# Thread: Ok I am drawing a blank

1. ## Ok I am drawing a blank

$\displaystyle integral$(e^(-x²))...from there I went to e^(-x²)≡$\displaystyle sigma$[(-x²)^n/n!]=$\displaystyle sigma$[(-1)^n*x^(2n)/n!] so$\displaystyle integral$(e^(-x²))=$\displaystyle integral$[!]=$\displaystyle sigma$[(-1)^n*x^(2n)/n!] dx]...but I am stuck on where to go from here if there is route

2. ## Sorry

It should read $\displaystyle integral$(e^(-x²))=$\displaystyle sigma$[(-1)^n*x^(2n)/n!] dx]...

3. Well it's just a direct application of Maclaurin series for $\displaystyle e^x.$

4. ## This is embarrassing

I've learned it but forgotten could you elaborate

5. Originally Posted by Mathstud28
$\displaystyle integral$(e^(-x²))...from there I went to e^(-x²)≡$\displaystyle sigma$[(-x²)^n/n!]=$\displaystyle sigma$[(-1)^n*x^(2n)/n!] so$\displaystyle integral$(e^(-x²))=$\displaystyle integral$[!]=$\displaystyle sigma$[(-1)^n*x^(2n)/n!] dx]...but I am stuck on where to go from here if there is route
$\displaystyle \int \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{n!(2n+1)}$

6. ## Just checking

but that holds true because n! and (-1)^n are constants and you just yanked them out evaluated your integral and then reintroduced them?