Integration of a gaussian function

[JHans]

Hey, all. I need help computing a particular integral that's come up in my quantum mechanics homework (Griffith 1.3 for anyone who's interested). It involves integration of a gaussian function:

\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {x}{e^{{-\lambda}{(x-a)^2}}} \, dx

Where \lambda and a are positive real constants.


For the life of me I cannot figure out how to compute that integral... u-substitution doesn't seem to be an option, and integration by parts just makes things more confusing. Can someone please help me?

 

[Ben Niehoff]

Hey, all. I need help computing a particular integral that's come up in my quantum mechanics homework (Griffith 1.3 for anyone who's interested). It involves integration of a gaussian function:

\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {x}{e^{{-\lambda}{(x-a)^2}}} \, dx

Where \lambda and a are positive real constants.


For the life of me I cannot figure out how to compute that integral... u-substitution doesn't seem to be an option, and integration by parts just makes things more confusing. Can someone please help me?

You could do it easily if you had

\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {(x-a)}{e^{{-\lambda}{(x-a)^2}}} \, dx
right? Because then it would be a simple substitution. So write

\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {x}{e^{{-\lambda}{(x-a)^2}}} \, dx = \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {(x-a)}{e^{{-\lambda}{(x-a)^2}}} \, dx + \sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {a}{e^{{-\lambda}{(x-a)^2}}} \, dx
and think about how to do the second piece. It should look familiar.

 

[JHans]

I hadn't thought of that! That's a clever trick. Am I correct in thinking the second integral simplifies to:

{\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {a}{e^{{-\lambda}{(x-a)^2}}} \, dx}={a}

And so the integral:


\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {x}{e^{{-\lambda}{(x-a)^2}}} \, dx

is simply the first integral you wrote down (solvable by u-substitution) + a?

 

[gb7nash]

{\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {a}{e^{{-\lambda}{(x-a)^2}}} \, dx}={a}

How did you obtain a from that? I'm not saying that's wrong, but you need to show how you got it.

 

[JHans]

I'm sorry! Here's the process I went through:

{\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {a}{e^{{-\lambda}{(x-a)^2}}} \, dx}

Because a is a constant, it can be taken out of the integral:

{{a}\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {e^{{-\lambda}{(x-a)^2}}} \, dx}

Then I performed a u-substitution:

{u ={x-a}},{du = dx}

{{a}\sqrt{\frac{\lambda}{\pi}}\int_{-\infty}^{+\infty} \! {e^{{-\lambda}{u^2}}} \, du}

I then used a second u-substitution:

{v = {\sqrt{\lambda}u}}, {\frac{dv}{\sqrt{\lambda}} = du}

{{a}\sqrt{\frac{1}{\pi}}\int_{-\infty}^{+\infty} \! {e^{{-\lambda}{v^2}}} \, dv}


The integral is the well-known Gaussian integral:

{\int_{-\infty}^{+\infty} \! {e^{{-\lambda}{v^2}}} \, dv}={\sqrt{\pi}}

And so I have:

{{a}{\sqrt{\frac{1}{\pi}}}\sqrt{\pi}}={a}

 

[JHans]

Now I'm a bit confused, though... I came across an integral on Wikipedia's list of integrals for exponential functions, and they have this:

\int_{-\infty}^{\infty} \! {x}{e^{-a{(x-b)^2}}} \ dx={b\sqrt{\frac{\pi}{a}}}

How does the integral we've broken into two parts become that?

 

[JHans]

Is anyone able to help? This is seriously stumping me.

 

[Ray Vickson]

Is anyone able to help? This is seriously stumping me.

You have already been given a complete solution, but it does require some final work on your part.RGV

 

[JHans]

I'm sorry, but I'm having difficulty "seeing" this complete solution and how to work it.

I understand what was said earlier, that:

{\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {x}{e^{-\lambda{(x-a)^2}}} \ dx} = {\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {(x-a)}{e^{-\lambda{(x-a)^2}}} \ dx} + {\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {a}{e^{-\lambda{(x-a)^2}}} \ dx}

It seems to me that the first integral is solvable by u-substitution:

{u=(x-a)}, {du=dx}
{\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {(x-a)}{e^{-\lambda{(x-a)^2}}} \ dx} = {\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {u}{e^{-\lambda{u^2}}} \ dx}

The antiderivative of that integral is:
\frac{-e^{-\lambda{u^2}}}{2\lambda}

Evaluating that antiderivative at the limits of negative and positive infinity seems to yield 0.

Meanwhile, as I've shown about, the second integral can be solved by two round of u-substitution, yielding the constant a. Altogether, then:

{\sqrt{\frac{\lambda}{\pi}} \int_{-\infty}^{+\infty} \! {x}{e^{-\lambda{(x-a)^2}}} \ dx}=a

However, that doesn't mesh with the definite integral identity I found online (shown in an earlier reply). Have I gone awry somewhere in my calculations, or is the identity I found incorrect?

EDIT: Nevermind... I realized that in examining the integral identity I found in an integral table, I'd been neglecting to apply the constant in front of the integral. Once I did that, it all clicked and became equivalent to the answer I derived (with help) above. Thank you all so much. I apologize for being a bit dense toward the end.