Taylor expansion of the Error function

Question

The error function $\operatorname{erf}(z)$ is defined by the integral $$ \operatorname{erf}(z)=\frac{2}{\sqrt{\pi}} \int_0^z e^{-t^2}\,dt,\quad t\in\mathbb R$$

Find the Taylor expansion of $\operatorname{erf}(z)$ around $z_0=0$.

How to expand a function, when it is defined by the integral sign? Can somebody explain the steps to compute this?

Answer 1

For functions obeying just some very fairly modest restrictions, you differentiate under the integral sign by evaluating the function being integrated at the endpoint of the integral; this is the fundamental theorem of integral calculus.

So $$ \frac{d}{dz} \frac{2}{\pi} \int_0^z e^{-t^2}dt =\frac{2}{\pi} e^{-z^2} $$ At $z = 0$, this is $\frac{2}{\pi}$ And then the next derivative is easy: $$ \left. \frac{d}{dz} \frac{2}{\pi} e^{-z^2} \right|_{z=0}= -\frac{2}{\pi} \left. 2ze^{-z^2} \right|_{z=0} = 0 $$ And then $$ \left. \frac{d^2}{dz^2} \frac{2}{\pi} e^{-z^2} \right|_{z=0} = \frac{2}{\pi} \left( \left. \left[ 4z^2-2\right]e^{-z^2} \right|_{z=0} \right) =\frac{4}{\pi} $$ And so on from there. Note that all the odd derivatives are zero.

Answer 2

As $$e^t=\sum_{n=0}^\infty\frac{t^n}{n!},$$ we have $$e^{-t^2}=\sum_{n=0}^\infty\frac{(-1)^nt^{2n}}{n!}.$$ Integrating term to term (why is possible?), $$\frac{\sqrt\pi}{2}\text{erf}(z)=\sum_{n=0}^\infty\frac{(-1)^nt^{2n+1}}{(2n+1)n!}.$$