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I Integrals of expressions

  1. Jun 2, 2016 #1
    I know that [tex] \sqrt{f(x)^2} = |f(x)| [/tex] However...
    I've just noticed that integrals of expressions like this are usually assumed to be equal to the integral of f(x) without the absolute value. I'd like to know how that's possible.

    Is weird for me to consider those expressions; specially because of what Minkowski's inequality says about it:
    [tex] \int{|f(x)|\,dx} \leq \left| \;{\int{f(x)dx}}\; \right| [/tex]

    In general, I would expect the integral without the absolute value to be different from the "absolute-valued" one.
    Check, for example:

    [tex] \int_\pi^{3\pi/2} |\sin(x)| \, dx [/tex] shouldn't give a negative value as a result, however, that's what WolframAlpha gives me. [Just like if it were ignoring the absolute value and just integrating sin(x) ]
     
  2. jcsd
  3. Jun 2, 2016 #2
    I don't think this is correct. Shouldn't you take the inequality the other way around?
     
  4. Jun 2, 2016 #3

    nrqed

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    as tommyxu3 pointed out, your inequality is in the wrong direction (just a typo, I am sure).

    If Wolfram alpha does it as if there was no absolute (and gives a negative answer for that integral), then it is simply wrong (or there is something incorrect in th away you input the integrand). You are correct that on cannot ignore the absolute sign if we integrate over a region where the integrand may take negative values.
     
  5. Jun 2, 2016 #4
    You're right, sorry for that.
    That's what common sense says to me too, but, there are two specific cases where I can't do that:

    1. Trigonometric substitutions:
    Let's say this one: https://www.physicsforums.com/threads/difficult-integral-for-trig-substitution.301301/#post-2126338
    There, sqrt{4tan^2(x)} is assumed to be 2tan(x).

    2. Integrals of volume:
    Any volume results from the integration of the product of the corresponding "scale factors" in the specific orthogonal system that you're working on.
    If you have spherical polar coordinates:
    x = r sin(t) cos(p)
    y = r sin(t) sin(p)
    z = r cos(t)
    The first scale factor (h1) corresponds to the square root of sum of squares of the derivatives of each variable {x,y,z} with respect to r. The second one is the same but, with respect to t. The last one is with respect to p. Taking that into account, h1=1, h2=|r|, h3=|r sin(t)|. Clearly different from:
    https://en.wikipedia.org/wiki/Orthogonal_coordinates#Table_of_orthogonal_coordinates

    Probably there are another examples but those are pretty illustrative.
     
  6. Jun 2, 2016 #5

    nrqed

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    Let's talk about this last case first. Note that ##r## is always larger or equal to zero so ##|r| = r##. Note also that ##\theta## is between 0 and ##\pi## radians so ##| \sin \theta | = \sin \theta ##. Or maybe I am misunderstanding your point?
     
  7. Jun 2, 2016 #6
    That was in fact, what I was missing, thanks for pointing it out. :wink:
    For sure, a really important thing to keep in mind.
     
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