Question

When I integrate using power series....

Answer 1

I have to find: \(\large\color{black}{ \displaystyle \int e^{x^2}~dx }\)

I know that: \(\large\color{black}{ \displaystyle e^{x}=\sum_{n=0}^{\infty}\frac{x^n}{n!} }\)

therefore, \(\large\color{black}{ \displaystyle e^{x^2}=\sum_{n=0}^{\infty}\frac{(x^2)^n}{n!} }\) which simplifies to \(\large\color{black}{ \displaystyle e^{x^2}=\sum_{n=0}^{\infty}\frac{x^{2n}}{n!} }\)


then, after integrating both sides with respect to x

\(\large\color{black}{ \displaystyle \int e^{x^2} dx=\int \sum_{n=0}^{\infty}\frac{x^{2n}}{n!}dx }\)

and the right side becomes,

\(\large\color{black}{ \displaystyle \int e^{x^2} dx= \sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)n!}dx +C }\)

Answer 2

Is this correct, or the +C shouldn't be there?

Answer 3

oh, I have the dx in my last expression, it shoudn't be there

Answer 4

Looks good so far. C should be there.

Answer 5

\(\large\color{black}{ \displaystyle \int e^{x^2} dx= \sum_{n=0}^{\infty}\left(\frac{x^{2n+1}}{(2n+1)n!} +C \right)}\)

Answer 6

or is +C not included in the series?

Answer 7

\(\large\color{black}{ \displaystyle \int e^{x^2} dx= \sum_{n=0}^{\infty}\left(\frac{x^{2n+1}}{(2n+1)n!} +C \right)}\)

\(\large\color{black}{ \displaystyle \int e^{x^2} dx= \sum_{n=0}^{\infty}\left(\frac{x^{2n+1}}{(2n+1)n!} \right)+C}\)

it is the second one I guess?

Answer 8

No C shouldn't be included because otherwise this series will diverge.

Answer 9

So yeah it's second one.

Answer 10

oh, true than (adding C's forver) .. thx for the help

Answer 11

Unless C=0, but we don't know, so yeah. Welcome.