How to compute $\lim_{A\to\infty}\int_{-A}^A {\sin(x)e^{itx}\over x} \, dx$.

Question

for $t\in R$, how to compute

$$\lim_{A\to\infty}\int_{-A}^A {\sin(x) e^{itx} \over x}dx.$$

Is there some way to compute this integral? I am trying to use complex analysis to compute it, but still don't know how to do it.

Answer 1

$$\begin{eqnarray*}\int_{-A}^{A}\frac{\sin(x)}{x}e^{itx}\,dx &=& \int_{0}^{A}\frac{\sin x}{x}\left(e^{itx}+e^{-itx}\right)dx = \int_{0}^{A}\frac{2\sin(x)\cos(tx)}{x}\,dx\\&=&\int_{0}^{A}\frac{\sin((1-t)x)}{x}\,dx+\int_{0}^{A}\frac{\sin((1+t)x)}{x}\,dx\tag{1}\end{eqnarray*}$$ So, once you prove that $$ \lim_{A\to +\infty}\int_{0}^{A}\frac{\sin(\beta x)}{x}\,dx =\frac{\pi}{2}\text{Sign}(\beta)\tag{2} $$ (for instance, by exploiting the parity of the integrand function and the residue theorem)
you have that

$$ \lim_{A\to +\infty}\int_{-A}^{A}\frac{\sin(x)}{x}e^{itx}\,dx = \color{red}{\frac{\pi}{2}\left(\text{Sign}(1-t)+\text{Sign}(1+t)\right)}\tag{3} $$ where the RHS is a function of the $t$ variable supported on the interval $[-1,1]$.

Answer 2

We have that $$\begin{align}I\left(t\right)= & \lim_{A\rightarrow\infty}\int_{-A}^{A}\frac{\sin\left(x\right)}{x}e^{itx}dx=\frac{1}{2i}\int_{0}^{\infty}\frac{\left(e^{ix}-e^{-ix}\right)\left(e^{itx}+e^{-itx}\right)}{x}dx \\= &\frac{1}{2i}\int_{0}^{\infty}\frac{e^{ix\left(1+t\right)}+e^{ix\left(1-t\right)}-e^{-ix\left(1-t\right)}-e^{-ix\left(1+t\right)}}{x}dx \\ = & \frac{1}{2i}\int_{0}^{\infty}\frac{e^{ix\left(1+t\right)}-e^{-ix\left(1+t\right)}}{x}dx+\frac{1}{2i}\int_{0}^{\infty}\frac{e^{ix\left(1-t\right)}-e^{-ix\left(1-t\right)}}{x}dx \\= & \lim_{\epsilon\rightarrow0^{+}}\left(\frac{1}{2i}\int_{0}^{\infty}\frac{e^{\epsilon+ix\left(1+t\right)}-e^{\epsilon-ix\left(1+t\right)}}{x}dx+\frac{1}{2i}\int_{0}^{\infty}\frac{e^{\epsilon+ix\left(1-t\right)}-e^{\epsilon-ix\left(1-t\right)}}{x}dx\right) \end{align} $$ so using the Frullani's theorem for complex parameters we get $$\begin{align}I\left(t\right)= & \frac{1}{2i}\lim_{\epsilon\rightarrow0^{+}}\log\left(\frac{\epsilon+i\left(1+t\right)}{\epsilon-i\left(1+t\right)}\right)+\frac{1}{2i}\lim_{\epsilon\rightarrow0^{+}}\log\left(\frac{\epsilon+i\left(1-t\right)}{\epsilon-i\left(1-t\right)}\right) \\ = & \color{red}{\frac{\pi}{2}\left(\textrm{sgn}\left(1+t\right)+\textrm{sgn}\left(1-t\right)\right)}. \end{align} $$

Answer 3

Working in the Principal Value sense, since $\frac{\sin(x)\sin(tx)}x$ is odd, we get $$ \begin{align} \int_{-\infty}^\infty\frac{\sin(x)e^{itx}}{x}\,\mathrm{d}x &=\int_{-\infty}^\infty\frac{\sin(x)\cos(tx)}{x}\,\mathrm{d}x\\ &=\int_{-\infty}^\infty\frac{\sin((t+1)x)-\sin((t-1)x)}{2x}\,\mathrm{d}x\\[3pt] &=\frac\pi2\left[\operatorname{sgn}(t+1)-\operatorname{sgn}(t-1)\right]\tag{1} \end{align} $$ The last step in $(1)$ is due to the fact that the following is odd in $t$ and independent of $t$ for $t\gt0$: $$ \begin{align} \int_{-\infty}^\infty\frac{\sin(tx)}{x}\,\mathrm{d}x &=\int_{-\infty}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x\tag{2a}\\ &=\int_{-\infty-i}^{\infty-i}\frac{\sin(x)}{x}\,\mathrm{d}x\tag{2b}\\ &=\int_{-\infty-i}^{\infty-i}\frac{e^{ix}-e^{-ix}}{2ix}\,\mathrm{d}x\tag{2c}\\ &=\int_{\gamma_U}\frac{e^{ix}}{2ix}\,\mathrm{d}x-\int_{\gamma_L}\frac{e^{-ix}}{2ix}\,\mathrm{d}x\tag{2d}\\[6pt] &=\pi-0\tag{2e} \end{align} $$ Explanation:
$\text{(2a)}$: when $t\gt0$, substitute $x\mapsto x/t$
$\text{(2b)}$: no singularities in $[-R,R]\cup[R,R-i]\cup[R-i,-R-i]\cup[-R-i,-R]$
$\text{(2c)}$: $\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$\text{(2d)}$: the integral along the curved parts vanishes
$\phantom{\text{(2d)}: }\gamma_U=[-R-i,R-i]\cup Re^{i[0,\pi]}-i$
$\phantom{\text{(2d)}: }\gamma_L=[-R-i,R-i]\cup Re^{-i[0,\pi]}-i$
$\text{(2e)}$: $\gamma_U$ contains a singularity and $\gamma_L$ does not

Answer 4

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\int_{\infty}^{\infty}{\sin\pars{x}\expo{\ic tx} \over x}\,\dd x} & = \int_{\infty}^{\infty}\expo{\ic tx}\ \overbrace{\half\int_{-1}^{1}\expo{-\ic kx}\,\dd k} ^{\ds{\sin\pars{x} \over x}}\ \,\dd x = \pi\int_{-1}^{1}\ \overbrace{\int_{-\infty}^{\infty} \expo{-\ic\pars{k - t}}\,{\dd x \over 2\pi}}^{\ds{\delta\pars{k - t}}}\ \,\dd k \\[5mm] & = \pi\int_{-1}^{1}\delta\pars{k - t}\,\dd k = \color{#f00}{\pi\bracks{\vphantom{\large A}\verts{t} < 1}}\qquad \pars{~\bracks{\cdots}:\ Iverson\ Bracket~} \\[5mm] & = \color{#f00}{\left\{\begin{array}{rcc} \ds{\pi} & \mbox{if} & \ds{-1 < t < 1} \\[2mm] \ds{0} & \mbox{if} & \ds{\verts{t} > 1} \end{array}\right.} \end{align}