What is the Integral of x^{3}e^{-x^2}dx?

[amacinho]

Hi,

I'm trying to show the following equation is correct:

$$\int_{-\infty}^{\infty}x^{3}e^{-x^2}dx = 0$$

I obtained the result as 0 using Mathematica but couldn't figure out a way to evaluate the integral.

I am just an unfortunate computer scientist who happens to follow a graduate course on statistical mechanics :)

Also, I would appreciate if some one tells me how to query the forums to find an answer for a particular integration before posting it here.

Thanks,

Amac

 

[arildno]

Hint:
Is the integrand an even or odd function?

 

[amacinho]

Unfortunately $$x$$ can take negative values if that's what you ask.

 

[amacinho]

I see, I am expected to show my own work when posting a question here. Actually that integral is just a small part of an answer to calculate the first four moments of a gaussian distribution with direct integration of the pdf.

I tried to re-write the integral as
$$\frac{1}{2}\int_{-\infty}^{\infty}t e^{-t}dt$$
by using the substitution
$$t=u^2$$, $$dt=2u du$$
so
$$u du=\frac{dt}{2}$$.
By this approach I can say that t is positive and evaluate the integral as a Gamma function by carrying it from 0 to infinity and multiplying by 2. The problem is the result is not 0 which I am expected to get.

 

[arildno]

No, my hint concerned the type of integrand you're having.
With an EVEN function f, we mean that for any x, f(-x)=f(x)

With an ODD function, we mean that for any x, we have f(-x)=-f(x)

Suppose you are to integrate an odd function between the values -a and a.
What should that integral be?

 

[amacinho]

I got it arildno, thanks very much.

Could you also tell me how to search for the integrals before posting them here? I tried to use the latex code of the integral but it didn't help.

 

[Gib Z]

its quite difficult to search for a specific integral, best bet is it search integral. no latex won't work with the searches.

Basically when you have a function like arildno described, odd, then on the postive side of x, x>0, the values are the same as the values on the negative side, just with a different sign! so they cancel each other out.

 

[theperthvan]

functions.wolfram.com has good functions search

 

[HallsofIvy]

Arildno's point was that you don't need to find an anti-derivative. The fact that that integrand is an odd function tells you that its graph is anti-symmetric. Any "area under the curve" for for x> 0 is canceled by the "area above the curve" for x< 0. More specifically, the anti-derivative of any odd function is an even function. Evaluating it at any A and -A give the so subtracting results in 0.

However there are other ways of integrating functions than "looking them up"! In this case you can write the integral as
$$\int x^2 e^{-x^2} (xdx)$$
and make the substitution u= x². That reduces the integral to
$$2\int u e^{-u}du$$
which can be done by a simple integration by points.

However, arildno's point, as I said, is that you don't need to actually do the integral to get the answer!

 

[Gib Z]

Arildno's point was that you don't need to find an anti-derivative. The fact that that integrand is an odd function tells you that its graph is anti-symmetric. Any "area under the curve" for for x> 0 is canceled by the "area above the curve" for x< 0. More specifically, the anti-derivative of any odd function is an even function. Evaluating it at any A and -A give the so subtracting results in 0.

However there are other ways of integrating functions than "looking them up"! In this case you can write the integral as
$$\int x^2 e^{-x^2} (xdx)$$
and make the substitution u= x². That reduces the integral to
$$2\int u e^{-u}du$$
which can be done by a simple integration by points.

However, arildno's point, as I said, is that you don't need to actually do the integral to get the answer!

That $$2\int u e^{-u} du$$ should be a $$\frac{1}{2}\int u e^{-u} du$$ I think halls.

 

[amacinho]

Yeah, as I said I got the idea of integrating odd functions between -a and a, thank you very much. I solved my problem.

That $$2\int u e^{-u} du$$ should be a $$\frac{1}{2}\int u e^{-u} du$$ I think halls.

I tried to tell that I already did that substitution in my posting number 4. But I couldn't evaluate that integral either. I mistakenly argued that since $$u=x^2$$, I can say that u is always positive and the integral is equal to $$\int_{0}^{\infty}u e^{-u}du$$ (which happens to be the $$\Gamma$$ function). Then I realized that although $$u$$ is positive, $$du$$ is not. So it is still an odd function. Everything is fine now.

 

[HallsofIvy]

That $$2\int u e^{-u} du$$ should be a $$\frac{1}{2}\int u e^{-u} du$$ I think halls.

Yes, of course it is. I think I'll pretend the LaTex messed up!

Yeah, as I said I got the idea of integrating odd functions between -a and a, thank you very much. I solved my problem.



I tried to tell that I already did that substitution in my posting number 4. But I couldn't evaluate that integral either. I mistakenly argued that since $$u=x^2$$, I can say that u is always positive and the integral is equal to $$\int_{0}^{\infty}u e^{-u}du$$ (which happens to be the $$\Gamma$$ function). Then I realized that although $$u$$ is positive, $$du$$ is not. So it is still an odd function. Everything is fine now.

In fact, if you had converted the limits of integration to u when you changed variables, you would have found you were integrating from infinity to infinity! What is the integral of any function from a to a?

 

[amacinho]

In fact, if you had converted the limits of integration to u when you changed variables, you would have found you were integrating from infinity to infinity! What is the integral of any function from a to a?

Hmm, beautiful :)