2
$\begingroup$

I know that the imaginary error function, $\mathrm{Erfi}(x)=(2/\sqrt{\pi})\int_0^x \exp{t^2} \mathrm{d}t$, has the asymptotic expansion given in the answer to this question: Asymptotic order of $\frac{\mathrm{erfi}(\sqrt{x})}{\exp(x)\sqrt{x}}$ .

My question is about deriving this expansion, which I think is based on integration by parts. For example, if we use integration by parts once we can rewrite the function as $\mathrm{Erfi}(x) = \frac{2}{\sqrt{\pi}} \left[\frac{1}{2t}\mathrm{e}^{t^2}|_0^x+\frac{1}{2}\int_0^x\frac{\mathrm{e}^{t^2}}{t^2} \mathrm{d}t\right]$,

and then keep on doing this procedure for the leftover integral. This indeed produces the terms in the expansion mentioned above. However, what happens to the terms evaluated at $t=0$? For example, in the first term of the RHS above, we get a blow-up when we plug $t=0$.

How do we formally justify ignoring these blowing-up terms?

$\endgroup$
0
$\begingroup$

Note that $$ \operatorname{erfi} (\mathrm{i} \infty) = \frac{2}{\sqrt{\pi}} \int \limits_0^{\mathrm{i} \infty} \mathrm{e}^{t^2} \, \mathrm{d}t = \frac{2 \mathrm{i}}{\sqrt{\pi}} \int \limits_0^\infty \mathrm{e}^{-s^2} \, \mathrm{d}s = \mathrm{i} \, . $$ Therefore we can write $$ \operatorname{erfi}(x) = \mathrm{i} + \frac{2}{\sqrt{\pi}} \int \limits_{\mathrm{i} \infty}^x \mathrm{e}^{t^2} \, \mathrm{d} t \, .$$ Now we can integrate by parts without any problems and since $\mathrm{i}$ is completely negligible compared to $\operatorname{erfi}(x)$ in the limit $x \to \infty$ , we obtain the desired expansion (with $(-1)!! \equiv 1$): $$ \operatorname{erfi}(x) \sim \frac{\mathrm{e}^{x^2}}{\sqrt{\pi} \, x} \sum \limits_{k=0}^\infty \frac{(2k-1)!!}{(2 x^2)^k} \, , \, x \to \infty \, . $$

$\endgroup$
  • $\begingroup$ I know you are saying that $i$ is negligible compared to erfi($x$), but how can erfi($x$) have an imaginary part for real $x$? $\endgroup$ – David S Jul 26 '18 at 14:24
  • $\begingroup$ It does not have one. In every step of the derivation the imaginary part of the remaining integral is $-\mathrm{i}$, so the exact expression for $\operatorname{erfi}(x)$ remains real at all times. $\endgroup$ – ComplexYetTrivial Jul 26 '18 at 20:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.