Series integration

• Nov 5th 2006, 12:50 AM
jmytil
Series integration
We have to calculate Integrate[e^(-x^2),{x,0,0.1}] using series theory. Any help?
Thanks!
• Nov 5th 2006, 03:43 AM
CaptainBlack
Quote:

Originally Posted by jmytil
We have to calculate Integrate[e^(-x^2),{x,0,0.1}] using series theory. Any help?
Thanks!

$\displaystyle \int_0^{0.1} e^{-x^2}dx$

We will do this by expanding the integrand as a power series and then integrating term by term.

$\displaystyle e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!}$

so:

$\displaystyle e^{-x^2}=\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{n!}$

Hence:

$\displaystyle \int_0^{0.1} e^{-x^2}dx=\int_0^{0.1} \left[\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{n!}\right] dx$

Now we will interchange the order of the integration and summation, which
is valid in this case but I will leave the justification for you to provide:

$\displaystyle \int_0^{0.1} e^{-x^2}dx= \sum_{n=0}^{\infty} \left[ \frac{(-1)^n}{n!} \int_0^{0.1}x^{2n} dx \right]$

so:

$\displaystyle \int_0^{0.1} e^{-x^2}dx= \sum_{n=0}^{\infty} \left[ \frac{(-1)^n}{n!(2n+1)} (0.1)^{2n+1} \right]$

Summing the first two terms on the right gives an estimate for the integral of 0.0996666667
compared to a fairly accurate numerical integration which gives: 0.0996676643, and the
sum of the first five terms of the series give: 0.0996676643.

RonL