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

1. Jan 7, 2010

Gregg

1. The problem statement, all variables and given/known data

$$\int_{-\infty}^{\infty} e^{x\over 2}e^{-x^2\over 2} dx$$

2. Relevant equations

$$\int_{-\infty}^{\infty}e^{-x^2\over a} dx = \sqrt{\pi\over a}$$ $$a>0$$

3. The attempt at a solution

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

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

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

2. Jan 7, 2010

rock.freak667

$$e^{\frac{x^2}{2}}e^{\frac{-x^2}{2}}=e^{\frac{x^2-x^2}{2}}=e^0=1$$

Are you sure you typed the integrand properly?

EDIT: sorry I am blind.

Last edited: Jan 7, 2010
3. Jan 7, 2010

Gregg

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

4. Jan 7, 2010

Gregg

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

5. Jan 7, 2010

Pengwuino

Try completing the square in the exponent.

6. Jan 7, 2010

Gregg

$$\int_{-\infty}^{\infty} e^{x\over 2}e^{{-x^2}\over 2} dx = \int_{-\infty}^{\infty} e^{-\frac{1}{2}(x-\frac{1}{2})^2+\frac{1}{8}} dx$$

$$= e^{\frac{1}{8}} \int_{-\infty}^{\infty} e^{\frac{u^2}{-2}} du = e^{\frac{1}{8}}\sqrt{\frac{\pi}{2}}$$

Thanks!

7. Jan 8, 2010

t!m

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

8. Jan 8, 2010

Gregg

That was a typo by me!