Assume that $r > 0$.
Two more integrals that evaluate to zero are $$\int_{0}^{\infty} \sin(a k) J_{1}(rk) \, \mathrm dk, \quad a >r,$$ and $$\int_{0}^{\infty} \frac{\cos (ak)}{k} J_{1}(rk) \, \mathrm dk, \quad a \ge r. $$
In a previous answer I showed that $$\int_{0}^{\infty} \sin(ax) J_{n}(bx) \, dx = \begin{cases}
\frac{\sin \left(n \arcsin \left(\frac{a}{b} \right) \right)}{\sqrt{b^{2}-a^{2}}} \, & \quad 0 < a < b \\
\frac{\cos \left(\frac{\pi n}{2} \right) b^{n} }{ \sqrt{a^{2}-b^{2}} \left(\sqrt{a^{2}-b^{2}}+a \right)^{n}} & \quad a > b >0
\end{cases}$$
For $n=1$, $b=r$, and $a > r \ge 0$ we have $$ \int_{0}^{\infty} \sin(ak) J_{1}(rk) \, \mathrm dk = 0$$ since $\cos \left(\frac{\pi}{2} \right)=0$.
To show that the second integral evaluates to zero, notice that $$\begin{align} \int_{0}^{\infty} \frac{e^{-sk}}{k}J_{1}(rk) \, \mathrm dk &\overset{(1)}{=} \int_{s}^{\infty} \mathcal{L} \{J_{1}(rk) \} (p) \, dp \\ &\overset{(2)}{=} \int_{s}^{\infty} \frac{\sqrt{p^2+r^{2}}-p}{r\sqrt{p^2+r^{2}}} \, \mathrm dp \\ &= \frac{\sqrt{s^{2}+r^{2}}-s}{r}. \end{align}$$
Now let $s = \beta + ia $, where $\beta, a >0$.
Since $\int_{0}^{\infty} \frac{e^{-iak}}{k} J_{1}(rk) \, \mathrm dk$ is absolutely convergent, it follows from the dominated convergence theorem that $$\begin{align} \int_{0}^{\infty} \frac{e^{-ia k}}{k}J_{1}(rk) \, \mathrm dk &= \lim_{\beta \to 0^{+}} \int_{0}^{\infty} \frac{e^{-(\beta + ia) k}}{k}J_{1}(rk) \, \mathrm dk \\ &= \lim_{\beta \to 0^{+}} \frac{\sqrt{(\beta +ia)^{2}+r^{2}}-\beta -ia}{r} \\ &= \frac{\sqrt{r^{2}-a^{2}}-ia}{r} \\ &= \frac{i \left(\sqrt{a^{2}-r^{2}}-a \right)}{r}. \end{align}$$
Equating the real parts on both sides of the equation, we get $$\int_{0}^{\infty} \frac{\cos (ak)}{k} J_{1}(rk) \, \mathrm dk =0, \quad a \ge r. $$
As a bonus, we also get $$\int_{0}^{\infty} \frac{\sin (ak)}{k} J_{1}(rk) \, \mathrm dk = \frac{a-\sqrt{a^{2}-r^{2}}}{r}, \quad a \ge r . $$
$(1)$ https://en.wikipedia.org/wiki/Laplace_transform#Evaluating_improper_integrals
$(2)$ Laplace transform of the Bessel function of the first kind
Hint for $\int_0^\infty e^{-ax}J_v\left(bx\right)dx=\frac{\left(\sqrt{a^2+b^2}-a\right)^v}{b^v\sqrt{a^2+b^2}}$
We can use contour integration to create other similar integrals, like $$\int_{0}^{\infty} \frac{\sin(ak) J_{2}(bk)}{k^{2}} J_{1}(rk) \, \mathrm dk = 0, \quad a \ge |b|+r. $$
