Improper integral with e^(-x/2)e^(-x^2/2). Realling annoying.

[Gregg]

Homework Statement

[tex]\int_{-\infty}^{\infty} e^{x\over 2}e^{-x^2\over 2} dx [/tex]

Homework Equations

[tex] \int_{-\infty}^{\infty}e^{-x^2\over a} dx = \sqrt{\pi\over a} [/tex] [tex] a>0 [/tex]

The Attempt at a Solution

Can't seem to penetrate it, I thought about trying to isolate the second term with integration by parts.

[tex]\int_{-\infty}^{\infty} e^{x\over 2}e^{-x^2\over 2} dx = e^{x\over 2}\int e^{-x^2\over 2}dx - \int \frac{d}{dx}e^{x\over 2} \left[ \int e^{-x^2\over 2} dx \right] dx [/tex]

But I don't think there's any sensible way to put limits in on the RHS to eliminate those factors.

 

[rock.freak667]

[tex]e^{\frac{x^2}{2}}e^{\frac{-x^2}{2}}=e^{\frac{x^2-x^2}{2}}=e^0=1[/tex]

Are you sure you typed the integrand properly?EDIT: sorry I am blind.

 

[Gregg]

Yeah re-read it! One is squared one isn't.

 

[Gregg]

My last comment could look like "yeah re-read it [past tense]..." I still can't do this.

 

[Pengwuino]

Try completing the square in the exponent.

 

[Gregg]

[tex] \int_{-\infty}^{\infty} e^{x\over 2}e^{{-x^2}\over 2} dx = \int_{-\infty}^{\infty} e^{-\frac{1}{2}(x-\frac{1}{2})^2+\frac{1}{8}} dx [/tex]

[tex] = e^{\frac{1}{8}} \int_{-\infty}^{\infty} e^{\frac{u^2}{-2}} du = e^{\frac{1}{8}}\sqrt{\frac{\pi}{2}} [/tex]

Thanks!

 

[t!m]

Careful, your 'Relevant equation' is wrong. For gaussian integrals,
[tex]\int_{-\infty}^\infty e^{-ax^2}dx = \sqrt{\frac{\pi}{a}}[/tex]

 

[Gregg]

That was a typo by me!