Distributional convergence of integrals

Question

Hi :) I'm studying distributions and I would like to know if I'm doing this right. Let's take an ordinary function $$f(x)=\int_0 ^\infty dk\ \sin(kx)$$ and we want to study the functional on a test function $$ \left<T_f,\phi\right> = \int dx\ f(k)\phi(x) = \int _{-\infty} ^\infty dx \int _0 ^\infty dk\ \sin(kx)\phi(x) $$ To study the functional we add the cut off $e^{-\alpha k}$, then evaluate the limit $\alpha\rightarrow 0$. So $$ \left<T_f,\phi\right> = \int _{-\infty} ^\infty dx \int _0 ^\infty dk \sin(kx)\phi(x)e^{-\alpha k} $$ Since the double integral converges uniformly in $k$ and $x$ we can swtich the integration order and study $$f_\alpha(x)=\int_0 ^\infty dk \sin(kx) e^{-\alpha k}=\frac{x}{x^2 + \alpha^2}$$ Substituting we get $$ \left<T_{f_\alpha},\phi\right> = \int _{-\infty} ^\infty dx \frac{x}{x^2 + \alpha ^2}\phi(x) = \left<P.V. \frac{x}{x^2+\alpha^2},\phi\right> \xrightarrow{\alpha \to 0}\left<P.V. \frac{1}{x}, \phi\right>$$ So, in a distributional sense, I have $$f(x)=\int_0 ^\infty dk \sin(kx)=P.V. \frac{1}{x}$$ where $P.V.$ denotes the principal value. Is this correct?