1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    [tex]\int_{0}^{\infty}\frac{1}{\sqrt{x}e^{\sqrt{x}}}dx[/tex]



    2. Relevant equations

    [tex]\int_{c}^{\infty}\frac{dx}{x^{p}}[/tex]

    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.

    Thanks!
     
  2. jcsd
  3. Dec 22, 2014 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    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

    Zondrina

    User Avatar
    Homework Helper

    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

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member


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

    Zondrina

    User Avatar
    Homework Helper

    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

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Limit for a problem of convergence
  1. Limit convergence (Replies: 10)

  2. Converge limit (Replies: 1)

  3. Limit and Convergence (Replies: 1)

Loading...