Does an improper integral from 0 to infinity of xe^-x converge to zero?

[lcam2]

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.

 

[Mark44]

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.

 

[CompuChip]

If you mean that
$$\lim_{x \to \infty} x e^{-x} = 0$$
then yes. But that does not mean that
$$\int_0^\infty x e^{-x} \, dx = 0$$
In fact it is not, because $x e^{-x} \not\equiv 0$.

Try integration by parts?

 

[lcam2]

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

 

[CompuChip]

You forgot the lower boundaries in your partial integration. You get
$$\lim_{b \to 0} \left. (-x e^{-x} - e^{-x} \right|_{x = 0}^\infty$$
which is
$$(-b e^{-b} + 0 e^{-0}) - (e^{-b} - e^{-0})$$

 

[lcam2]

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'hospital's, my new limits is 1
Is that correct?? Can i separate the limits?? I am not sure if i can do that.
Thanks for all the help

 

[Mark44]

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
$$(-b e^{-b} - e^{-b} ) - (-0e^{-0} - e^{-0})$$

$$\lim_{b \to \infty} (-b e^{-b} - e^{-b} ) - (-0e^{-0} - e^{-0}) = 1$$

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

separating the limits and using l'hospital's, my new limits is 1
Is that correct?? Can i separate the limits?? I am not sure if i can do that.
Thanks for all the help

 

[CompuChip]

? What does this mean?

In particular, what is the = sign doing inside the brackets?

 

[rjs123]

? What does this mean?

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

$$\lim_{b \to \infty} (-b e^{-b} - e^{-b} ) - (-0e^{-0} - e^{-0}) = 1$$

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:
$$-b e^{-b}$$
when you plug in infinity for b...would you get -infinity * 0...which is undefined or an indeterminate form?

 

[Char. Limit]

I'm confused with this term in the above expression:
$$-b e^{-b}$$
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.

 

[lcam2]

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??

 

[Mark44]

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 $\Sigma$ button that pops open a list of commonly used LaTeX tags.