Convergence of an improper integral

[NickMusicMan]

Homework Statement

For what values of r does \int(from 0 to infinity) x^re^-x dx converge?

I assume that the problem refers to r as any real number.

2. The attempt at a solution

I have given this a try but I am really not confident that I did it right...

First i used integration by parts to try to discover a pattern:

\intx^re^-xdx = -x^re^-x-\intrx^r-1(-e^-x)dx

I won't write out the whole thing since I can't find all the appropriate summation/product symbols, but carrying out this exact integration by parts an infinite number of times gives:

-e^-x(a polynomial in x with an infinite number of terms)

(Note: the polynomial will have a finite # of terms if r happens to be a natural number)

Now I know that the -e^-x term will always go to zero as x gets larger.

So I consider the x^r, x^r-1, x^r-2,... terms:

If r\leq0, then those terms will all go to zero, but if r>0 then some of those terms will go to infinity.

From this I concluded that the integral converges iff r\leq 0.


Does this solution make any sense? Thanks in advance to anyone who is able to help me out. I apologize if its not clear; I wrote it much more clearly (and in more detail) on paper but this is my first try at typing math on a computer so i couldn't figure out how to express some things.

-Nick

 

[Char. Limit]

Actually, try a value like r=3 and you'll find that the integral converges for at least some r>0.

 

[SammyS]

\lim_{x\to\infty}x^re^{-x}=0 for all real r

So, the problem is convergence at 0.