How to solve an integral of a gaussian distribution

[DragonPetter]

Homework Statement

integrate

[tex]\int_{-\infty}^\infty\! e ^{(x-a)^{2}}\, dx[/tex]

Homework Equations

[tex]\int \! e^u\, du = e^u + C[/tex]

The Attempt at a Solution

i just know that du = 2(x-a), but there is no x to make use of substitution, so I am confused on how to go about solving this since I cannot use substitution

 

[foxjwill]

EDIT: I'm assuming you mean

[tex]
\int_{-\infty}^\infty\! e ^{-(x-a)^{2}}\, dx
[/tex]​

since what you wrote doesn't converge.Anyway, this is a very tricky integral that requires a trick. By far the best way starts as follows:

First, letting I denote the original integral (after replacing x-a with x), notice that

[tex]I = 2\int_{0}^\infty\! e ^{-x^{2}}\, dx[/tex]​

by symmetry. Next comes the first trick:

[tex]I^2 = 4 \int_{0}^\infty\! e ^{-x^{2}}\, dx \int_{0}^\infty\! e ^{-x^{2}}\, dx = 4 \int_{0}^\infty\! e ^{-x^{2}}\, dx \int_{0}^\infty\! e ^{-y^{2}}\, dy[/tex]​

since the x in the integral is just a dummy variable. But then

[tex]I^2 = 4\int_0^\infty\int_0^\infty\! e ^{-(x^{2}+y^{2})}\,dx\,dy.[/tex]​

The traditional method is to at this point switch to polar coordinates, but a better way (I think) involves the substitution y=xu (so dy=xdu):

[tex]I^2 = 4\int_0^\infty\int_0^\infty\! e^{-x^2(1+u^2)}x\,dx\,du.[/tex]​

From here, I think you should be able to figure this out on your own.

 

[DragonPetter]

hmm my textbook is certainly not nice as it leaves this as an exercise with just telling me "(look up any integrals you need)"
and yes, i forgot to include the - sign for the exponential

 

[foxjwill]

hmm my textbook is certainly not nice as it leaves this as an exercise with just telling me "(look up any integrals you need)"

That's wierd.

 

[statdad]

On a wild hunch: this seems to be a traditional type problem in a first mathematical stat or calc-based probability course. If you've seen this (normal density integrating to 1)

[tex]
\int_{-\infty}^\infty \frac 1 {\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{\sigma^2}} \, dx = 1
[/tex]

the "trick" is to identify your integrand as part of some normal density and go from there.

of course, the item i have posted could be what the hint "look up any integrals needed" refers to as well.

 

[foxjwill]

[tex]
\int_{-\infty}^\infty \frac 1 {\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{\sigma^2}} \, dx = 1
[/tex]

You mean

[tex]
\int_{-\infty}^\infty \frac 1 {\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \, dx = 1
[/tex]​

of course

 

[statdad]

You mean

[tex]
\int_{-\infty}^\infty \frac 1 {\sqrt{2\pi} \sigma} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \, dx = 1
[/tex]​

of course

Yes - many thanks for correcting my typing, and more for possibly clearing up a problem for the original poster.