Unusual Limit: Understanding the Discrepancy in the Integral of xe^-x

[flux!]

This was just very basic, I have accepted it in just a heartbeat, but when I tried to chopped it and examined one by one, somethings fishy is happening, this just involved \int_{0}^{\infty}x e^{-x}dx=1.

Well, when we do Integration by parts we will have let u = x du = dx dv = e^{-x}dx v = -e^{-x} so that would give us \int_{0}^{\infty}x e^{-x}dx= - x e^{-x} + \int_{0}^{\infty} e^{-x}dx evaluating the second term will just yield 1, so, \int_{0}^{\infty}x e^{-x}dx= - x e^{-x} + 1, the second term should yield zero by employing limits as x approaches infinity, \lim_{x\rightarrow \infty }xe^{-x}=0 but whoa, it didn't. I will yield instead 1/e Where did I went wrong?

 

[Mark44]

This was just very basic, I have accepted it in just a heartbeat, but when I tried to chopped it and examined one by one, somethings fishy is happening, this just involved \int_{0}^{\infty}x e^{-x}dx=1.

Well, when we do Integration by parts we will have let u = x du = dx dv = e^{-x}dx v = -e^{-x} so that would give us \int_{0}^{\infty}x e^{-x}dx= - x e^{-x} + \int_{0}^{\infty} e^{-x}dx evaluating the second term will just yield 1, so, \int_{0}^{\infty}x e^{-x}dx= - x e^{-x} + 1, the second term should yield zero by employing limits as x approaches infinity, \lim_{x\rightarrow \infty }xe^{-x}=0 but whoa, it didn't. I will yield instead 1/e Where did I went wrong?

It's not clear what you're asking.
$\lim_{x \to \infty} xe^{-x} = 0$ is correct. Are you getting 1/e for this limit? If so, that's where you went wrong.

 

[flux!]

I get it now, I am imputing the wrong variable from my computer, indeed lim xe^{-x} as x approaches infinity indeed gives 0.