How can I calculate the following integral:
$$\int_0^{\infty}\frac{x^{1/2}}{1-x^2}\sin(ax)\sin[a(1-x)] dx$$
where $a>0$.
It seems like the integrand is well defined without any singularities, but I don't have any clue how to proceed.
Can anyone show me how to do it? Thank you
Edit
Following exactly what @Maxim suggested in the comment
$$\int_0^\infty f(x) dx = \operatorname {Re} \operatorname {v. \! p.} \int_0^\infty g(x) dx = \operatorname {Re} \left(\int_0^{i \infty} g(x) dx + \pi i \operatorname* {Res}_{x = 1} g(x) \right).$$
taking $x=iu$
\begin{align}
\int_0^{i\infty}g(x) dx &=\frac{1}{2}ie^{i\frac{\pi}{4}-a}\int_0^{\infty}\frac{\sqrt{u}e^{-2au}}{1+u^2} du-\frac{1}{2}ie^{i\frac{\pi}{4}}\cos (a) \int_0^{\infty}\frac{\sqrt{u}}{1+u^2} du\\
&=\frac{1}{2}ie^{i\frac{\pi}{4}-a}I-i\frac{\sqrt{2}\pi}{4} \cos (a)e^{i\frac{\pi}{4}}
\end{align}
The integral $I=\int_0^{\infty}\frac{\sqrt{u}e^{-2au}}{1+u^2} du$ can be calculated using the method in the comment of this question.
Let $f_1(u)=\sqrt{u}$ and $g_1(u)=e^{-2au}/(1+u^2)$
\begin{align}
I &=\int_0^{\infty} f_1(u)g_1(u)du\\
&=\int_0^\infty \mathcal L[f](u) \mathcal L^{-1}[g](u) dx\\
&=\frac{\sqrt{\pi}}{2}\int_{2a}^{\infty}\frac{\sin (u-2a)}{u^{3/2}}
\end{align}
then let $u=x^2$
\begin{align}
I &=\sqrt{\pi}\int_{\sqrt{2a}}^{\infty} \frac{\sin (x^2-2a)}{x^2}dx\\
&=\sqrt{\pi}\cos(2a)\int_{\sqrt{2a}}^{\infty}\frac{\sin (x^2)}{x^2}dx-\sqrt{\pi}\sin(2a)\int_{\sqrt{2a}}^{\infty}\frac{\cos (x^2)}{x^2}dx\\
&=\sqrt{\pi}\cos(2a)I_1-\sqrt{\pi}\sin(2a)I_2
\end{align}
Making use of the parameterization technique, we have:
$$I_1=\frac{\sqrt{2\pi}}{2}-2C(\sqrt{2a})+\frac{\sin^2 (\sqrt{2a})}{\sqrt{\pi a}}$$
$$I_2=-\frac{\sqrt{2\pi}}{2}+2S(\sqrt{2a})+\frac{\cos^2 (\sqrt{2a})}{\sqrt{\pi a}}$$
Hence, we can obtain the desired result by substituting $I_1, I_2$ and $I$ into original integral.
My qustion is:
If I am only interested in the asymptotic behavior of the original integral as $a\to +\infty$, is there any simplier way to do it without going through all these steps?
Best Answer
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\infty} {x^{1/2} \over 1-x^{2}} {\sin\pars{ax}\sin\pars{a\bracks{1 - x}} \over 2}\,\dd x\,\right\vert_{\ a\ >\ 0}} \\[5mm] = &\ \int_{0}^{\infty} {x^{1/2} \over 1 - x^{2}} {\cos\pars{2ax - a} - \cos\pars{a} \over 2}\,\dd x \\[5mm] = &\ {1 \over 2}\,\Re\int_{0}^{\infty} {x^{1/2} \over 1 - x^{2}} \bracks{\expo{\ic\pars{2ax - a}} - \expo{\ic a}}\,\dd x \\[5mm] = &\ {1 \over 2}\,\Re\bracks{\expo{-\ic a}\int_{0}^{\infty} {x^{1/2} \over 1 - x^{2}} \pars{\expo{2\ic ax} - \expo{2\ic a}}\,\dd x} \\[5mm] \stackrel{x\ \mapsto\ x^{2}}{=}\,\,\, &\ \Re\bracks{\expo{-\ic a}\int_{0}^{\infty} {x^{2} \over 1 - x^{4}} \pars{\expo{2\ic ax^{2}} - \expo{2\ic a}} \,\dd x} \end{align} Now,I'll "close" the integration along a pizza-slice $\ds{\mathcal{P}_{s}}$ in the first quadrant. Namely, $\ds{\mathcal{P}_{s} = \pars{0,R}\cup R\expo{\ic\pars{0,\pi/4}}\cup \pars{R,0}\expo{\ic\pi/4}}$ with $\ds{R \to \infty}$. The integration along the arc $\ds{R\expo{\ic\pars{0,\pi/4}}}$ vanishes out as $\ds{R \to \infty}$.
Then, \begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\infty} {x^{1/2} \over 1-x^{2}} {\sin\pars{ax}\sin\pars{a\bracks{1 - x}} \over 2}\,\dd x\,\right\vert_{\ a\ >\ 0}} \\[5mm] = &\ -\Re\bracks{\expo{-\ic a}\int_{\infty}^{0} {\ic r^{2} \over 1 + r^{4}} \pars{\expo{-2ar^{2}} - \expo{2\ic a}} \expo{\ic\pi/4}\dd r} \\[5mm] = &\ \sin\pars{a - {\pi \over 4}}\ \underbrace{\int_{0}^{\infty}{r^{2}\expo{-2ar^{2}} \over r^{4} + 1}\dd r} _{\ds{\equiv \mathcal{I}}} \\[2mm] + &\ \sin\pars{a + {\pi \over 4}}\ \underbrace{\int_{0}^{\infty}{r^{2} \over r^{4} + 1}\dd r} _{\ds{{\root{2} \over 4}\pi}}\label{1}\tag{1} \end{align} Lets evaluate $\ds{\mathcal{I}}$: \begin{align} \mathcal{I} & \equiv \int_{0}^{\infty}{r^{2}\expo{-2ar^{2}} \over r^{4} + 1}\dd r = \Re\int_{0}^{\infty}{\expo{-2ar^{2}} \over r^{2} + \ic}\dd r \\[5mm] & \stackrel{2ar^{2}\ \mapsto\ r^{2}}{=} \,\,\, \Re\int_{0}^{\infty}{\expo{-2ar^{2}} \over r^{2} + \ic}\dd r = \root{2a}\Re\int_{0}^{\infty}{\expo{-r^{2}} \over r^{2} + 2a\ic}\dd r \\[5mm] & = \root{2a} \Re\int_{0}^{\infty}\expo{-r^{2}} \\[2mm] & \pars{% {1 \over r - \root{2a}\expo{3\pi\ic/4}} - {1 \over r + \root{2a}\expo{3\pi\ic/4}}}{1 \over 2\root{2a}\expo{3\pi\ic/4}}\dd r \\[2mm] & = {1 \over 2}\Re\braces{\expo{-3\pi\ic/4} \bracks{\on{G}\pars{-\root{2a}\expo{3\pi\ic/4}} - \on{G}\pars{\root{2a}\expo{3\pi\ic/4}}}} \end{align} where $\ds{\on{G}}$ is the Goodwin-Staton Integral.
Finally ( see \ref{1} ), \begin{align} &\bbox[5px,#ffd]{\left.\int_{0}^{\infty} {x^{1/2} \over 1-x^{2}} {\sin\pars{ax}\sin\pars{a\bracks{1 - x}} \over 2}\,\dd x\,\right\vert_{\ a\ >\ 0}} \\[5mm] = &\ {1 \over 2}\sin\pars{a - {\pi \over 4}}\ \times \\[2mm] &\ \Re\braces{\expo{-3\pi\ic/4} \bracks{\on{G}\pars{-\root{2a}\expo{3\pi\ic/4}} - \on{G}\pars{\root{2a}\expo{3\pi\ic/4}}}} \\[2mm] + &\ {\root{2} \over 4}\pi \sin\pars{a + {\pi \over 4}} \end{align}