$ \text{ Let } I(\alpha) = \int_0^\infty \frac{e^{-\beta x} sin\alpha x}{x} dx \\ $

$ \frac{dI}{d\alpha} = \int_0^\infty \frac{\partial}{\partial a} \frac{e^{-\beta x} sin\alpha x} { x} dx \\ $

$ = \frac{e^{-\beta x} cos\alpha x} { x} dx \\ $

$ = \left [ \frac{e^{-\beta x} } {\alpha ^2 + \beta ^ 2 } (-\beta cos\alpha x + \alpha sin\alpha x ) \right ] _0^\infty \\ $

$ \frac{dI}{d\alpha} = \frac{\beta}{\alpha ^2 + \beta ^ 2 } \\ $

$ I (\alpha ) = tan^{-1}\frac{\alpha}{\beta} + c \\ $

$ \alpha = 0 , I(0) = tan^{-1}(0) + c \\ $

$ c = 0 \\ $

$ I (\alpha ) = tan^{-1}\frac{\alpha}{\beta} \\ $

$ \text{To Prove } \int_0^\infty \frac{sin\alpha x}{x} dx = \frac{\pi}{2} \\ $

$ \text{Put } \beta = 0 \\ $

$ \int_0^\infty \frac{sin\alpha x}{x} dx = tan^{-1}(\infty) = \frac{\pi}{2} \\ $

$ \alpha \lt 0, \int_0^\infty \frac{sin\alpha x}{x} dx = tan^{-1}(\infty) =\frac{\pi}{2} \\ $

$ \alpha = 0, \int_0^\infty \frac{sin\alpha x}{x} dx = \frac{\pi}{2} \\ $

$ \alpha \gt 0, \int_0^\infty \frac{sin\alpha x}{x} dx = \frac{\pi}{2} \\ $