## Integrals over an infinite interval.

For the discussion of math. Duh.

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xepher
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### Integrals over an infinite interval.

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

MidsizeBlowfish
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### Re: Integrals over an infinite interval.

Check out the wikipedia article on improper integrals.

minno
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### Re: Integrals over an infinite interval.

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.
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Dopefish
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### Re: Integrals over an infinite interval.

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.

Eebster the Great
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### Re: Integrals over an infinite interval.

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)).

Dopefish
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### Re: Integrals over an infinite interval.

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.

Dason
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### Re: Integrals over an infinite interval.

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.
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Dopefish
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### Re: Integrals over an infinite interval.

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.

jestingrabbit
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### Re: Integrals over an infinite interval.

\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*}

That's not the greatest LaTeX, there are nicer ways to do the dx stuff.
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