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Find the integral $$\int ^\infty_0e^{-xu}\frac{\sin u}{u}du$$

my idea

$L\left(\frac{1}{t}f(t)\right)=\int_{s}^{\infty}L(f(t))dt$

$L(\sin bt)=t=\frac{1}{1+s^2}$

$\int ^\infty_0e^{-xu}\frac{\sin u}{u}du$ ?? how we processed

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  • $\begingroup$ Does this involve complex numbers? $\endgroup$ – Yip Jung Hon May 1 '18 at 5:35
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \left.\int_{0}^{\infty}\expo{-xu}{\sin\pars{u} \over u}\,\dd u \,\right\vert_{\ x\ >\ 0} & = \int_{0}^{\infty}\expo{-xu}\bracks{{1 \over 2}\int_{-1}^{1}\expo{\ic ku} \dd k}\,\dd u = {1 \over 2}\int_{-1}^{1}\int_{0}^{\infty}\expo{-\pars{x - \ic k}u} \dd u\,\dd k \\[5mm] & = {1 \over 2}\int_{-1}^{1}{\dd k \over x - \ic k} = x\int_{0}^{1}{\dd k \over k^{2} + x^{2}} = \int_{0}^{1/x}{\dd k \over k^{2} + 1} \\[5mm] & = \bbx{\arctan\pars{1 \over x}} \end{align}

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In honour of Cleo:

$$ \tan^{-1}\Big(\frac{1}{x}\Big) $$

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  • $\begingroup$ Out of curiosity, has anybody seen Cleo of late. Haven't seen her in MSE for the last two years ... $\endgroup$ – Nilotpal Kanti Sinha May 1 '18 at 6:39
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In honor of Feynman:

Let us define (with $x>0$) $$I(x) = \int_0^\infty e^{-x u} \frac{\sin u}{u} du\;.$$

Differentiating (under the integral sign), we obtain $$I'(x) = -\int_0^\infty e^{-x u} \sin u\;du = -\operatorname{Im}\int_0^\infty e^{-xu +i u}\;du =\operatorname{Im} \frac1{i-x} = - \frac{1}{1+x^2}\;. $$

We thus obtain $$I(x) = -\arctan(x) + c \;.$$ It is easy to see that $I(u) \to 0$ for $x \to \infty$, this sets $c=\pi/2$ and thus $$ I(x) = \frac{\pi}{2} -\arctan(x) = \arctan(1/x)\;.$$

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  • $\begingroup$ Good day, I've not seen integrating techniques like these before. May I ask what are these integrating techniques you've used here called? $\endgroup$ – Yip Jung Hon May 1 '18 at 6:13
  • $\begingroup$ @YipJungHon: it is called Differentiation Under the Integral Sign. $\endgroup$ – Fabian May 1 '18 at 12:53
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Sure your idea works: Let $f(t)=\sin(t)$, then \begin{align} \int_0^{\infty}e^{-st} \frac{\sin(t)}{t} &= \mathcal{L}\left\{\tfrac{1}{t} f(t)\right\}(s) = \int_s^{\infty} \mathcal{L}\left\{f(t)\right\}(s')\;ds'\\ &= \int_s^{\infty} \frac{ds'}{1+s'^2} = \left[\tan^{-1}(s')\right]_s^{\infty}\\ &= \tfrac{\pi}{2} - \tan^{-1}(s) \end{align}

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