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Limits - Improper Integration

  1. Jun 4, 2005 #1
    hello all

    well this is the first time Iv used latex it really took me along time to write, is it suppose to take that long or is there a better way of doing it?

    anyway this is a small part of a bigger problem everytime i tried i always show that the limite does not exist, but when i chuck it into mathematica the limite does exist, can anybody help, its really awkward check it out


    [tex] \lim_{\epsilon \rightarrow 0} (\int_\epsilon^1\frac{\Pi\coth\Pi\times\x}{2}-\frac{1}{2\times\x}dx)[/tex]
    then eventually i get this
    [tex] \lim_{\epsilon \rightarrow 0} (\frac{\log[\epsilon]}{2}+\frac{\log[\sinh\Pi]}{2}-\frac{\log[\sinh\Pi\epsilon]}{2})[/tex]
    but no matter how much i tried i cannot get it to equal
  2. jcsd
  3. Jun 4, 2005 #2
    [tex]\int_0^1\frac{\pi\coth\pi x}{2}-\frac{1}{2x}dx[/tex]
    [tex]= [\frac{log(sinh\pi x)}{2} - \frac{log(x)}{2}]_0^1 [/tex]
    [tex]= [\frac{log(\frac{sinh\pi x}{x})}{2}]_0^1 [/tex]
    [tex]= \frac{log(sinh\pi)}{2} - \lim_{x \rightarrow 0}\frac{log(\frac{sinh\pi x}{x})}{2}[/tex]
    [tex]= \frac{log(sinh\pi)}{2} - \frac{log(\pi)}{2}[/tex]

    -- AI
    Please use \pi with small p and use x instead of \times when referring to variable x. \PI and \times look absolutely awful in your given expression :smile:
  4. Jun 4, 2005 #3
    hello there

    thanxs for the advice about latex i sure hope i will improve in the near future
    anyway i dont really understand how you did this step

    [tex] \lim_{x \rightarrow 0}\log(\frac{sinh\pi x}{x})=log(\pi)[/tex]

    thats exactly where im getting concused am i missing something i am suppose to know?

  5. Jun 4, 2005 #4
    You can take lim inside log (since the log function is continuous)

    -- AI
  6. Jun 4, 2005 #5
    well this is what i have done so far but i dont see how thats going to get me to [tex]\pi[/tex], i tried plotting it on mathematica, it wouldnt plot

    [tex] \lim_{x \rightarrow 0}\log(\frac{sinh\pi x}{x})[/tex]
    [tex]=\log(\lim_{x \rightarrow 0}\frac{sinh\pi x}{x})[/tex]
    [tex]=\log(\lim_{x \rightarrow 0}\frac{e^{\pi x}+e^{-\pi x}}{2x})[/tex]

    where am i going wrong?
  7. Jun 4, 2005 #6
    [tex]sinh\pi x = \frac{e^{\pi x}-e^{-\pi x}}{2}[/tex]
    Apply L' Hospital.

    -- AI
  8. Jun 4, 2005 #7
    wow this L' Hospital stuff is great thanxs for the directions muchly appreciated

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