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Properly speaking, since sin(x) and cos(x) don't go to zero as x \rightarrow \infty, the following integrals are undefined:
\int_0^{\infty} cos(kx) dk
\int_0^{\infty} sin(kx) dk
However, in the handwavy way of physicists, we can often pretend that the cosine integral "converges" to \delta(x), where \delta(x) is defined via:
\int dx \delta(x) f(x) = f(0)
This interpretation is sort-of justified because for nicely-behaved functions f, we can prove:
\int_{0}^{+\infty} dk (\int_{-\infty}^{+\infty} f(x) cos(kx) dx) = f(0)
If we blithely switch the order of integration, then we can write this as:
\int_{-\infty}^{+\infty} dx f(x) (\int_{0}^{+\infty} cos(kx) dk) = f(0)
which sort of justifies identifies the inner integral with \delta(x).
My question is: Is there a related, equally hand-wavy interpretation of the sine integral?
\int_0^{\infty} cos(kx) dk
\int_0^{\infty} sin(kx) dk
However, in the handwavy way of physicists, we can often pretend that the cosine integral "converges" to \delta(x), where \delta(x) is defined via:
\int dx \delta(x) f(x) = f(0)
This interpretation is sort-of justified because for nicely-behaved functions f, we can prove:
\int_{0}^{+\infty} dk (\int_{-\infty}^{+\infty} f(x) cos(kx) dx) = f(0)
If we blithely switch the order of integration, then we can write this as:
\int_{-\infty}^{+\infty} dx f(x) (\int_{0}^{+\infty} cos(kx) dk) = f(0)
which sort of justifies identifies the inner integral with \delta(x).
My question is: Is there a related, equally hand-wavy interpretation of the sine integral?