# integral of exp(-x^2)

• November 29th 2011, 11:18 AM
moonnightingale
integral of exp(-x^2)
what is integral of
• November 29th 2011, 01:20 PM
TheEmptySet
Re: integral of exp(-x^2)
Quote:

Originally Posted by moonnightingale
what is integral of

$e^{-x^2}$ does not have an elementry antiderivative. The integral can be calulated using Taylor series.

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

So we can integrate this term by term to get

$\int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{1}x^{2n}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}$

This can be used to approximate the value to any accuracy you wish. The error bound is simple because the series is alternating.
• November 30th 2011, 04:02 AM
moonnightingale
Re: integral of exp(-x^2)
Quote:

Originally Posted by TheEmptySet
$e^{-x^2}$ does not have an elementry antiderivative. The integral can be calulated using Taylor series.

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

So we can integrate this term by term to get

$\int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{1}x^{2n}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}$

This can be used to approximate the value to any accuracy you wish. The error bound is simple because the series is alternating.

Hi if i solve the question by simpsons rule or by matlab i get answer .745
but i am unable to get this answer by manual method
plz explain me manual method by puttiing values
• November 30th 2011, 04:16 AM
HallsofIvy
Re: integral of exp(-x^2)
What do you mean by "manual method"?
• November 30th 2011, 05:34 AM
moonnightingale
Re: integral of exp(-x^2)
i mean how to solve this by hand
by standard integration method
• November 30th 2011, 05:35 AM
moonnightingale
Re: integral of exp(-x^2)
Kindly tell me how u will solve this integration question on paper
• November 30th 2011, 06:20 AM
CaptainBlack
Re: integral of exp(-x^2)
Quote:

Originally Posted by moonnightingale
Kindly tell me how u will solve this integration question on paper

You have already been told that this is not expressible a closed form in terms of elementary functions. You have been given a series approximation which will give the numerical value to any precision you require.

$\int_{0}^{1}\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{n!} dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{n!}\int_{0}^{1}x^{2n}dx = \sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)n!}\\ \\ \phantom{SSSSSSS}=\sum_{n=0}^{N-1}\frac{(-1)^n}{(2n+1)n!} +R_N$

Where $|R_N|<\frac{1}{(2N+1)N!}$

CB
• November 30th 2011, 08:28 AM
moonnightingale
Re: integral of exp(-x^2)
can u tell me answer of my question
with simpson rule it comes to be
0.7455
• November 30th 2011, 08:49 AM
HallsofIvy
Re: integral of exp(-x^2)
You have been told twice now, by TheEmptySet and Captain Black, that [itex]e^{-x^2}[/tex] does not have an "elementary" anti-derivative.

If you insist upon such an answer then it is $\frac{\sqrt{\pi}}{2} erf(1)$ where "erf" is the Gauss error function. It is defined as $\frac{2}{\sqrt{\pi}}\int_0^x e^{-x^2}dx$. Does that help?
• November 30th 2011, 08:53 AM
Opalg
Re: integral of exp(-x^2)
Quote:

Originally Posted by moonnightingale
can u tell me answer of my question
with simpson rule it comes to be
0.7455

If you multiply it by $2/\sqrt\pi$, this integral becomes the error function, for which detailed tables of values are available (for example on the Wikipedia page that I just linked to). This gives erf(1) = 0.8427008. So your integral is that number times $\sqrt\pi/2$, or approximately 0.7468241.
• November 30th 2011, 09:16 AM
Amer
Re: integral of exp(-x^2)
cant we use that method suppose

$I = \int_{0}^{1} e^{-x^2} \cdot dx$

$I^2 = \int_{0}^{1} \int_{0}^{1} e^{-x^2}\cdot e^{-y^2} \; dx \;dy$

then transfer to polar ??
• November 30th 2011, 09:19 AM
TheEmptySet
Re: integral of exp(-x^2)
Quote:

Originally Posted by Amer
cant we use that method suppose

$I = \int_{0}^{1} e^{-x^2} \cdot dx$

$I^2 = \int_{0}^{1} \int_{0}^{1} e^{-x^2}\cdot e^{-y^2} \; dx \;dy$

then transfer to polar ??

The problem is you need to integrate over the unit square in polar coordinates and that is very difficult. When we go to infinity we can just integrate over a large circle of radius r.