# Convergence of an improper integral

## Homework Statement

For what values of r does $$\int$$(from 0 to infinity) xre-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 im really not confident that I did it right....

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

$$\int$$xre-xdx = -xre-x-$$\int$$rxr-1(-e-x)dx

I wont write out the whole thing since I cant 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 xr, xr-1, xr-2,... terms:

If r$$\leq$$0, 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 couldnt figure out how to express some things.

-Nick

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

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