a complex infinity to $x$. $\endgroup$ – alecbo Mar 31 at 15:08

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