Integrals over an infinite interval.

For the discussion of math. Duh.

Moderators: gmalivuk, Moderators General, Prelates

xepher
Posts: 83
Joined: Tue Mar 09, 2010 1:42 am UTC

Integrals over an infinite interval.

Postby xepher » Mon Oct 11, 2010 12:47 pm UTC

How would one go about evaluating them? For example, a Fourier transform. Or a normal distribution curve.

MidsizeBlowfish
Posts: 77
Joined: Tue Aug 11, 2009 9:17 pm UTC
Location: Champaign, IL

Re: Integrals over an infinite interval.

Postby MidsizeBlowfish » Mon Oct 11, 2010 1:54 pm UTC

Check out the wikipedia article on improper integrals.

User avatar
minno
Posts: 152
Joined: Fri Jul 31, 2009 11:33 pm UTC
Location: Where you least expect it.

Re: Integrals over an infinite interval.

Postby minno » Wed Oct 13, 2010 2:59 am UTC

Essentially, you can take the limit as one or both bounds approach infinity, and as long as it comes out to be finite you can consider that to be the area under the infinite interval.
If you fight fire with fire, you'll get twice as burned.

User avatar
Dopefish
Posts: 855
Joined: Sun Sep 20, 2009 5:46 am UTC
Location: The Well of Wishes

Re: Integrals over an infinite interval.

Postby Dopefish » Wed Oct 13, 2010 3:06 am UTC

Improper integrals are the way to go, followed by limit taking.

That doesn't mean you can always explicitly evaluate it though. Trying to integrate the normal curve for example is gonna spawn an error function, which you can only approximate numerically, which is why tables exist when dealing with normal distributions.

User avatar
Eebster the Great
Posts: 3484
Joined: Mon Nov 10, 2008 12:58 am UTC
Location: Cleveland, Ohio

Re: Integrals over an infinite interval.

Postby Eebster the Great » Wed Oct 13, 2010 6:18 am UTC

Dopefish wrote:Improper integrals are the way to go, followed by limit taking.

That doesn't mean you can always explicitly evaluate it though. Trying to integrate the normal curve for example is gonna spawn an error function, which you can only approximate numerically, which is why tables exist when dealing with normal distributions.

That is perhaps not the best example, as some improper integrals of the gaussian function do have a closed form (e.g. pi^(1/2)).

User avatar
Dopefish
Posts: 855
Joined: Sun Sep 20, 2009 5:46 am UTC
Location: The Well of Wishes

Re: Integrals over an infinite interval.

Postby Dopefish » Wed Oct 13, 2010 7:33 pm UTC

I only used that since they mentioned it in their post. If they were looking to be able to work out the area from -infinity to 'a' for example by hand, they're not going to be able to do it exactly.

In the case of the normal distribution, it's normalised to begin with anyway, so you know the total area under that curve to be 1 right from the start, so you don't really need the -infinity to infinity case. Although I suppose that might provide some insight on how it was normalised to begin with...

Although that does remind me, integrating a gaussian draws on some trickery by trying to find the integral squared and then jumping over to polar, and only then can you deal with the improper integral, so that's just another example of how you sometimes need to jump through some extra hoops to integrate over an infinite region.

User avatar
Dason
Posts: 1311
Joined: Wed Dec 02, 2009 7:06 am UTC
Location: ~/

Re: Integrals over an infinite interval.

Postby Dason » Wed Oct 13, 2010 8:33 pm UTC

Dopefish wrote:In the case of the normal distribution, it's normalised to begin with anyway,

But how do you know that? Somebody told you it was a pdf? Are you just going to take their word for it? They're probably lying to you.
double epsilon = -.0000001;

User avatar
Dopefish
Posts: 855
Joined: Sun Sep 20, 2009 5:46 am UTC
Location: The Well of Wishes

Re: Integrals over an infinite interval.

Postby Dopefish » Wed Oct 13, 2010 10:16 pm UTC

If it wasn't normalised, then by definition it wouldn't be the normal distrubution. It'd just be a gaussian that potentially looks similar.

As I went on to say though, the ability to integrate gaussians (from -infinity to infinity at least) can allow you to verify that what someone claims is the normal distribution is, and I semi-sketched out how one would actually do so. (If I spoke TeX sufficiently well I'd probably derive the pi^(1/2) result here just to demonstrate, along with explicitly showing the limit taking, but TeX is a skill I presently lack.) In actual practice though, when you deal with the normal distribution, you're not going to deal with integrals, but tables instead.

User avatar
jestingrabbit
Factoids are just Datas that haven't grown up yet
Posts: 5967
Joined: Tue Nov 28, 2006 9:50 pm UTC
Location: Sydney

Re: Integrals over an infinite interval.

Postby jestingrabbit » Wed Oct 13, 2010 10:31 pm UTC

[math]\begin{align*} \left( \int_{-\infty}^\infty e^{-x^2} dx\right)^2 &= \left( \int_{-\infty}^\infty e^{-x^2} dx\right)\left( \int_{-\infty}^\infty e^{-y^2} dy\right) \\
&= \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-x^2}\cdot e^{-y^2} dx\ dy\\
&= \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} dx\ dy\\
&= \int_0^{2\pi} \int_0^\infty e^{-r^2} r dr\ d\theta\\
&= \left(\int_0^{2\pi} d\theta\right) \left[ \frac{-1}{2}e^{-r^2} \right]_0^\infty \\
&= 2\pi ( 0 - (-1/2) )\\
&= \pi\end{align*}[/math]

That's not the greatest LaTeX, there are nicer ways to do the dx stuff.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.


Return to “Mathematics”

Who is online

Users browsing this forum: No registered users and 11 guests