Show that the integral does not converge Show that $\int_0^\infty \sin(x) \,$ does not converge.
$$\int_0^k \sin(x) \leq \int_0^k \, dx = k$$
$$\lim_{k \to \infty} k = \infty$$
So this integral does not converge.
Is this correct?
 A: big hint
$$\int_0^k\sin(x)\mathrm d x=1-\cos(k)$$
Then $\int_0^\infty \sin(x)\mathrm d x$ converge if and only if $\lim_{k\to \infty }\cos(k)$ exist. 
A: It has already been shown in another posted solution that $\int_0^\infty \sin(x)\,dx$ fails to converge as an improper Riemann Integral.  
And clearly we have $\int_0^\infty |\sin(x)|\,dx=\infty$ and hence $\int_0^{\infty}\sin(x)\,dx$ fails to exist as a Lebesgue integral.
However, if we interpret the integral as a Fourier Sine Transform then writing the sign function as $\text{sgn}$ and the Heaviside Function as $H$ gives
$$\begin{align}
\int_0^\infty \sin(x)\,dx&=\frac1{2i}\int_{-\infty}^{\infty}H(x)(e^{ix}-e^{-ix})\,dx\\\\
&=\frac1{2i}\int_{-\infty}^{\infty} \text{sgn}(x) e^{ix}\,dx \\\\
&=\frac1{2i} \mathscr{F}\{\text{sgn}\}(1)\\\\
&=1
\end{align}$$
So, while $\int_0^\infty \sin(x)\,dx$ fails to exist as either a Riemann Integral or a Lebesgue Integral, if we more broadly interpret the integral as a DISTRIBUTION, then we can write
$$\int_0^\infty \sin(x)\,dx\sim1$$
