Problem when trying to find the following maclaurin series because of the integral $ f(x) = \int_0^{x} e^{-t^2} dt $?

Question

I'm trying to find the maclaurin series of:

$ f(x) = \int_0^{x} e^{-t^2} dt $

So far I've got:

$P_n(x) = \frac{f(0)}{0!} + \frac{f'(0)x}{1!} + \frac{f''(0)x^2}{2!} + ...+\frac{f^{(n)}(0)x^n}{n!}$

But I do not know how to find the value of $f(0)$

Could I solve this by finding the maclaurin series of $e^{-t^2}$ and then integrating? I think I probaly would have to do the same integral.

Answer 1

Note that we have $f(0)=\int_0^0 e^{-t^2}\,dt=0$. And from the Fundamental Theorem of Calculus, $f'(x)=e^{-x^2}$ so that $f'(0)=1$.

Answer 2

$$ \int_{0}^x e^{-t^2} dt = \sum_{k=0}^\infty \frac{(-1)^k}{k!}\int_{0}^x t^{2k} dt = \sum_{k=0}^\infty \frac{(-1)^k}{k!}\frac{x^{2k+1}}{2k+1} $$