How do you express the integral of e^(-x^(2)) as an infinite series??

Question

anonymous · Answer

$$e^{-x^2}$$ ?

anonymous · Answer

Yes

anonymous · Answer

if so , 
$$e^x=\sum _{k=0}^{\infty } \frac{x^k}{k!}$$
replace x with -x^2
$$\sum _{k=0}^{\infty } \frac{\left(-x^2\right)^k}{k!}$$

anonymous · Answer

That's all?? Oh wow xD

anonymous · Answer

yes

zarkon · Answer

what about the integral part?

anonymous · Answer

I don't know about that

zarkon · Answer

integrate the series above term by term

zarkon · Answer

$$\int e^{-x^2}dx=\int\sum_{k=0}^{\infty}\frac{(-x^2)^k}{k!}dx=\sum_{k=0}^{\infty}\int\frac{(-x^2)^k}{k!}dx$$

anonymous · Answer

inf inf is sqrt pi...

zarkon · Answer

we don't have limits of integration

zarkon · Answer

$$\sum_{k=0}^{\infty}\int\frac{(-x^2)^k}{k!}dx=\sum_{k=0}^{\infty}\int\frac{(-1)^kx^{2k}}{k!}dx$$

$$\sum_{k=0}^{\infty}\int\frac{(-x^2)^k}{k!}dx=\sum_{k=0}^{\infty}\frac{(-1)^k}{k!}\int x^{2k}dx$$

anonymous · Answer

I remeber doing this in a probability course..

zarkon · Answer

really....i doubt that...maybe this

$$\int\limits_{-\infty}^{\infty}e^{-x^2}dx$$

but that is a different problem than what we have here

anonymous · Answer

Yes, that one..root pi, right...

zarkon · Answer

sure...but the answer  to the original problem is a function not a number

anonymous · Answer

You don't need limits of integrations of this. the final answer should be an integral of a series. leaving you with a final series but I figured it out anyways. It equals to the $$\sum_{n=0}^{\infty} ((-1)^{n}x ^{2n+1})/n!(2n+1)$$