Convergence of an improper integral

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SUMMARY

The integral \(\int_{0}^{\infty} x^r e^{-x} dx\) converges for values of \(r \leq 0\). The discussion highlights the use of integration by parts to analyze the behavior of the integral as \(x\) approaches infinity and zero. It was established that while the term \(-e^{-x}\) approaches zero as \(x\) increases, the polynomial terms \(x^r, x^{r-1}, x^{r-2}, \ldots\) dictate convergence behavior. Specifically, for \(r > 0\), these terms diverge, confirming that the integral converges only when \(r\) is less than or equal to zero.

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  • Understanding of improper integrals
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  • Knowledge of exponential decay functions
  • Basic polynomial behavior in limits
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NickMusicMan
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Homework Statement



For what values of r does [tex]\int[/tex](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 I am really not confident that I did it right...

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

[tex]\int[/tex]xre-xdx = -xre-x-[tex]\int[/tex]rxr-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 xr, xr-1, xr-2,... terms:

If r[tex]\leq[/tex]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[tex]\leq[/tex] 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
 
Last edited:
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Actually, try a value like r=3 and you'll find that the integral converges for at least some r>0.
 
[tex]\lim_{x\to\infty}x^re^{-x}=0[/tex] for all real r

So, the problem is convergence at 0.
 

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