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:(adsbygoogle = window.adsbygoogle || []).push({});

[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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# I Sine integral

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**