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I have to compute$$\int_0^\infty \dfrac{e^{-tx}\cdot \sin(x)}{x}dx$$

This is following a helping problem $$\int_0^\infty e^{-tx}\cdot \sin(x)dx$$ which using IPB two times turned out to be $$\dfrac{1}{1+t^2}$$

I think there must be a substitution to get to the first problem, but I just cannot see it. Any hint appreciated.

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Define $$ I = \int \limits_0 ^\infty e^{-tx}\frac{\sin(x)}{x}dx $$ then $$ \dfrac{dI}{dt}=-\int \limits_0 ^\infty e^{-tx}\sin(x)dx = -\frac{1}{1+t^2} $$ hence $$ I = -\arctan(t) + D $$

We fix $D$ using $I(0)=D=\dfrac\pi 2$, easily obtainable through complex analysis.

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  • $\begingroup$ You don't need to use complex analysis. Simply note that $I(\infty)=0$ to find $D$. (Now you can plug in $t=0$, giving $I(0)=\frac\pi2$ without complex analysis.) $\endgroup$ – Akiva Weinberger Sep 28 '14 at 23:41
  • $\begingroup$ @columbus8myhw That's true, easily obtained from $ 0\leq I \leq \int\limits_0^{+\infty}e^{-tx}dx $ $\endgroup$ – Frédéric Sep 29 '14 at 7:50
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We can compute $\int_0^{\infty}e^{-tx}\sin x/x dx$ for a positive $t$. Indeed, define $$F(t,x):=\frac{\sin x}xe^{-tx}.$$ If $\delta$ is a positive number, then for any non-negative $x$, $$\sup_{s\geqslant \delta}\partial_tF(s,x)|\leqslant e^{-\delta x}.$$ Since the map $x\mapsto e^{-\delta x}$ is integrable on $[0,\infty)$, we can take the derivative under the integral.

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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\large\tt\mbox{With}\quad t > 0}$:

\begin{align} &\color{#00f}{\large\int_{0}^{\infty}{\expo{-tx}\sin\pars{x} \over x}\,\dd x} =\int_{0}^{\infty}\expo{-tx}\pars{\half\int_{-1}^{1}\expo{\ic kx}\,\dd k} =\half\int_{-1}^{1}\int_{0}^{\infty}\expo{\pars{\ic k - t}x}\,\dd k \\[3mm]&=\half\int_{-1}^{1}{1 \over t - \ic k}\,\dd k =t\int_{0}^{1}{\dd k \over t^{2} + k^{2}} =\int_{0}^{1/t}{\dd k \over 1 + k^{2}}= \arctan\pars{1 \over t} \\[3mm]&=\color{#00f}{\Large{\pi \over 2} - \arctan\pars{t}} \end{align}

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$$ \begin{aligned} \int_{0}^{\infty} e^{-tx} \frac{\sin x}{x} \, \mathrm{d}x &= \int_{0}^{\infty} e^{-tx} \sin x \int_{0}^{\infty} e^{-zx} \,\mathrm{d}z\,\mathrm{d}x\\ &= \int_{0}^{\infty}\!\!\int_{0}^{\infty} e^{-x\left(z+t\right)} \sin x\,\mathrm{d}x\,\mathrm{d}z\\ &= \int_{0}^{\infty} \frac{\mathrm{d}z}{\left(z+t\right)^2 +1}\\ &= \left[\lim_{N\to\infty}\arctan(N+t)\right] - \arctan t \\ &= \frac{\pi}{2} - \arctan t\\ &= \operatorname{arccot} t \end{aligned} $$

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  • $\begingroup$ Note: We can say that $ \displaystyle{\frac{1}{x} = \int_{0}^{\infty} e^{-zx}\,\mathrm{d}z} $ because we are integrating over the interval $(0, \infty)$, i.e., $x>0$, so the integral converges. $\endgroup$ – user149844 Sep 28 '14 at 21:24

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