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Improper Integral

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Integral from negative infinity to positive infinity of (1/(sqrt(1+x^2))dx

    2. The attempt at a solution
    Using trig substitution I got the integral equal to ln|sqrt(1+x^2) + x| Finding this was not the difficult part. Evaluating it is.

    I set it up like this: lim b --> infinity and lim a --> neg infinity [(ln(sqrt(1+b^2)) + b) - (ln(sqrt(1+a^2)) + a)]

    the 'b portion' goes to infinity. For the 'a portion' I rewrote it as ln|1/sqrt(1+x^2) - a| Plug in negative infinity and it is ln|1/infinity|. This is where im not sure what it is. If 1/infinity = 0, then isnt it indeterminate because you cannot take the ln(0)? If it is simply the ln(extremely small number) then it would be negative infinity, which means the overall answer is infinity, correct?
     
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  3. Feb 22, 2009 #2

    djeitnstine

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    From my analysis the second part seems to be ln(0). Without rewriting it [tex]Lim_{x \rightarrow -\infty} ln(\sqrt{1+x^{2}}+x)[/tex] boils down to [tex]Lim_{x \rightarrow -\infty} ln(|x|+x)[/tex] so a positive plus its negative is always 0.
     
  4. Feb 22, 2009 #3

    djeitnstine

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    Also I forgot to mention, rewriting it like that would give division by 0
     
  5. Feb 22, 2009 #4

    lanedance

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    do youknow about hyperbolic trig functions? could be useful here
    woops... see you've probably already used them...
     
    Last edited: Feb 22, 2009
  6. Feb 22, 2009 #5

    lanedance

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    you could also look at the symmetry of the integral to re-write it as 2 times the integral from 0 to inf, though this is essentially ln(inf) so infinite itself

    but yeah in the limit, the integral tends toward +infinity, as x gets big, the integrand looks like 1/x which is known not to converge
     
  7. Feb 23, 2009 #6

    Dick

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    Like lanedance said, it doesn't converge. Use a comparison test to show it. Did you misstate the problem?
     
  8. Feb 23, 2009 #7
    I definitely stated the problem correctly. In my class we have only gone through all the techniques for integration (u sub, parts, partial fractions, trig sub, etc). We have not done convergence tests or hyperbolic functions.

    I see that you can take twice the integral from 0 to infinity since its symmetrical, I just dont understand how you do the limit as a goes to neg infinity of ln(sqrt(1+a^2)) + a)
     
  9. Feb 23, 2009 #8

    Dick

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    That limit doesn't exist. Neither does the original integral. The area under the curve is infinite. Or lim a->-infinity=-infinity and lim a->+infinity=+infinity. Same thing.
     
    Last edited: Feb 23, 2009
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