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Improper integral limit

  1. Jul 22, 2012 #1
    \int_0^1 \frac{1}{\sqrt{x}}\,\mathrm{d}x

    \lim_{\varepsilon \to 0+}\int_\varepsilon^1 \frac{1}{\sqrt{x}}\,\mathrm{d}x

    My question is about the usage of 0+ in the limit.(I evaluated the integrals and arrived at the part where I substitute upper and lower limits.)
    Did the author deliberately choose to use [tex]\lim_{\varepsilon \to 0+} [/tex] instead of 0 or 0- so that any imaginary numbers arising from the expression [tex]2\sqrt{x}[/tex] do not arise?
    Or is there any other reason?
  2. jcsd
  3. Jul 22, 2012 #2

    I don't know what author you're talking about, but taking that limit is what has to be done simply by the definition of improper

    integral with one of the limits being a point of unboundness for the function...

  4. Jul 22, 2012 #3
    Yes,i understand this case.But,what if I had a case of a function best described by limit->0-?
    Will I not have a problem when I substitute 0- into the square root?

    I can't exactly describe the function,I mean it for some arbitrary function have variable 'x' under the square root and me having to apply limit->0-.Wont this give rise to a imaginary number?
  5. Jul 22, 2012 #4
    Taking the limit from below would result in the expression being undefined.
  6. Jul 22, 2012 #5

    I'm not completely sure I follow you, but the function in the integral is defined only for positive real numbers: whatever you want

    to do with it will have to comply with this restriction. Thus, there is not meaning to the expression
    [tex]\lim_{x\to 0^-}\sqrt x[/tex]
    as it assumes the existence of the square roots of negative numbers within the real numbers, which is absurd.

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