Computing $\int_0^\infty \frac{\sin(u)}{u}e^{-u^2 b} \, du$

Question

I want to compute

$\int_0^\infty u^{-1}(1-e^{\frac{-u^2 t}{2}})\sin(u(|x|-r))\,du$ and so ,as shown below, I want to compute

$$\int_0^\infty \frac{\sin(u)}{u}e^{-u^2 b} \, du$$

Attempt

We split the first integral into two:

• $\displaystyle \int_0^\infty u^{-1} \sin(u(|x|-r))\, du=\frac{\pi}{ 2}$

• $\displaystyle \int_0^\infty u^{-1} e^{\frac{-u^2 t}{2}}\sin(u(|x|-r))\,du = \int_0^\infty \frac{\sin(z)}{z} e^{-z^2 b} \, dz$

where $b=\dfrac{t}{2(|x|-r)^2}$ is a positive real constant.

any suggestions?

How to show that the value is $\frac{\pi}{2}erf(\frac{1}{2\sqrt{b}})$?

Given that I can swap integral and sum we have

$\sum_{-1}^{\infty}\frac{(-i)^{n}+(i)^{n}}{2(1+n)}\int_{0}^{\infty}x^{n}e^{-bx^{2}}dx=\sum_{-1}^{\infty}\frac{(-i)^{n}+(i)^{n}}{2(1+n)} \frac{\Gamma(\frac{n+1}{2})}{2b^{\frac{n+1}{2}}}.$

Answer 1

Hint. Here is an approach.

Let $b$ be a real number such that $b>0$ and set $$ f(b):=\int_0^\infty \frac{\sin{u}}{u}e^{-u^2 b}du. \tag1 $$ We are allowed to write that $$ \begin{align} f'(b) &=-\int_0^\infty u\sin{u}\: e^{-u^2 b}du \tag2 \\\\ &=\frac{1}{2b}\left[\sin{u}\:e^{-u^2 b}\right]_{0}^{\infty} -\frac{1}{2b}\int_0^\infty \cos{u}\: e^{-u^2 b}du\\\\ &=0-\frac{1}{4b}\Re \int_{-\infty}^\infty e^{-u^2 b-iu}du\\\\ &=-\frac{e^{-\frac{1}{4b}}}{4b}\Re \int_{-\infty}^\infty e^{-\left(\sqrt{b}\:u+\dfrac{i}{\sqrt{b}}\right)^2}du\\\\ &=-\frac{e^{-\frac{1}{4b}}}{4b} \int_{-\infty}^\infty e^{-U^2}\frac{dU}{\sqrt{b}}\\\\ &=-\frac{e^{-\dfrac{1}{4b}}}{4b}\frac{\sqrt{\pi}}{\sqrt{b}} \tag3\\\\ \end{align} $$ and, since from $(1)$ we have $\displaystyle \lim\limits_{b\to +\infty}f'(b)=0 $, then from $(3)$ we obtain $$ \begin{align} f(b) &=-\frac{\sqrt{\pi}}{4}\int_b^\infty \frac{e^{\large -\frac{1}{4\:t}}}{t\sqrt{t}}dt \tag4 \\\\ &=\sqrt{\pi}\int_0^{\large \frac{1}{2\sqrt{b}}} e^{-x^2}dx \qquad \left(x:=\frac{1}{2\sqrt{t}},\quad dx:=-\frac{1}{4\:t\sqrt{t}}dt\right)\\\\ &=\frac{\pi}{2}{\rm{erf}}\left(\frac{1}{2\sqrt{b}}\right) \end{align} $$ as desired.

Answer 2

Are you wanting just an answer or the full solution. Assuming $b$ is positive and Real, Mathematica says: $$ \int_0^\infty \frac{\sin{u}}{u}e^{-u^2 b}du\quad= \quad\frac{1}{2}\pi \,Erf(\frac{1}{2\sqrt{b}}) $$ Where $Erf$ is the error function.

Answer 3

Let $I(a)=\displaystyle\int_0^\infty\cos(au)~e^{-bu^2}~du$, and evaluate it using Euler's formula, completing the square, and the value of the Gaussian integral. Then express your integral in terms of $\displaystyle\int I(a)~da$ and let $a=1$. The final result will be a simple error function in b.

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align}&\color{#66f}{\large% \int_{0}^{\infty}{\sin\pars{u} \over u}\,\expo{-u^{2}b}\,\dd u} =\half\int_{-\infty}^{\infty}\expo{-u^{2}b}\,{\sin\pars{u} \over u}\,\dd u =\half\int_{-\infty}^{\infty}\expo{-u^{2}b}\ \overbrace{\half\int_{-1}^{1}\expo{\ic k u}\,\dd k} ^{\ds{=\ \dsc{\sin\pars{u} \over u}}}\ \,\dd u \\[5mm]&={1 \over 4}\int_{-1}^{1} \int_{-\infty}^{\infty}\exp\pars{-b\bracks{u^{2} - {\ic k \over b}u}} \,\dd u\,\dd k \\[5mm]&={1 \over 4}\int_{-1}^{1} \int_{-\infty}^{\infty} \exp\pars{-b\braces{\bracks{u - \ic\,{k \over 2b}}^{2} + {k^{2} \over 4b^{2}}}} \,\dd u\,\dd k \\[5mm]&={1 \over 4}\int_{-1}^{1}\exp\pars{-\,{k^{2} \over 4b}} \int_{-\infty - \ic k/\pars{2b}}^{\infty - \ic k/\pars{2b}}\exp\pars{-bu^{2}} \,\dd u\,\dd k \\[5mm]&=\half\,\pars{2\root{b}}\int_{0}^{1/\pars{\root{2}b}}\exp\pars{-k^{2}} {1 \over \root{b}}\ \overbrace{\int_{-\infty}^{\infty}\exp\pars{-u^{2}}\,\dd u} ^{\dsc{\root{\pi}}}\ \,\dd k \\[5mm]&=\dsc{\root{\pi}}{\root{\pi} \over 2}\bracks{{2 \over \root{\pi}} \int_{0}^{1/\pars{\root{2}b}}\exp\pars{-k^{2}}\,\dd k} =\color{#66f}{\large{\pi \over 2}\,\,{\rm erf}\pars{1 \over \root{2b}}} \end{align}