# Integral of x^{3}e^{-x^2}dx

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

Last edited:

arildno
Homework Helper
Gold Member
Dearly Missed
Hint:
Is the integrand an even or odd function?

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

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
Homework Helper
Gold Member
Dearly Missed
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?

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
Homework Helper
its quite difficult to search for a specific integral, best bet is it search integral. no latex wont 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.

functions.wolfram.com has good functions search

HallsofIvy
Homework Helper
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 cancelled 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
Homework Helper
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 cancelled 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.

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
Homework Helper
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?

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 :)