How do I compute the integral of x/(e^x) with limits 0 and infinity?

[steelphoenix]

Im doing a chapter on Improper integrals, and one of my problems is i need to compute an integral if it converges. I really don't know where to start with this one. Anyone point me in the right direction?

integral of x/(e^x) where the lower limit is 0 and the upper is infinite

really any help would be appreciated, I got a bit of a cold and seem to be forgetting every bit of math

 

[steelphoenix]

ook wait, i think i figured it out

i stat off by calculating the integral, which turns out to be

-( (x+1)/e^x )

then you end up getting 1 after you plug in the limits, because when its infinity it becomes zero and when 0 it becomes -1, so when you subtract negetive 1 it becomes 1, and 1 is the answer

 

[flebbyman]

You have x/(e^x) = $$xe^{-x}$$

So $$\int_{0}^{\infty}xe^{-x}dx$$

Integration by parts gives $${\lim }\limits_{x \to \infty } -e^{-x}(x+1) + 1$$

$${\lim }\limits_{x \to \infty } \frac{-(x+1) }{e^x} + 1$$

Applying l'Hôpital's rule gives $${\lim }\limits_{x \to \infty } \frac{-1 }{e^x} + 1$$ = 0 + 1
= 1

 

[genneth]

As an aside, in case anyone cares:

$$\int_0^\infty x^n e^{-x}\,dx = n!$$

For integer n. However, the integral form is also well-defined for real, and even complex n, except at the negative integers where it diverges. Look up "gamma function" for more details.

 

[flebbyman]

I somehow recognized it before I read the rest of your post. Math is fascinating