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Euler Constant

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data
    I need to verify an integral representation of the Euler constant:

    [itex]\int^{1}_{0}[/itex][itex]\frac{1-e^{-t}}{t}[/itex]dt-[itex]\int^{\infty}_{1}[/itex][[itex]\frac{e^{-t}}{t}[/itex]dt=[itex]\gamma[/itex]


    2. Relevant equations



    3. The attempt at a solution
    OK, I'm supposed to use this fact (which I have already proved):

    [itex]\sum^{N}_{n=1}[/itex][itex]\frac{1}{n}[/itex]=[itex]\int^{1}_{0}[/itex][itex]\frac{1-(1-t)^{n}}{t}[/itex]dt.

    Then I am supposed to rescale t so that I can apply the follow definition of the exponential function:

    lim as n-->infinity of (1+[itex]\frac{z}{n}[/itex])[itex]^{n}[/itex]=[itex]\sum^{\infty}_{k=0}[/itex][itex]\frac{z^{k}}{k!}[/itex]=e[itex]^{z}[/itex]

    I'm not seeing how I can use the first fact at all...
     
  2. jcsd
  3. Oct 1, 2012 #2
    Someone has to have an idea on this one...
     
  4. Oct 1, 2012 #3

    jbunniii

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    I assume that should be a capital N in your integral expression for [itex]\sum_{n=1}^{N} \frac{1}{n}[/itex]:
    [tex]\sum_{n=1}^{N} \frac{1}{n} = \int_{0}^{1} \frac{1 - (1-t)^N}{t}dt[/tex]
    Assuming that is the case, you could let [itex]u = Nt[/itex]; then your integral becomes
    [tex]\sum_{n=1}^{N}\frac{1}{n} = \int_{0}^{N} \frac{1 - \left(1 - \frac{u}{N}\right)^N}{u} du[/tex]
    and you could break this apart as [itex]\int_{0}^{1} + \int_{1}^{N}[/itex]. Dunno if this will get you anywhere but it seems promising.
     
  5. Oct 1, 2012 #4
    Hmm... OK, this makes sense. What do you mean with the last part about splitting the integral up?
     
  6. Oct 2, 2012 #5

    jbunniii

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    Well, I suggested that because one of the integrals you'll eventually need has 0 and 1 as its endpoints:
    [tex]\int_{0}^{1}\frac{1 - e^{-t}}{t} dt[/tex]
    So if you can justify interchanging the order of limit and integration,
    [tex]\lim_{N \rightarrow \infty} \int_{0}^{1}\frac{1 - \left(1 - \frac{u}{N}\right)^N}{u}du[/tex]
    will give you what you need.

    That leaves you with the part from 1 to N:
    [tex]\int_{1}^{N}\frac{1 - \left(1 - \frac{u}{N}\right)^N}{u}du[/tex]
    which you could further split up as
    [tex]\int_{1}^{N}\frac{1}{u}du - \int_{1}^{N}\frac{\left(1 - \frac{u}{N}\right)^N}{u}du[/tex]
    Obviously the term on the left will be useful, given the definition of [itex]\gamma[/itex]. And the term on the right looks like it might converge as [itex]N \rightarrow \infty[/itex] to one of the other integrals you need, although that will also need to be proved.
     
  7. Oct 13, 2012 #6
    OK, I've been able to work this out successfully. I'm stuck with two small things:

    (1) How can I justify switching the order of integration and summation, as you mentioned?

    (2) How do I show that the last integral converges?
     
  8. Oct 13, 2012 #7
    Never mind, I was able to get the integral to converge. Now my only questions is what justifies switching the order? I know this is a subtle point, but I'd like to understand it.
     
  9. Oct 13, 2012 #8

    jbunniii

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    A sufficient condition for switching the order is if the convergence is uniform, i.e., if
    [tex]\lim_{N \rightarrow \infty} \frac{1 - \left(1 - \frac{u}{N}\right)^N}{u}[/tex]
    converges uniformly in [itex][0,1][/itex]. This in turn will be true if and only if
    [tex]\lim_{N \rightarrow \infty} \left(1 - \frac{u}{N}\right)^N[/tex]
    converges uniformly in [itex][0,1][/itex].
     
  10. Oct 13, 2012 #9
    Hmm, ok, great. That makes sense. How can I show that this is true? Weirestrass M-test?
     
  11. Oct 13, 2012 #10

    jbunniii

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