Providing the maclaurin series for $\int_0^x te^{(-t)^2} dt$?

Question

Find the Maclaurin series for $\int_0^x te^{-t^2} dt$ ?

I know it has something to do with the Maclaurin series for $\int_0^x e^{-t^2} dt$, but that's about it

Answer 1

Given

$$f(x)=\displaystyle\int_0^xt\,e^{-t^2}\,dt=\left[-\frac{1}{2}e^{-t^2}\right]_0^x =\frac{1}{2}\left(1-e^{-x^2}\right)$$

We know that Maclaurin's series for $e^{-x}$ is

\begin{equation} e^{-x}=\sum_{k=0}^\infty (-1)^k\,\dfrac{x^k}{k!} \end{equation}

therefore the series for $e^{-x^2}$ is

\begin{equation} e^{-x^2}=\sum_{k=0}^\infty (-1)^k\,\dfrac{x^{2k}}{k!} \end{equation}

Therefore

\begin{eqnarray} \frac{1}{2}\left(1-e^{-x^2}\right)&=&\frac{1}{2}\left(1-\sum_{k=0}^\infty (-1)^k\,\dfrac{x^{2k}}{k!}\right)\\ &=&\sum_{k=1}^\infty(-1)^{k+1}\,\dfrac{x^{2k}}{2k!} \end{eqnarray}