show $\int^\infty_0 e^{-sx} x^{-1} \sin{x} dx = \frac14 \log{(1+4s^{-2})}$ for $s>0$

Question

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.

Answer 1

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.

Answer 2

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

$$\int_0^{\infty} dx \, e^{-s x} \frac{\sin{x}}{x} = \arctan{\frac1{s}} $$

Answer 3

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.

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{0}^{\infty}\expo{-sx}x^{-1}\sin\pars{x}\,\dd x =\arctan\pars{1 \over s}:\ {\large ?}.\quad s\ >\ 0}$.

\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}