- #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?
[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?