calculating the taylor series when there is an integral involved one of the exercises is to calculate the taylor expansion at x=0 and degree 4 for some function. For example:
$$\int_{0}^{x} e^{-t^{2}} dt$$
I actually have no clue how to get started. I know how to approximate a function with taylor theorem but the integral sign confuses me. Can someone maybe briefly explain or point to a good resource for this?
Thanks in advance! 
 A: Hint: define $f(x) = \int_0^x e^{-t^2} dt$. What are $f'(x)$, $f''(x)$, etc?
A: Look up the Taylor Series expansion for $e^{x}$ but instead of $x$ you have $-t^{2}$. It is a power series.
A: Let 
$$f(x) = \int_{0}^{x} e^{-t^{2}} dt$$
and use Leibniz rule to obtain
\begin{align}
f^{'}(x) &= e^{- x^{2}} \\
f^{''}(x) &= - 2 x \, e^{- x^{2}} = (-1)^{1} \, H_{1}(x) \, e^{- x^{2}} \\
f^{'''}(x) &= (4 x^{2} - 2) \, e^{- x^{2}} = (-1)^{2} \, H_{2}(x) \, e^{- x^{2}}. 
\end{align}
The general formula is
\begin{align}
f^{(n+1)}(x) = (-1)^{n} \, H_{n}(x) \, e^{- x^{2}} \hspace{5mm} n \geq 0
\end{align}
where $H_{n}(x)$ are the Hermite polynomials. Now, 
$f(0) = 0$, $f^{(2n+2)}(0) = 0$, $f^{(2n+1)}(0) = (-1)^{n} ((2n)!/n!)$ for which, by use of Taylor's theorem,
\begin{align}
f(x) &= \sum_{m=0}^{\infty} a_{m} \, \frac{x^{m}}{m!} = \sum_{m=0}^{\infty} D^{m}(f(x))|_{x=0} \, \frac{x^{m}}{m!} \\
&= f^{(1)}(0) \, \frac{x^{1}}{1!} + f^{(3)}(0) \, \frac{x^{3}}{3!} + \cdots \\
&= \sum_{m=0}^{\infty} \frac{(-1)^{m} \, x^{2m+1}}{m! \, (2m+1)}.
\end{align}
