TITLE: Evaluate an integral using Laplace transform
QUESTION [12 upvotes]: 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.

REPLY [10 votes]: 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
$$
\int_0^\infty\frac{e^{-ax}\sin(bx)}{x}\,\mathrm{d}x=\tan^{-1}\left(\frac ba\right)\tag{5}
$$