# Improper integral

## 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

## 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
Mentor

## 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

## 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
Homework Helper
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?

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
Homework Helper
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
$$(-b e^{-b} + 0 e^{-0}) - (e^{-b} - e^{-0})$$

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

Mark44
Mentor
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'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

CompuChip
Homework Helper
??? What does this mean?

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

??? 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?

Last edited:
Char. Limit
Gold Member
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.

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
Mentor
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.