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Limit of an integral is from a to a does it always equal to zero?

  1. Dec 15, 2011 #1
    ∫f(x).dx from a(lower limit) to a(upper limit) =0
    Is this always true
     
    Last edited: Dec 15, 2011
  2. jcsd
  3. Dec 15, 2011 #2

    HallsofIvy

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    Yes, for any integrable function f, [itex]\int_a^a f(t)dt= 0[/itex].
     
  4. Dec 15, 2011 #3
    I have a question which has really let me think about this. Let me present it.
    i want to integrate e-x^2 from 0 to infinity. let me say it I.
    How i did it.
    (You need a copy pen so that you can write it mathematically.)
    step 1 replace x by -x
    limits will change from -infinity to zero. rest of the function inside integral will remain same.
    step 2 now add I you obtain in step 2 with I in original question.
    Now it's function is same but limit is from -infinity to infinity and it is equal to 2I
    now replace x by 1/x. limits will from zero to zero.
    No matter what the function inside the integral. Now it should zero(i mean I).
    so finally i came to know that I=0.
    It's really confusing result because if we draw rough curve of e-x^2 then we will find that it is always above the x axis. so how it's integral can be zero since definite integral denote geometrical area under under the curve.
     
  5. Dec 15, 2011 #4

    mathman

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    when you replace x by 1/x, you have a term 1/x2 in the integrand, which is not integrable near 0. Furthermore the limits are not [0,0] but [0,∞] and [-∞,0].
     
  6. Dec 15, 2011 #5
    why not you replying guys. who is wrong here?
    I think wolframalpha try to say it is infinity
     
    Last edited: Dec 16, 2011
  7. Dec 16, 2011 #6

    Mute

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    The problem is that when you make a change of variables in an integral, the new variable has to be defined on every point of the interval of the old variable. If you have an integral on the interval [itex]t \in (-\infty,\infty)[/itex], you cannot simply make the change [itex]x = 1/t[/itex], because the new variable x is not defined when t = 0, which exists on that interval. Hence, you must split the integral up first, before you can make that change of variables:

    [tex]\int_{-\infty}^{\infty} dt~f(t) = \int_{-\infty}^{0} dt~f(t) + \int_{0}^{\infty} dt~f(t).[/tex]

    If you change variables now to t=1/x, you still have to be careful, because in the first term (from -infinity to 0), the variable is negative, whereas it is positive in the second term. If you just haphazardly let t = 1/x, then in the first integral you might mistakenly think the upper zero is +infinity, when really it should be -Infinity. You need to account for this, and this is what mathman was telling you. So, after letting t = 1/x, the integrals are really

    [tex]-\int_{0}^{-\infty} dx~\frac{f(1/x)}{x^2} - \int_{\infty}^{0} dx~\frac{f(1/x)}{x^2} = \int_{-\infty}^{\infty} \frac{dx}{x^2}~f(1/x).[/tex]

    Wolframalpha will confirm that

    [tex]\int_{-\infty}^\infty dx~\frac{\exp(-1/x^2)}{x^2} = \sqrt{\pi} = \int_{-\infty}^\infty dt~\exp(-t^2).[/tex]

    Any time your change of variables is undefined or multivalued on the interval of the original variable, you need to be careful. Consider the following integral as an example:

    [tex]\int_{-a}^a dx~x^2.[/tex]

    You know the result is just [itex](2/3)a^3[/itex], but suppose you changed variables to [itex]t = x^2[/itex], giving [itex]dt = 2x dx = 2\sqrt{t}dx[/itex]. If you blindly apply this change of variables you might think you get

    [tex]\frac{1}{2}\int_{a^2}^{a^2} dt~\sqrt{t} = 0.[/tex]
    The problem, of course, is that the change of variable is different on the [-a,0] and [0,a] intervals. On the negative interval, you need to select the negative root, x = -t1/2, while on the positive interval you need to select the positive root, x = +t1/2. Up above, I 'mistakenly' used the positive root on both intervals. If I am careful, this integral is really

    [tex]-\frac{1}{2}\int_{a^2}^0 dt~\sqrt{t} + \frac{1}{2}\int_0^{a^2} dt~\sqrt{t} = \int_0^{a^2} dt~\sqrt{t} = \frac{2}{3}(a^2)^{3/2} = (2/3)a^3,[/tex]
    as expected.
     
    Last edited: Dec 16, 2011
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