1. ## Integration question

I would like to know if I have the correct answer concerning:

$\int e^{x^2} dx = xe^{x^2} -2xe^{x^2} + e^{x^2}$ ?

and expressing it as a Maclaurin series would be:

$e^{x^2} = \sum^{\infty}_{n=0} \frac{(x^2)^n}{n!}$

$\int e^{x^2} dx = \int \sum^{\infty}_{n=0} \frac{(x^2)^n}{n!} dx= \int \left( 1+ x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} +\frac{x^8}{4!} + ... + \frac{(x^2)^n}{n!} \right)dx$ $= \left( x + \frac{x^3}{3} + \frac{x^5}{5 \cdot 2!} + \frac{x^7}{7 \cdot 3!} + \frac{x^9}{9 \cdot 4!} + ... + \frac{x^{2n+1}}{(2n+1) \cdot n!}\right)$ $= \left( x + \frac{x^3}{3} + \frac{x^5}{10} + \frac{x^7}{42} + \frac{x^9}{216} + ... + \frac{x^{2n+1}}{(2n+1) \cdot n!}\right)$

2. Originally Posted by lllll
I would like to know if I have the correct answer concerning:

$\int e^{x^2} dx = xe^{x^2} -2xe^{x^2} + e^{x^2}$ ? Mr F says: The right hand side is dead >wrong<. No closed form in terms of elementary functions exists.

and expressing it as a Maclaurin series would be:

$e^{x^2} = \sum^{\infty}_{n=0} \frac{(x^2)^n}{n!}$

$\int e^{x^2} dx = \int \sum^{\infty}_{n=0} \frac{(x^2)^n}{n!} dx$

Mr F asks: What's happened to the infinite sum in the expressions below?? It's gone AWOL. You need to show using appropriate notation that these sums go on forever.

$= \int \left( 1+ x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} +\frac{x^8}{4!} + ... + \frac{(x^2)^n}{n!} \right)dx$ $= \left( x + \frac{x^3}{3} + \frac{x^5}{5 \cdot 2!} + \frac{x^7}{7 \cdot 3!} + \frac{x^9}{9 \cdot 4!} + ... + \frac{x^{2n+1}}{(2n+1) \cdot n!}\right)$ $= \left( x + \frac{x^3}{3} + \frac{x^5}{10} + \frac{x^7}{42} + \frac{x^9}{216} + ... + \frac{x^{2n+1}}{(2n+1) \cdot n!}\right)$
..

3. I have a question: why you need to give a series representation for that integral? (You forgot the constant too.)