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I Sine integral

  1. Mar 6, 2016 #1


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    Properly speaking, since [itex]sin(x)[/itex] and [itex]cos(x)[/itex] don't go to zero as [itex]x \rightarrow \infty[/itex], the following integrals are undefined:

    [itex]\int_0^{\infty} cos(kx) dk[/itex]
    [itex]\int_0^{\infty} sin(kx) dk[/itex]

    However, in the handwavy way of physicists, we can often pretend that the cosine integral "converges" to [itex]\delta(x)[/itex], where [itex]\delta(x)[/itex] is defined via:

    [itex]\int dx \delta(x) f(x) = f(0)[/itex]

    This interpretation is sort-of justified because for nicely-behaved functions [itex]f[/itex], we can prove:

    [itex]\int_{0}^{+\infty} dk (\int_{-\infty}^{+\infty} f(x) cos(kx) dx) = f(0)[/itex]

    If we blithely switch the order of integration, then we can write this as:

    [itex]\int_{-\infty}^{+\infty} dx f(x) (\int_{0}^{+\infty} cos(kx) dk) = f(0)[/itex]

    which sort of justifies identifies the inner integral with [itex]\delta(x)[/itex].

    My question is: Is there a related, equally hand-wavy interpretation of the sine integral?
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  3. Mar 6, 2016 #2
    how you can say that? we know the function oscillates!
  4. Mar 6, 2016 #3


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    That's why the word "converges" is in scare-quotes. The integral doesn't converge. However, for certain purposes, we can often act as if it converges to the delta function. (And for certain purposes, we can act as if the delta function is actually a function).
    Last edited: Mar 6, 2016
  5. Mar 8, 2016 #4
    That's typical answer of Mathematicians a century ago !
  6. Mar 8, 2016 #5


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    By substituting ##\sin(kx)=\cos(kx-\pi/2)## into the integral of sine and then use the substitution technique to compute the integral as well as the hand-wavy definition of the cosine integral, I got an extremely unintuitive answer of ##\pi/(2x)##.
  7. Mar 8, 2016 #6


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    And still the answer in the ongoing war between mathematicians and physicists!
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