Use a known power series expansion to find the power series representation of the integral function $g(x) =\int_0^x e^{-t^2}dt$ centered at $a=0$

My approach

Note that $g'(x) = e^{-x^2}$. We also know the Maclaurin Series for $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$.

Then, $$g'(x)=e^{-x^2} = \sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}$$ $$g(x) = \int \bigg[\sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!}\bigg] dx$$ $$g(x) = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \int x^{2n}dx $$ $$g(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{n!(2n+1)} +K, \text{ where $K$ is unknown}$$

Consider $x=0$, we figured out that $g(0) = 0 +K = 0$, hence $K=0$. Hence $$g(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{n!(2n+1)}$$

Is this right? I cannot differentiate between power series and Maclaurin series or stuffs like that. Can anybody clarify on this too?


1 Answer 1


It is correct, although I would have integrated each term between $0$ and $x$ instead of the indefinite integral.

Power series are series of the form $\sum_{n=0}^\infty a_n(x-x_0)^n$. A Taylor series is a series where $a_n=f^{(n)}(x_0)/n!$ and $f$ is a $C^\infty$ on a neighborhood of $x_0$. If $x_0=0$, it is called a Maclaurin series.

  • $\begingroup$ thanks! so does that mean the Maclaurin/Taylor Series considered to be power series? $\endgroup$
    – Joshua
    Dec 14, 2014 at 22:40
  • $\begingroup$ They ARE power series. $\endgroup$ Dec 14, 2014 at 22:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .