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.