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How do I prove the error function $$ \mbox{erf}(z) = \frac{2}{\sqrt{\pi}} \int_{0}^{z} e^{-t^{2}} dt. $$ is entire?

Could you give me some scratch proof?

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A related problem. First make the change of variables $ t=zy $, then advance with the proof as in this answer. Changing the variables results in the following integral

$$ \text{erf}(z) = \frac{2}{\pi}\int_{[0,z]} e^{-t^2}\,dt=\frac{2}{\sqrt{\pi}}\,{\int _{0}^{1}\!{z\,{\rm e}^{-{z}^{2}{y}^{2}}}{dy}}

\,.$$

Let $[0,z]$ denote the straight-line path from $0$ to $z$. Then we define $$\text{erf}(z) = \frac{2}{\pi}\int_{[0,z]} e^{-t^2}\,dt.$$

Now note, using Cauchy’s theorem (and the analyticity of $e^{-t^2}$), that

$$\frac{\text{erf}(z+h) – \text{erf}(z)}{h} = \frac{2}{\pi h}\int_{[0,z+h] – [0,z]} e^{-t^2}\,dt = \frac{2}{\pi h}\int_{[z,z+h]} e^{-t^2}\,dt.$$ Finally, this last expression tends to $(2/\pi)e^{-z^2}$ as $h\to 0$, so that $\text{erf}$ is differentiable with derivative $\text{erf}^\prime(z) = 2e^{-z^2}/\pi$. Indeed, let $\epsilon>0$. Choose $h$ small enough that $e^{-t^2}$ differs from $e^{-z^2}$ by less than $\epsilon$ as long as $|t-z|\leq|h|$. Then $$\left|\frac{1}{h}\int_{[z,z+h]} e^{-t^2}\,dt – e^{-z^2}\right| = \left|\frac{1}{h}\int_{[z,z+h]} (e^{-t^2} – e^{-z^2})\,dt\right| \leq \frac{1}{h} \int_{[z,z+h]} |e^{-t^2} – e^{-z^2}|\,dt \leq \epsilon.$$

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