show $\int^\infty_0 e^{-sx} x^{-1} \sin{x} dx = \frac14 \log{(1+4s^{-2})}$ for $s>0$ This is problem 2.6.58 of Folland's Real Analysis book:
show $\int^\infty_0 e^{-sx} x^{-1} \sin{x} dx = \frac14 \log{(1+4s^{-2})}$ for $s>0$ by integrating $e^{-sx} \sin{(2xy)}$ over x and y. 
I get the general gist of the problem, if I integrate $e^{-sx} \sin{(2xy)}$ first with respect to $y$, and make the change of variables $z = 2xy$, I can get an integral that has an $x^{-1}$ term. I am not sure how to choose what are the boundaries I am integrating over though, or what comes next. Any help would be appreciated.
 A: Alternately, evaluate $I(s)=\displaystyle\int_0^\infty\sin x\cdot e^{-sx}$ using the fact that $\sin x=\Im(e^{ix})$, then integrate both sides with regard to s.
A: The result stated in the OP's question is not correct.  I know $s=0$ is not in the domain, but the limit as $s \to 0$ should produce an answer of $\pi/2$; the answer given by the OP blows up.
The problem lies in the fact that
$$\int_0^1 dy \, \sin{2 x y} = \frac{\sin^2{x}}{x}$$
not $\sin{x}/x$.  Thus, the integral you seek is
$$\int_0^{\infty} dx \, e^{-s x} \frac{\sin^2{x}}{x} = \frac14 \log{\left ( 1+\frac{4}{s^2}\right )} $$
The integral you posted is, instead
$$\int_0^{\infty} dx \, e^{-s x} \frac{\sin{x}}{x} = \arctan{\frac1{s}} $$
A: Your integral is the Laplace transform of the function $f(x)=\frac{\sin(x)}{x}$ where this transform is defined as 
$$\mathcal{L}(f(x))(s)=\int^{\infty}_{0}f(x)e^{-sx}\,dx$$
Laplace transform of your function $f(x)=\frac{\sin(x)}{x}$ is 
$$\mathcal{L}(\frac{\sin(x)}{x})(s)=\int^{\infty}_{0}\frac{\sin(x)}{x}e^{-sx}\,dx=\arctan(\frac{1}{s})$$ 
Therefore you should seek for another integrand to get your result.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{\infty}\expo{-sx}x^{-1}\sin\pars{x}\,\dd x
     =\arctan\pars{1 \over s}:\ {\large ?}.\quad s\ >\ 0}$.

\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}\expo{-sx}x^{-1}\sin\pars{x}\,\dd x}
=\int_{0}^{\infty}\expo{-sx}\,\ \overbrace{%
\half\int_{-1}^{1}\expo{\ic k x}\,\dd k}^{\ds{\color{#c00000}{x^{-1}\sin\pars{x}}}}\,\ \dd x
\\[5mm]&=\half\int_{-1}^{1}\int_{0}^{\infty}\expo{\pars{-s + \ic k}x}\,\dd x\,\dd k
=\half\int_{-1}^{1}{-1 \over -s + \ic k}\,\dd k
=\half\int_{-1}^{1}{s + \ic k \over k^{2} + s^{2}}\,\dd k
=\int_{0}^{1}{s\,\dd k \over k^{2} + s^{2}}
\\[5mm]&=\int_{0}^{1/s}{\dd k \over k^{2} + 1}
=\color{#66f}{\large\arctan\pars{1 \over s}}
\end{align}

