What is the Integral of x^{2} e^{-x^2} dx in Signal Processing?

In summary, the conversation discussed different approaches to solving the integral \int_{-\infty}^{\infty} x^{2} e^{-x^{2}} dx. Suggestions included using the error function, differentiating with respect to a, and using Leibniz's rule. Ultimately, the simplest approach was to recognize the integral as \sqrt{\pi} times the variance of a Gaussian random variable.
  • #1
niks
3
2
I ran into an integral while working on response of a signal processing filter, it looks like:

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

While trying integration by parts u = [tex]x^{2}[/tex] we get du = 2xdx but can't proceed with dv = [tex]e^{-x^{2}}[/tex] because then
v = [tex]\int e^{-x^{2}}[/tex]
can't be integrated unless we use the limits.

Can anyone suggest an approach for this?

Thanks,
Niks
 
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  • #2
What exactly are you trying to do? As you point out, your v is not any elementary function, and that tells you that neither is
[tex]\int x^2e^{-x^2} dx[/tex]
You might be able to do that in terms of the "error function", Erf(x), which is defined to be
[tex]\int e^{-x^2} dx[/tex]
 
  • #3
If you just want to calculate the definite integral, I don't see why you wouldn't want to include the limits when integrating by parts?
 
  • #4
His point, about the limits of integration, was that it is well known that
[tex]\int_{-\infty}^{\infty} e^{x^2}dx= 2\sqrt{\pi}[/itex]
while the anti-derivative is not an elementary function. in other words, he could, theoretically, do it as a definite integral with the right limits but not as an indefinite integral.
 
  • #5
Was that adressed to me?

Anyway, Maple tells me that:
[tex]
\int_{-\infty}^{\infty} x^{n} e^{-x^{2}}\rm{d} x=\frac{1}{2}\Gamma\left(\frac{n}{2}+\frac{1}{2}\right)\left(1+(-1)^n\right)
[/tex] ,
which should be possible to prove by induction.

PS. HallsofIvy: You have forgotten the minus-sign in your integrand.
 
  • #6
A cute way to solve this is to recall that

[tex]\int_{-\infty}^{\infty}e^{-ax^2}=\frac{\sqrt{\pi}}{\sqrt{a}}[/tex]

Then use Feynman's favorite trick and differentiate both sides with respect to a, and evaluate at a = 1.
 
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  • #7
HallsofIvy said:
His point, about the limits of integration, was that it is well known that
[tex]\int_{-\infty}^{\infty} e^{x^2}dx= 2\sqrt{\pi}[/itex]
while the anti-derivative is not an elementary function. in other words, he could, theoretically, do it as a definite integral with the right limits but not as an indefinite integral.
For one, it's [itex]e^{-x^2}[/itex] and for two, it's [itex] \sqrt{\pi}[/itex].
 
  • #8
As you point out, your v is not any elementary function,
I think I should use this as a guideline for future problems. Both u and v should be elementary functions otherwise integration by parts becomes too messy(perhaps impossible).

while the anti-derivative is not an elementary function. in other words, he could, theoretically, do it as a definite integral with the right limits but not as an indefinite integral.
Yes, that was what I had in mind. That's why I got stuck there.

Anyway, Maple tells me that:
[tex]

\int_{-\infty}^{\infty} x^{n} e^{-x^{2}}\rm{d} x=\frac{1}{2}\Gamma\left(\frac{n}{2}+\frac{1}{2}\r ight)\left(1+(-1)^n\right)

[/tex]
which should be possible to prove by induction.
Thanks! That will help me move forward.

Thanks to everyone who replied, I learned a lot from this thread.

-Niks
 
  • #9
Using lots of substitutions and integration by parts I get this:

[tex]\int x^{2}e^{-{x^2}}dx=xe^{x^{2}}\left[1-\sum_{n=1}^{\infty}\frac{\prod_{k=2}^{n}\left(2k-3\right)}{2^{n}x^{2n}}\right][/tex]

I would go over the derivation but LaTex is killing me.
 
  • #10
Quite frankly, I think using Leibniz' rule, as suggested by Nicksauce, would by far be the simplest method in this case.
 
  • #11
Quite frankly, I think using Leibniz' rule, as suggested by Nicksauce, would by far be the simplest method in this case.

Quite true.

But, if doing by parts, then the proper selection of u an dv is

[tex]u=x[/tex]

[tex]dv= x*e^{-x^2}dx[/tex]

and then things won't be so messy - however, it will involve the definite integral [itex]\int^{\infty}_{-\infty}{ e^{-{x^2}}dx}[/itex] which we know equals [itex]\sqrt{\pi}[/itex].
 
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  • #12
umm mathematica gives me [tex]\frac{\sqrt{\pi}}{2}[/tex]

and for the indefinite :

[tex]
\frac{1}{4} \sqrt{\pi } \text{erf}(x)-\frac{1}{2} e^{-x^2} x
[/tex]
 
  • #13
Well, perhaps the very simplest approach is to recognize that the integral is [itex]\sqrt{\pi}[/itex] times the variance of a Gaussian random variable with mean 0 and standard deviation [itex]\frac{1}{\sqrt{2}}[/itex]. That's certainly all I'd bother doing in the signal processing context the OP mentioned.
 

1. What is the formula for the integral of x^{2} e^{-x^2} dx?

The formula for the integral of x^{2} e^{-x^2} dx is ∫x^{2} e^{-x^2} dx = -\frac{1}{2} e^{-x^2} (x^2 + 1) + C, where C is the constant of integration.

2. How do you solve the integral of x^{2} e^{-x^2} dx?

To solve the integral of x^{2} e^{-x^2} dx, you can use integration by parts. Let u = x and dv = xe^{-x^2} dx. Then, du = dx and v = -\frac{1}{2} e^{-x^2}, and the integral becomes ∫x^{2} e^{-x^2} dx = -\frac{1}{2} x e^{-x^2} - \frac{1}{2} ∫e^{-x^2} dx. Using the substitution method, the second term can be evaluated as -\frac{1}{4} \sqrt{\pi} erf(x) + C, where erf(x) is the error function. Combining the two terms, the final solution is ∫x^{2} e^{-x^2} dx = -\frac{1}{2} e^{-x^2} (x^2 + 1) + C.

3. What is the value of the integral of x^{2} e^{-x^2} dx at a specific point?

The value of the integral of x^{2} e^{-x^2} dx at a specific point x=a can be found by plugging in the value of a into the solution of ∫x^{2} e^{-x^2} dx = -\frac{1}{2} e^{-x^2} (x^2 + 1) + C and evaluating the expression. For example, if a=2, the value of the integral is ∫x^{2} e^{-x^2} dx = -\frac{1}{2} e^{-4} (4 + 1) + C = -\frac{5}{2e^{4}} + C.

4. What is the graph of the integral of x^{2} e^{-x^2} dx?

The graph of the integral of x^{2} e^{-x^2} dx is a bell-shaped curve with a maximum value of approximately 0.443. The curve starts at the origin and approaches 0 as x increases. The graph also has a point of inflection at x = \frac{1}{\sqrt{2}} and is symmetric about the y-axis.

5. What is the significance of the integral of x^{2} e^{-x^2} dx in mathematics?

The integral of x^{2} e^{-x^2} dx is significant in mathematics because it is a special case of the Gaussian integral, which has many applications in statistics, physics, and other fields. It is also a common example used in calculus courses to illustrate the technique of integration by parts. The integral also has a closed form solution, making it a useful tool for solving more complex integrals involving exponential functions.

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