If a function $E(z)$ is defined on $\mathbb{C}$ by $$ E(z) = \int_0^z e^{-w^2} dw,$$ find a power series expansion for $E(z)$ about $0$. What does this power series converge?

I know how this should go. I need to find a power series expansion of $e^{-w^2}$, and then integrate term by term to find the final result.

In class, we usually used some kind of trick to transform the integrand into a geometric series, but I can't quite see how to do that here. Any hints or suggestions as to this first step would be most appreciated.


Do you know the power series for $e^x$? Try plugging $-w^2$ into it...

  • $\begingroup$ We have $e^x = \sum_{k=0}^\infty \frac{z^k}{k!}$ Thus, $e^{-w^2} = \sum_{k=0}^\infty \frac{(-w^2)^k}{k!}$. This would lead to $E(z) = \sum_{k=0}^\infty \int_0^z \frac{(-w^2)^k}{k!}dw$ Evaluating the integral, we get $E(z)=\sum_{k=0}^\infty \frac{1}{k!}\left(\frac{z(-z^2)^k}{2k+1}\right)$. Does this seem correct (i.e. is it in the form of a power series $\sum_{k=0}^\infty c_k(z-z_0)^k$ where $z_0=0$?) $\endgroup$ – Hildegarde Nov 1 '14 at 20:39
  • $\begingroup$ That all looks right. Note that you can simplify $z(-z^2)^k$ by using the standard exponential laws, and then it'll look more like a power series... $\endgroup$ – Micah Nov 1 '14 at 20:55
  • $\begingroup$ Ah, I see. Great, thanks for your help. $\endgroup$ – Hildegarde Nov 1 '14 at 20:58

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