I want to calculate asymptotics of this integral $B(r) = \int_0^\infty \frac{2}{\pi} \frac{k ^2 \sin(k r )}{k^3 + a} dk$ for small and big values of parameter $r$, assuming it is positive.
$a$ is also a positive constant.
I was able to calculating large $r$ asymptotic by applying substitution $z = k r$ and then leaving in denominator only leading term by $r$. The result is $B(r) \sim -\frac{1}{r^3}$
However i can't find small $r$ expansion. Numerical calculating says it is linear ,starting from 1: $B(r) = 1 – \beta r$,
Best Answer
$$B(r)=\frac{2}{\pi}\int_0^\infty \frac{k^2 \sin(k r )}{k^3 + a} dk\overset{k=a^{\frac13}t}{=}\frac{2}{\pi}\int_0^\infty \frac{t^2 \sin(a^{\frac13}rt)}{t^3 + 1} dt$$ Denoting for a while $\,a^{\frac13}r=b\ll1$ $$B(r)=\frac2\pi\int_0^\infty t^2\sin(bt)\left(\frac1{t^3+1}-\frac1{t^3}+\frac1{t^3}\right)dt=1-\frac{2b}\pi\int_0^\infty\frac{\sin(bt)}{bt}\frac{dt}{1+t^3}$$ The function $\bigg|\frac{\sin(bt)}{bt}\bigg|\frac1{1+t^3}$ is dominated by $\frac1{1+t^3}$. Taking the limit $b\to0$ under the integral sign $$B(r)\sim1-\frac{2b}\pi\int_0^\infty\frac{dt}{1+t^3}\overset{t^3=x}{=}1-\frac{2b}{3\pi}\int_0^\infty\frac{x^{-\frac23}}{1+x}dx\overset{t=\frac1{1+x}}{=}1-\frac{2b}{3\pi}\int_0^1(1-t)^{-\frac23}t^{-\frac13}dt$$ $$=1-\frac{2b}{3\pi}\Gamma\Big(\frac13\Big)\Gamma\Big(\frac23\Big)=1-\frac{2b}{3\pi}\frac\pi{\sin\frac\pi3}=1-\frac{4b}{3\sqrt3}$$ Coming back to $a$ $$\boxed{\,\,B(r)=1-\frac{4a^{\frac13}}{3\sqrt3}+o\big(a^{\frac13}\big)\,\,}$$ To get the asymptotics for $b\gg1$ you can use integration by parts several times: $$B(r)=\frac{2}{\pi}\int_0^\infty \frac{t^2 \sin(bt)}{t^3 + 1} dt=-\frac2{\pi b}\frac{t^2 \cos(bt)}{t^3 + 1}\,\bigg|_{t=0}^\infty+\frac2{\pi b}\int_0^\infty\cos(bt)\left(\frac{t^2}{1+t^3}\right)'dt$$ $$=\frac2{\pi b^2}\sin(bt)\left(\frac{t^2}{1+t^3}\right)'\,\bigg|_{t=0}^\infty-\frac2{\pi b^2}\int_0^\infty\sin(bt)\left(\frac{t^2}{1+t^3}\right)''dt$$ $$=\frac2{\pi b^3}\cos(bt)\left(\frac{t^2}{1+t^3}\right)''\,\bigg|_{t=0}^\infty-\frac2{\pi b^3}\int_0^\infty\cos(bt)\left(\frac{t^2}{1+t^3}\right)'''dt$$ $$=-\frac4{\pi\,b^3}+o\Big(\frac1{b^3}\Big)$$