<s>[INST] How to calculate the integral $\int_0^{\infty}\frac{x^{1/2}}{1-x^2}\sin(ax)\sin[a(1-x)] dx$ 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?
 [/INST] 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_1](u) \mathcal L^{-1}[g_1](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:
$$I_1=\frac{\sqrt{2\pi}}{2}-2C(\sqrt{2a})+\frac{\sin (2a)}{\sqrt{2a}}$$
$$I_2=-\frac{\sqrt{2\pi}}{2}+2S(\sqrt{2a})+\frac{\cos (2a)}{\sqrt{2a}}$$
Substituting $I_1$ and $I_2$ into the expression of $I$ yields
$$I=\sqrt{\pi}\cos(2a)\left[\frac{\sqrt{2\pi}}{2}-2C(\sqrt{2a})+\frac{\sin (2a)}{\sqrt{2a}}\right]-\sqrt{\pi}\sin(2a)\left[-\frac{\sqrt{2\pi}}{2}+2S(\sqrt{2a})+\frac{\cos (2a)}{\sqrt{2a}}\right]$$
Finally, the desired integral can be written as:
$$\int_0^{\infty}f(x)dx=\frac{\sqrt{2}}{4}(\sin a -\cos a)I + \frac{\pi}{4}\sin a + \frac{\pi}{4}\cos a$$
</s>