Improper integral depending on three parameters

Note that: $$\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}x^{2n+1}$$

Therefore:

$$\begin{aligned}\int_0^\infty \frac{e^{-\alpha x}-e^{-\beta x}}{x}\sin(\gamma x) \ dx &= \int_0^\infty \left( \frac{e^{-\alpha x}-e^{-\beta x}}{x}\right)\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}(\gamma x)^{2n+1} \ dx \\&= \sum_{n=0}^\infty \frac{(-1)^{n}\gamma^{2n+1}}{(2n+1)!} \left\{ \int_0^\infty e^{-\alpha x}x^{2n}dx-\int_0^\infty e^{-\beta x}x^{2n}dx\right\} \\ &= \sum_{n=0}^\infty\frac{(-1)^{n}\gamma^{2n+1}}{(2n+1)!} \left\{ \frac{\Gamma(2n+1)}{\alpha^{2n+1}}-\frac{\Gamma(2n+1)}{\beta^{2n+1}}\right\} \\&=\sum_{n=0}^\infty\frac{(-1)^{n}\gamma^{2n+1}}{(2n+1)!} \left\{ \frac{(2n)!}{\alpha^{2n+1}}-\frac{(2n)!}{\beta^{2n+1}}\right\} \\&=\sum_{n=0}^\infty \frac{(-1)^n \left( \frac{\gamma}{\alpha}\right)^{2n+1}}{2n+1} - \sum_{n=0}^\infty \frac{(-1)^n \left( \frac{\gamma}{\beta}\right)^{2n+1}}{2n+1} \\&=\arctan \left( \frac{\gamma}{\alpha}\right)-\arctan \left( \frac{\gamma}{\beta}\right)\end{aligned}$$

In the last step, I have used

$$\arctan(x) = \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}$$

Here is an approach, $$I(\gamma)=\int_0^\infty \frac {e^{-\alpha x}-e^{-\beta x}}{x} \sin(\gamma x) dx\implies \frac{dI(\gamma)}{d\gamma}=\int_0^\infty ({e^{-\alpha x}-e^{-\beta x}}) \cos(\gamma x) dx.$$

Now, the last integral is the Laplace transforms of $\cos(x)$ with respect to $\alpha$ and $\beta$. Can you finish the problem now?