It would seem that (one) definition of the Dirac Delta function implies the Sin function converges to zero at infinity.
The Dirac Delta function can be defined as:
$\delta(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ikx}dk$
Which follows from the Fourier transform of the delta function. The delta function evaluated at x=1 is zero by definition so:
$\delta(1) = 0 = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ik}dk$
But this would mean:
$0 = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ik}dk = \frac{1}{2\pi i }e^{ik} |_{-\infty}^{\infty}= \frac{1}{2\pi i }[cos(k)+isin(k)] |_{-\infty}^{\infty} = \frac{1}{2\pi i }[2isin(k)] |_{\infty} = \frac{1}{\pi }sin(\infty)$
So:
$sin(\infty) = 0$
What am I doing wrong here? Is it incorrect to assume this is a statement of convergence at infinity?