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Integration by parts when a limit is infinity.

  1. Mar 16, 2005 #1
    I'm having a tough time trying to do integration by parts with one of my limits being infinity. My Integral looks like:

    [tex] \int_0^\infty x^z e^{-x} dx[/tex] with [tex] z = \frac{-1}{\pi}[/tex]

    Now if I let [tex]u = e^{-x} [/tex] and [tex]dv = x^z dx[/tex],
    I will have: [tex]du = -e^{-x} dx [/tex] and [tex] v = \frac{1}{z + 1} x^{z + 1}
    [/tex] and so
    [tex] uv - \int_0^\infty v du
    = e^{-x} \frac{1}{z + 1} x^{z + 1} - \int_0^\infty \frac{1}{z + 1} x^{z + 1} -e^{-x} dx[/tex]

    However, since one of the limit is infinity, the term
    [tex] uv = e^{-x} \frac{1}{z + 1} x^{z + 1}[/tex] has a freakin infinity subbed in it. The answer is actually,
    [tex] \frac{1}{z + 1} \int_0^\infty x^{z + 1} e^{-x} dx[/tex], which is what my integral part looks like. What am I doing wrong? Why do I get a term with infinity at the front?
  2. jcsd
  3. Mar 16, 2005 #2
    What is [tex]\lim_{x\rightarrow \infty} x^ze^{-x}[/tex]?
  4. Mar 16, 2005 #3
    I think it is easier to look at the limit like this:

    [tex]\lim_{x\rightarrow\infty} \frac{1}{x^{\frac{1}{\pi}}e^{x}}[/tex]
  5. Mar 16, 2005 #4
    [tex]x^ze^{-x} = \frac{x^z}{e^x} \neq \frac{1}{x^{\frac{1}{z}}e^x}[/tex]
    Last edited: Mar 16, 2005
  6. Mar 16, 2005 #5
    Shinjo said in his first post that [tex]z = -\frac{1}{\pi}[/tex]
  7. Mar 16, 2005 #6

    Anyways the integral actually is just equal to [tex]\Gamma \left( \frac{\pi - 1}{\pi} \right)[/tex]
  8. Mar 16, 2005 #7
    But the way to calculate the gamma function is to do that actual integral...

    anywayz, do you see my problem though? when my [tex] v [/tex] becomes [tex]v = \frac{1}{z + 1} x^{z + 1}, [/tex]my [tex]x^{z + 1}[/tex] term gets moved to the top and turns into infinity.
  9. Mar 16, 2005 #8
    [tex] \lim_{x \rightarrow \infty} x^ye^{-x} = 0 \ \forall y \in \mathbb{R}[/tex]
  10. Mar 16, 2005 #9
    Hmm...maybe I should have mentioned this earlier, but I am not trying to solve the integral. I have to get it in a specific form so I can use Gaussian Quadrature. In order to do that I have to integrate by parts.

    The answer is supposed to be: [tex] \frac{1}{z + 1} \int_0^\infty x^{z + 1} e^{-x} dx[/tex], but I don't know how to get rid of the [tex] uv = e^{-x} \frac{1}{z + 1} x^{z + 1}[/tex] term, since one of the limits is infinity.

    Thank you for the help btw, it is much appreciated.
  11. Mar 16, 2005 #10
    [tex] \left[e^{-x}\frac{1}{z+1}x^{z+1}\right]_0^\infty = \lim_{x\rightarrow \infty} \left(\frac{e^{-x}x^{z+1}}{z+1}\right) - 0 = \frac{1}{z+1}\lim_{x\rightarrow \infty} e^{-x}x^{z+1}[/tex]

    then just apply the result in my last post.
  12. Mar 16, 2005 #11
    Ahh...I see what you're saying. Thank you.

    I don't mean to question your method, I know I might be becoming a nuisance by now, but can you tell me why [tex] \lim_{x \rightarrow \infty} x^ye^{-x} = 0 \ \forall y \in \mathbb{R}[/tex]? I mean, doesn't [tex]\lim_{x \rightarrow \infty} x^y = \infty[/tex]?
  13. Mar 16, 2005 #12
    That's not really relevant when you have a denominator that is also a function of x. For example, the trivial f(x) = x/x or g(x) = x/(x2 + 1), neither of which diverge as x increases without bound. Now refer to your problem f(x) = xz+1/ex. Note that both numerator and denominator diverge as x increases without bound, so we may apply L'Hospital's rule to examine the equivalent limit (z+1)!/ex which approaches zero as x increases without bound.
    Last edited: Mar 16, 2005
  14. Mar 16, 2005 #13
    I think he meant [tex]x^z[/tex] This gives you [tex]\lim_{x\rightarrow\infty} \frac{1}{x^{\frac{1}{\pi}}e^{x}} = 0[/tex]
  15. Mar 16, 2005 #14
    Ah, Thank you. I just noticed, when the x is an exponent, it increases the term much faster than when x is a base. So I can see how x^y / e^x will = 0 for all y.

    Thank you to all once again.
  16. Mar 16, 2005 #15
    No nuisance. If I thought it were a nuisance to justify my math, I wouldn't be on a math forum! :)

    Say [tex]y \in \mathbb{R}[/tex]. Note that if [tex]y \leq 0[/tex] the result is obvious, so assume [tex] y > 0 [/tex]. Then let [tex]\lfloor a \rfloor[/tex] and [tex] \lceil a \rceil[/tex] represent the floor and ceiling functions applied to [tex]a[/tex] respectively. Thus if [tex] m = \lfloor y \rfloor, \ M = \lceil y \rceil[/tex] we clearly have [tex]x^me^{-x} \leq x^ye^{-x} \leq x^Me^{-x} \ \forall x \geq 1 \Longrightarrow \lim_{x\rightarrow \infty}x^me^{-x} \leq \lim_{x \rightarrow \infty} x^ye^{-x} \leq \lim_{x \rightarrow \infty} x^Me^{-x}[/tex] (by a slightly generalized version of the squeeze theorem).

    Now, let [tex]a \in \mathbb{N}^+[/tex]. By l'Hopital's rule, applied [tex] a[/tex] times, we have

    [tex]\lim_{x \rightarrow \infty} x^ae^{-x} = \lim_{x \rightarrow \infty} \frac{x^a}{e^x} = \lim_{x \rightarrow \infty} \frac{ax^{a-1}}{e^x} = . \ . \ . = \lim_{x \rightarrow \infty} \frac{a!}{e^x} = a! \lim_{x \rightarrow \infty} e^{-x} = 0[/tex]

    and since for [tex] y \ge 0[/tex] clearly [tex] M \geq m \geq 0, \ M, m \in \mathbb{N}[/tex] we get [tex] 0 = \lim_{x \rightarrow \infty} x^m e^{-x} \leq \lim_{x\rightarrow \infty} x^ye^{-x} \leq \lim_{x \rightarrow \infty} x^M e^{-x} = 0 \Longrightarrow \lim_{x \rightarrow \infty} x^y e^{-x} = 0[/tex] as we wanted. QED.
    Last edited: Mar 16, 2005
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