Suppose $ \alpha, \beta>0 $. Compute: $ \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx $ 
Suppose $ \alpha, \beta>0 $. Compute:
  $$ \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx $$

Here is what I do:
$$\begin{align}
\int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx &= \int_{0}^{\infty}dx\int_{\alpha}^{\beta}\sin (yx)dy\\
&=\int_{\alpha}^{\beta}dy\int_{0}^{\infty}\sin(yx)dx\\
& \\
&   \qquad\text{let $ yx=u $}\\
& \\
&=\int_{\alpha}^{\beta}\frac{1}{y}dy\int_0^{\infty}\sin u du\\
&=\int_{\alpha}^{\beta}\frac{1}{y}dy\left( -\cos u|_{\infty}+\cos u|_0 \right)\\
&=\log\frac{\beta}{\alpha}(-\cos(\infty)+1)\\
&=\log\frac{\beta}{\alpha}-\cos(\infty)\log\frac{\beta}{\alpha}
\end{align}$$
But $ \cos(\infty) $ does not exist right? Does it mean the integral actually diverse?
Edit: The question comes from https://math.uchicago.edu/~min/GRE/files/week1.pdf
Who can point out my mistake in the above deduction?

\begin{align}
\int_{0}^{\infty}e^{-tx}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx &= \int_{0}^{\infty}dx\int_{\alpha}^{\beta}e^{-tx}\sin (yx)dy\\
&=\int_{\alpha}^{\beta}dy\int_{0}^{\infty}e^{-tx}\sin(yx)dx\\
&=\int_{\alpha}^{\beta}dy\dfrac{y}{t^2+y^2}\\
&=\dfrac12\ln\dfrac{t^2+\beta^2}{t^2+\alpha^2}
\end{align}
now let $t=0$.
