# Ok I am drawing a blank

• Mar 31st 2008, 05:15 PM
Mathstud28
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
• Mar 31st 2008, 05:16 PM
Mathstud28
Sorry
It should read $\displaystyle integral$(e^(-x²))=$\displaystyle sigma$[(-1)^n*x^(2n)/n!] dx]...
• Mar 31st 2008, 05:17 PM
Krizalid
Well it's just a direct application of Maclaurin series for $\displaystyle e^x.$
• Mar 31st 2008, 05:18 PM
Mathstud28
This is embarrassing
I've learned it but forgotten (Giggle) could you elaborate
• Mar 31st 2008, 05:20 PM
TheEmptySet
Quote:

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

$\displaystyle \int \sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!}dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{n!(2n+1)}$
• Mar 31st 2008, 05:21 PM
Mathstud28
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?