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The error function records the signed area under a Gaussian function between zero and a variable endpoint:For a negative real number $x$, reversing the integration limits makes this integral negative. The normalization uses the Gaussian integral: the full area of $e^{-t^2}$ is $\sqrt{\pi }$, so its positive half contributes one after multiplication by $2/\sqrt{\pi }$. The same definition extends to a complex number $z$ along any path from zero to $z$, giving an entire function.^[ref]

For example, $\mathrm{erf}(0)=0$, $\mathrm{erf}(1)\approx 0.842701$, and $\mathrm{erf}(-1)\approx -0.842701$. A finite positive endpoint captures only part of the positive Gaussian integral; $\mathrm{erf}(x)$ approaches one as $x\to +\infty$ but never reaches it at a finite real $x$.

The useful idea is to name a recurring integral once instead of carrying it through every calculation. The error function and the normal distribution express probabilities and quantiles through this area, while an error-function diffusion front describes how diffusion smooths a sharp jump. The complementary error function retains the small area left in the tail.

The fundamental theorem of calculus givesThus the real-axis error function is strictly increasing, with values in $(-1,1)$. Its integrand is an even function, so the error function is an odd function, $\mathrm{erf}(-x)=-\mathrm{erf}(x)$. The derivative also gives ${\mathrm{erf}}^{\prime \prime }(x)=-4x{e}^{-x^2}/\sqrt{\pi }$: the curve is concave for $x>0$ and convex for $x<0$. These real-axis properties do not imply boundedness throughout the complex plane; the existing quadrant asymptotics of the error function describe a different behavior in other directions.

The integrand is an entire function, so the defining contour is immaterial. A rectangular contour, closed far to the right where the Gaussian function decays, gives the exact horizontal-tail identityThe constant is the half-line Gaussian integral. Expand the endpoint amplitude $e^{-s^2}=1-s^2+\cdots$ in this Watson lemma. Alternatively, repeated integration by parts gives the same expansion, including purely imaginary $z$, because every derivative of $e^{-s^2}$ is integrable on the positive real axis. Thus, on each fixed ray in the stated quadrant,The remainder is measured on the exponential endpoint scale. When that scale grows, this formula does not resolve terms exponentially small relative to it.

Write $z=r{e}^{i\theta }$. The endpoint term has modulus $e^{-r^2\cos 2\theta }/(2r)$. For $0\le \theta <\pi /4$ it is exponentially small, and at $\theta =\pi /4$ it is oscillatory of order $r^{-1}$. For $\pi /4<\theta \le \pi /2$ it dominates the constant. The quadrant asymptotics of the error function therefore give the requested leading asymptotic expansionsThe boundary ray $\theta =\pi /4$ belongs to the constant-dominated case because of the extra factor $1/r$. On the imaginary axis this gives $f(ir)\sim i{e}^{r^2}/(2r)$, consistent with $f(ir)=i{\int }_0^r{e}^{s^2}ds$. The constant/exponential dominance exchange is an anti-Stokes line phenomenon; it should not be confused with dropping the constant everywhere in the quadrant.