Evaluate an integral using Laplace transform 
Evaluate the following integral using Laplace transform
  $$\int_0^\infty\frac{e^{-ax}\sin bx}{x}\,dx$$

I obtained  this partial result
$$=\int_0^\infty \frac{1}{p+1} \frac{b}{p^2+b^2}\,dp$$
and I am stuck here. I know that the final answer is $$\arctan\frac{b}{a}.$$ I would appreciate if someone could help me finish to attain the final answer. 
 A: Your partial result does not hold. According to Laplace transform properties 
$$\int_0^\infty\frac{e^{-ax}\sin bx}{x}\,dx=\cal{L}\left(\frac{\sin bx}{x}\right)(a)=\int_a^{+\infty}
\cal{L}\left(\sin bx\right)(p)dp=\int_a^{+\infty}
\frac{b}{p^2+b^2}\, dp.$$
Can you take it form here?
A: the standard form of LT is 
$$\mathcal{L}f(t)=\int_{0}^{\infty}e^{-st}f(t)\,dt$$
So for the given integral you just have to calculate LT of $\frac{\sin bx}{x}$ then put $s=a$
these two properties of LT may be useful 
$$\mathcal{L} \sin (bx)=\frac{b}{b^2+s^2}$$and $$\mathcal{L}\left\{\frac{f(t)}{t}\right\}=\int_{s}^{\infty}F(s)\,ds$$
A: This doesn't use the Laplace Transform, however, it is pretty simple.
$$
\begin{align}
\frac{\partial}{\partial a}\int_0^\infty\frac{e^{-ax}\sin(bx)}{x}\,\mathrm{d}x
&=-\int_0^\infty e^{-ax}\sin(bx)\,\mathrm{d}x\tag{1}\\
&=-\frac1b+\frac ab\int_0^\infty e^{-ax}\cos(bx)\,\mathrm{d}x\tag{2}\\
&=-\frac1b+\frac{a^2}{b^2}\int_0^\infty e^{-ax}\sin(bx)\,\mathrm{d}\tag{3}\\
&=-\frac{b}{a^2+b^2}\tag{4}
\end{align}
$$
Explanation:
$(1)$: differentiate inside the integral with respect to $a$
$(2)$: integrate by parts: $-e^{-ax}\sin(bx)\,\mathrm{d}x=\frac1be^{-ax}\,\mathrm{d}\cos(bx)$
$(3)$: integrate by parts: $e^{-ax}\cos(bx)\,\mathrm{d}x=\frac1be^{-ax}\,\mathrm{d}\sin(bx)$
$(4)$: $\frac{b^2}{a^2+b^2}$ times $(3)$ plus $\frac{a^2}{a^2+b^2}$ times $(1)$
Integrating $(4)$ with respect to $a$ gives
$$
\int_0^\infty\frac{e^{-ax}\sin(bx)}{x}\,\mathrm{d}x=\tan^{-1}\left(\frac ba\right)\tag{5}
$$
