Help with a tricky improper integration problem

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In summary: Use this (your previous result):\int_0^{\infty} x^2e^{-x^2}dx = y\int_0^{\infty}e^{-x^2}dxand make the substitution e^{-x^2} = y on the LHS. The result follows very quickly. Thanks!Thanks!
  • #1
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Show that:

∫(x^2)e^(-x^2)=(1/2)∫(e^(-x^2)

(both integrals are from 0 to ∞)



I don't even know if it's possible to integrate these functions, so it makes me think that what's interesting about the problem is that it only works with this integration limits.

I know improper integration, but don't know if prooving this "numerically" is the right path.

Thanks, I would appreciate any advice
 
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  • #2
stonecoldgen said:
Show that:

∫(x^2)e^(-x^2)=(1/2)∫(e^(-x^2)

(both integrals are from 0 to ∞)
I don't even know if it's possible to integrate these functions, so it makes me think that what's interesting about the problem is that it only works with this integration limits.

I know improper integration, but don't know if prooving this "numerically" is the right path.

Thanks, I would appreciate any advice

Hint:

[tex]\int_0^{\infty} x^2e^{-x^2}dx = \int_0^{\infty} x.(xe^{-x^2})dx[/tex]

Integrate that by parts (use [itex]u = x, dv = xe^{-x^2}dx[/itex]), then use L'Hopital's Rule to calculate the limit of one of the expressions.
 
Last edited:
  • #3
Curious3141 said:
Hint:

[tex]\int_0^{\infty} x^2e^{-x^2}dx = \int_0^{\infty} x.(xe^{-x^2})dx[/tex]

Integrate that by parts (use [itex]u = x, dv = xe^{-x^2}dx[/itex]), then use L'Hopital's Rule to calculate the limit of one of the expressions.

is this correct?

I integrated by pats (differentiating x and integrating x^2e^{-x^2} )

until I got to the point where (e^{-x^2})/2 (evaluated from o to infinity)=0

Using L'Hopital's Rule, I effectively got 0=0




Is this the way?
 
  • #4
stonecoldgen said:
is this correct?

I integrated by pats (differentiating x and integrating x^2e^{-x^2} )

until I got to the point where (e^{-x^2})/2 (evaluated from o to infinity)=0

Using L'Hopital's Rule, I effectively got 0=0




Is this the way?

No, I think you messed up the integration by parts. You should end up with:

[tex]\int_0^{\infty} x^2e^{-x^2}dx = xe^{-x^2}|_0^{\infty} +\frac{1}{2}\int_0^{\infty}e^{-x^2}dx[/tex]

and the first expression on the RHS vanishes when you take the limit by L' Hopital's.
 
  • #5
Curious3141 said:
No, I think you messed up the integration by parts. You should end up with:

[tex]\int_0^{\infty} x^2e^{-x^2}dx = xe^{-x^2}|_0^{\infty} +\frac{1}{2}\int_0^{\infty}e^{-x^2}dx[/tex]

and the first expression on the RHS vanishes when you take the limit by L' Hopital's.
It's the same thing (maybe my english isn't that good, I don't know) but Thanks!
 
  • #6
stonecoldgen said:
It's the same thing (maybe my english isn't that good, I don't know) but Thanks!

How is it the same thing? You said you got "0 = 0" which means nothing. You also said you got [itex]\frac{1}{2}\int_0^{\infty}e^{-x^2}dx = 0[/itex] which is clearly wrong.

Anyway, if you're confident you have the right solution, then it's fine.
 
  • #7
What about this one? Show:

∫e-x2=∫√(-lny)
(from 0 to ∞) (from 0 to 1)



I tried this one by solving for y in the right side of the equation, getting y=-ex2

However, when checking the solution graphically, it's not right.

So what the hell am I supposed to do? I also wonder if the integration limits changed when expressed in y or x.


Thanks
 
  • #8
You should do a substitution to change the limits and get the log into play, then integrate by parts. Your previous result might help here.
 
Last edited:
  • #9
stonecoldgen said:
What about this one? Show:

∫e-x2=∫√(-lny)
(from 0 to ∞) (from 0 to 1)



I tried this one by solving for y in the right side of the equation, getting y=-ex2

However, when checking the solution graphically, it's not right.

So what the hell am I supposed to do? I also wonder if the integration limits changed when expressed in y or x.


Thanks

Use this (your previous result):

[tex]\int_0^{\infty} x^2e^{-x^2}dx = \frac{1}{2}\int_0^{\infty}e^{-x^2}dx[/tex]

and make the substitution [itex]e^{-x^2} = y[/itex] on the LHS. The result follows very quickly.
 

1. How do I identify an improper integration problem?

An improper integration problem is one where the limits of integration are infinite or the function being integrated is undefined at certain points within the integration limits.

2. What are the common techniques used to solve improper integration problems?

The most common techniques used to solve improper integration problems are the limit comparison test, the direct comparison test, and the geometric test.

3. Can I use substitution to solve an improper integration problem?

Yes, substitution can be used to solve certain types of improper integration problems. However, it may not work for all cases and other techniques may be more effective.

4. How do I determine if an improper integration problem is convergent or divergent?

The behavior of the integrand near the boundaries of integration can be used to determine if an improper integration problem is convergent or divergent. If the integrand approaches a finite value as the boundaries approach infinity, the integral is convergent. If the integrand approaches infinity or oscillates as the boundaries approach infinity, the integral is divergent.

5. Are there any online resources that can help with solving improper integration problems?

Yes, there are many online resources such as forums, tutorials, and practice problems that can help with solving improper integration problems. Additionally, math tutoring services or online courses may also provide assistance with tricky integration problems.

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