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How to calculate the limit \begin{align*} \lim_{b\to \infty}\frac{\vert erf(\sqrt{a+\mathrm{i} b})\vert^{2}}{\sqrt{a^2+b^2}} \end{align*} where \begin{align*} erf(a+\mathrm{i} b)=\frac{2}{\sqrt{\pi}}\int_{0}^{a+\mathrm{i} b}e^{-t^2}~dt \end{align*} is the error function. The Mathematica shows that as $b$ grows, $\vert erf(\sqrt{a+\mathrm{i} b})\vert^{2} $ approaches to $1$ and consequently \begin{align*} \frac{\vert erf(\sqrt{a+\mathrm{i} b})\vert^{2}}{\sqrt{a^2+b^2}} \end{align*} becomes very small. If I try to simplify the error function, then I find an expression in terms of $\cos$ and $\sin$ and that diverge. How to proves that the above limit converges to $0$ (as Mathematica shows).

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  • $\begingroup$ Hint: try to, by substitution, take the $a+bi$ from the limit down to the integrand. $\endgroup$ – Szeto Jun 7 '18 at 9:12
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Use the asymptotic expansion for the error function,

$$ \text{erf}\,(z) \sim 1 - \frac{e^{-z^2}}{\sqrt{\pi\,z}} \text{ with } z=\sqrt{a+ib}=(a^2+b^2)^{1/4}\exp{\big(\frac{i}{2}\,\arctan{(b/a)} \big)}.$$

With $b \to \infty$ the argument of the exponential goes to $\pm i\pi/4,$ with the positive sign for positive $a$. There are more terms multiplying the exponential in the asymptotic expansion, but an asymptotic text will tell you that they are all bounded for $|ph(z)| \le \pi/4,$ and your case is right on the boundary, so the expansion works for you. Since $\exp(-r(1+i)) \to 0$ for positive $r=\sqrt{2(a^2+b^2)},$ all that remains is 1. The limit is zero entirely by the behavior of $\sqrt{a^2+b^2}$ in the denominator.

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  • $\begingroup$ Thanks. I did not understand the last part $exp(-r(1+i)\to 0$. When $r$ approaches $\infty$ and $\infty(1+i)$ has no value? $\endgroup$ – skorpion Jun 8 '18 at 18:51
  • $\begingroup$ @skorpion $|\exp(-r -ir)| = |\exp(-r)||\exp(-ir)| = |\exp(-r)|$, since $r$ is real. So indeed it goes to 0 as $r$ gets bigger. $\endgroup$ – Nikodem Dyzma Jun 19 '18 at 14:33

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