Improper integral

  • Thread starter lcam2
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  • #1
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Homework Statement


Find if the improper integral converges or diverges; If converges find its exact value.

Integral from 0 to infinity of (Xe-x)dX

Homework Equations





The Attempt at a Solution


Using limit properties, i found that the integral was zero.
Does it make sense that the integral converges to zero?? I don't understand.
 

Answers and Replies

  • #2
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Homework Statement


Find if the improper integral converges or diverges; If converges find its exact value.

Integral from 0 to infinity of (Xe-x)dX

Homework Equations





The Attempt at a Solution


Using limit properties, i found that the integral was zero.
Does it make sense that the integral converges to zero?? I don't understand.

No, the integral doesn't converge to zero. Show us what you did and we can steer you in the right direction.
 
  • #3
CompuChip
Science Advisor
Homework Helper
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If you mean that
[tex]\lim_{x \to \infty} x e^{-x} = 0[/tex]
then yes. But that does not mean that
[tex]\int_0^\infty x e^{-x} \, dx = 0[/tex]
In fact it is not, because [itex]x e^{-x} \not\equiv 0[/itex].

Try integration by parts?
 
  • #4
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0∫∞(xe^(x ))dx This is the original promblem

And this is what i did
0∫∞(xe^(x ))dx = lim┬(b→∞) 0∫b(xe^x )dx =lim┬(b→∞) (-be^(-b)-e^(-b) )= lim┬(b→∞) (-b)/e^b =0
 
  • #5
CompuChip
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You forgot the lower boundaries in your partial integration. You get
[tex]\lim_{b \to 0} \left. (-x e^{-x} - e^{-x} \right|_{x = 0}^\infty[/tex]
which is
[tex](-b e^{-b} + 0 e^{-0}) - (e^{-b} - e^{-0})[/tex]
 
  • #6
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Ok, i see.
Now evaluating the lower limits my new limit is

lim(b→∞) (-b=1)/(e^b)+1

separating the limits and using l'hopitals, my new limits is 1
Is that correct?? Can i separate the limits?? Im not sure if i can do that.
Thanks for all the help
 
  • #7
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Ok, i see.
Now evaluating the lower limits my new limit is

lim(b→∞) (-b=1)/(e^b)+1
??? What does this mean?

Your antiderivative turns out to be -xe-x - e-x, which you evaluate at b and at 0
[tex](-b e^{-b} - e^{-b} ) - (-0e^{-0} - e^{-0})[/tex]

[tex]\lim_{b \to \infty} (-b e^{-b} - e^{-b} ) - (-0e^{-0} - e^{-0}) = 1[/tex]

The first two terms approach zero in the limit, so you're left with 1.

separating the limits and using l'hopitals, my new limits is 1
Is that correct?? Can i separate the limits?? Im not sure if i can do that.
Thanks for all the help
 
  • #8
CompuChip
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??? What does this mean?

In particular, what is the = sign doing inside the brackets?
 
  • #9
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??? What does this mean?

Your antiderivative turns out to be -xe-x - e-x, which you evaluate at b and at 0
[tex](-b e^{-b} - e^{-b} ) - (-0e^{-0} - e^{-0})[/tex]

[tex]\lim_{b \to \infty} (-b e^{-b} - e^{-b} ) - (-0e^{-0} - e^{-0}) = 1[/tex]

The first two terms approach zero in the limit, so you're left with 1.

I'm confused with this term in the above expression:
[tex]
-b e^{-b}
[/tex]
when you plug in infinity for b...would you get -infinity * 0...which is undefined or an indeterminate form?
 
Last edited:
  • #10
Char. Limit
Gold Member
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I'm confused with this term in the above expression:
[tex]
-b e^{-b}
[/tex]
when you plug in infinity for b...would you get -infinity * 0...which is undefined or an indeterminate form?

Here's where we get the idea of growth and decay. e^(-b) decays to 0 much "faster" than b grows to infinity. I believe it's usually proven using L'Hopital's rule.
 
  • #11
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In particular, what is the = sign doing inside the brackets?

I'm sorry, it was supposed to be (-b-1), there should not be = inside the brackets.
Thanks, your explanation really helped me!

Actually I have a question, can i write equations in Physics forums, it was very hard to write the equation, Does physics forums have an equation bar?? or something like that??
 
  • #12
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7,033
There's LaTeX, but there isn't an equation bar. If you click Go Advanced, you get an extended menu bar with buttons for doing exponents (X2) and subscripts (X2). There's a [itex]\Sigma[/itex] button that pops open a list of commonly used LaTeX tags.
 

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