integration - Evaluate:$I=int^infty_0 frac {e^{-ax} sin bx}{x},dx$

How would I evaluate the integral:
$$I=\int^\infty_0 \frac {e^{-ax}\ \sin bx}{x}\,dx$$
I have started doing the problem by integration by parts but it seems to be more lengthy. Since it is definite integration, any properties may be there. I can't figure out the property basically. Any suggestion will be highly appreciated. Thanks!

Answer

Notice, using property of Laplace transform as follows
$$L\left(\frac{1}{t}f(t)\right)=\int_{s}^{\infty}L(f(t))dt$$
$$L(\sin bt)=\int_{0}^{\infty}e^{-st}\sin t dt=\frac{b}{b^2+s^2}$$

Now, we have
$$\int_{0}^{\infty} \frac{e^{-ax}\sin bx}{x}dx$$

$$=\int_{a}^{\infty} L(\sin bx)dx$$
$$=\int_{a}^{\infty}\frac{b}{b^2+x^2} dx$$
$$=b\int_{a}^{\infty}\frac{dx}{b^2+x^2} $$
$$=b\left[\frac{1}{b}\tan^{-1}\left(\frac{x}{b}\right)\right]_{a}^{\infty} $$
$$=\left[\tan^{-1}\left(\infty\right)-\tan^{-1}\left(\frac{a}{b}\right)\right] $$ $$=\frac{\pi}{2}-\tan^{-1}\left(\frac{a}{b}\right)$$ Hence, we have

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\int_{0}^{\infty} \frac{e^{-ax}\sin bx}{x}dx=\frac{\pi}{2}-\tan^{-1}\left(\frac{a}{b}\right)}}$$