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Limit for a problem of convergence

  1. Dec 22, 2014 #1
    1. The problem statement, all variables and given/known data

    I ultimately want to discuss convergence of the integral


    2. Relevant equations


    is convergent near x approaching infinity for p>1

    3. The attempt at a solution

    While I understand that the integral converges near 0, as x->infinity I find, for the integrand

    [tex]\lim_{x \rightarrow \infty }\frac{1}{\sqrt{x}e^{\sqrt{x}}}=\lim_{x \rightarrow \infty }\frac{1}{\sqrt{x}\left( 1+\sqrt{x} + \frac{x}{2} + \cdots \right) }=\lim_{x \rightarrow \infty }\frac{1}{\left( \sqrt{x} + x + \frac{\sqrt{x}x}{2} + \cdots \right) }[/tex]

    but the solution given to me says that I should end up with

    [tex]\frac{1}{\left(\frac{\sqrt{x}x}{2} + \cdots \right) }[/tex]

    which converges as p would be always greater than 3/2, but I don't see how the limit gets rid of the first two terms in the denominator.

  2. jcsd
  3. Dec 22, 2014 #2

    Ray Vickson

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    Much easier: change variables. There is a particularly "obvious" transformation, but if I tell you exactly what it is I will be violating PF rules.
  4. Dec 22, 2014 #3


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    The integral indeed converges because you know the integral below on the right converges:

    $$\int_0^{\infty} \frac{1}{\sqrt{x} e^{\sqrt{x}}} \space dx \leq \displaystyle \lim_{a \rightarrow \infty} \int_0^a \frac{1}{e^{\sqrt{x}}} \space dx$$

    So to determine what the integral on the left converges to, note that:

    $$\sqrt{x} e^{\sqrt{x}} = \sum_{n}^{\infty} \frac{x^{\frac{n+1}{2}}}{n!}$$

    So that:

    $$\int_0^{\infty} \frac{1}{\sqrt{x} e^{\sqrt{x}}} dx = \int_0^{\infty} \frac{1}{\sum_{n}^{\infty}\frac{x^{\frac{n+1}{2}}}{n!}} dx$$
  5. Dec 22, 2014 #4


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    I don't see how you get that inequality. Is that what you meant?
    Anyway, Ray's method is much simpler.
  6. Dec 22, 2014 #5


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    The integral is improper at ##0## and your inquality fails near ##0##.
  7. Dec 22, 2014 #6


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    Yes I don't know what I was doing using series, I wasn't fully awake when I wrote that, sorry. There is a much simpler way that has been mentioned already.
  8. Dec 27, 2014 #7
    There are two discontinuities here: x = 0 and x = infinity

    so rewrite as lim a→0 (lim b→∞ (∫dx/(√x e*√x) from a to b))
    u = √x
    2du = dx/√x
    lim a→0 (lim b→∞ (∫e^-u du from a to b))
    = lim a→0 (lim b→∞ -e^-b + e^-a)
    = 0 +1
    = 1
  9. Dec 27, 2014 #8


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    I don't think it means anything to say there's a discontinuity at infinity. The definition of an integral to infinity is the limit as the range tends to infinity.
    You later lost the 2.
  10. Dec 27, 2014 #9
    Yes you are right, it does not make sense to say that there's a discontinuity at infinity. I just meant that we can not treat as a normal limit of integration and we are gonna have to take the limit as that limit goes to infinity. Thank you from pointing that out.

    And yup, i must gave forgotten to write the two. so the correct answer is 2, not 1. Thanks again.
  11. Dec 30, 2014 #10
    The solution given to you says you should get something with x in it after taking limits?

    If you can find a value for the integral, then you know it converges. Taking the limit of the integrand as x goes to infinity will only tell you whether or not the integrand goes to zero. In this case, it does - however you still don't know if it converges or diverges. You can recognize that the integrand is just the derivative of the function [itex]-2e^{-\sqrt{x}}[/itex] and so you can use the fundamental theorem of calculus (part 2) and then evaluate the limits.
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