Calculating asymptotics of integral $B(r) = \int_0^\infty \frac{2}{\pi} \frac{k^2 \sin(k r )}{k^3 + a} dk$

Question

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: $B(r) = 1 - \beta r$,

Answer 1

$$B(r)=\frac{2}{\pi}\int_0^\infty \frac{k^2 \sin(k r )}{k^3 + a} dk\overset{k=a^{\frac13}t}{=}\frac{2}{\pi}\int_0^\infty \frac{t^2 \sin(a^{\frac13}rt)}{t^3 + 1} dt$$ 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: $$B(r)=\frac{2}{\pi}\int_0^\infty \frac{t^2 \sin(bt)}{t^3 + 1} dt=-\frac2{\pi b}\frac{t^2 \cos(bt)}{t^3 + 1}\,\bigg|_{t=0}^\infty+\frac2{\pi b}\int_0^\infty\cos(bt)\left(\frac{t^2}{1+t^3}\right)'dt$$ $$=\frac2{\pi b^2}\sin(bt)\left(\frac{t^2}{1+t^3}\right)'\,\bigg|_{t=0}^\infty-\frac2{\pi b^2}\int_0^\infty\sin(bt)\left(\frac{t^2}{1+t^3}\right)''dt$$ $$=\frac2{\pi b^3}\cos(bt)\left(\frac{t^2}{1+t^3}\right)''\,\bigg|_{t=0}^\infty-\frac2{\pi b^3}\int_0^\infty\cos(bt)\left(\frac{t^2}{1+t^3}\right)'''dt$$ $$=-\frac4{\pi\,b^3}+o\Big(\frac1{b^3}\Big)$$

Answer 2

This is probably an unsatisfying answer, but at least it should produce the correct answer.

Computational software (I used Mathematica) can evaluate the integral: $$ B(r) = \frac1{\pi^{3/2}\sqrt3} G_{1,7}^{5,1}\biggl( \genfrac{}{}{0pt}{}{1/2}{0, 1/6, 1/2, 1/2, 5/6, 1/3, 2/3\,} \;\bigg|\, \frac{a^2r^6}{46656} \biggr) $$ expressed in terms of the (previously unknown to me) Meijer $G$ function. The software correctly knows that the right-hand side tends to $1$ as $r\to0$.

The software can also calculate \begin{multline*} \frac d{dr} \frac1{\pi^{3/2}\sqrt3} G_{1,7}^{5,1}\biggl( \genfrac{}{}{0pt}{}{1/2}{0, 1/6, 1/2, 1/2, 5/6, 1/3, 2/3\,} \;\bigg|\, \frac{a^2r^6}{46656} \biggr) \\ = -\frac{a^2r^5}{7776\pi^{3/2}\sqrt3} G_{1,7}^{5,1}\biggl( \genfrac{}{}{0pt}{}{-1/2}{-5/6, -1/2, -1/2, -1/6, 0, -2/3, -1/3\,} \;\bigg|\, \frac{a^2r^6}{46656} \biggr), \end{multline*} as well as evaluate $$ \lim_{r\to0^+} -\frac{a^2r^5}{7776\pi^{3/2}\sqrt3} G_{1,7}^{5,1}\biggl( \genfrac{}{}{0pt}{}{-1/2}{-5/6, -1/2, -1/2, -1/6, 0, -2/3, -1/3\,} \;\bigg|\, \frac{a^2r^6}{46656} \biggr) = -\frac{4a^{1/3}}{3\sqrt3}. $$ Therefore the linear approximation in question is $B(r) \approx 1 - \dfrac{4a^{1/3}}{3\sqrt3}r$ for $r$ small. (The dependence on $a$ is appropriate since $B_a(r) = B_1(a^{1/3}r)$, if we add a subscript indicating the value of $a$ chosen.)

Answer 3

Let $$ h(t) = \sin t,\quad f(t) = \frac{2}{\pi }\frac{{t^2 }}{{t^3 + a}}. $$ Then $$ \mathscr{M}B(z) = \mathscr{M}h(z)\mathscr{M}f(1 - z), $$ where $\mathscr{M}$ denotes the Mellin transform. Now $$ \mathscr{M}h(z) = \sin \left( {\frac{\pi }{2}z} \right)\Gamma (z) $$ and $$ \mathscr{M}f(1 - z) = \frac{2}{3}a^{ - z/3} \csc \left( {\frac{\pi }{3}z} \right). $$ Thus, by Mellin inversion, $$ B(r) = \frac{1}{{2\pi {\rm i}}}\frac{2}{3}\int_{c - {\rm i}\infty }^{c + {\rm i}\infty } {(a^{1/3} r)^{ - z} \Gamma (z)\sin \left( {\frac{\pi }{2}z} \right)\csc \left( {\frac{\pi }{3}z} \right){\rm d}z} , \quad 0<c<3. $$ To obtain the asymptotic behaviour of $B(r)$ as $r\to 0^+$, we push the contour to the right through the poles at $z=0,-1,-3,-5,\ldots$. This yields the asymptotic expansion $$ B(r) \sim 1 - \frac{4}{{3^{3/2} }}(a^{1/3} r) - \frac{{6\log (a^{1/3} r) + 6\gamma - 11}}{{18\pi }}(a^{1/3} r)^3 + \frac{1}{{10 \cdot 3^{5/2} }}(a^{1/3} r)^5 + \ldots , $$ as $r\to 0^+$. If we push the contour to the left through the poles at $z=3,9,15,\ldots$, we obtain the asymptotic expansion \begin{align*} B(r) &\sim \frac{2}{\pi }\sum\limits_{n = 0}^\infty {( - 1)^{n + 1} \frac{{\Gamma (6n + 3)}}{{(a^{1/3} r)^{6n + 3} }}} \\ &= -\frac{4}{\pi }\frac{1}{{(a^{1/3} r)^3 }} + \frac{{80640}}{\pi }\frac{1}{{(a^{1/3} r)^9 }} - \frac{{174356582400}}{\pi }\frac{1}{{(a^{1/3} r)^{15} }} + \ldots , \end{align*} as $r\to+\infty$.