When is $\int_0^\infty dk \int_0^\infty dq \int_0^R dt \,f(k,q,t,r)\stackrel{?}{=}g(r)$ where both $f$ and $g$ are known functions

Question

I would like to know under which circumstances the following triple integral can be evaluated analytically as $$ \int_{k=0}^{k=\infty} \int_{q=0}^{q=\infty} \int_{t=0}^{t=R} f(k,q,t,r) \,\mathrm{d}t \,\mathrm{d}q \,\mathrm{d}k \stackrel{?}{=} g(r) \qquad\qquad (0 < t,r < R) \, , $$ where the two function $f$ and $g$ are given explicitely by $$ f(k,q,t,r) = \frac{4qk}{\pi \alpha^4} \, \left( Q-q\right) \left( K+k \right) \left( e^{-k} - e^{-K} \right) \sin(qt) \sin(kt) J_1(qr) \, , $$ and $$ g(r) = \frac{r}{s^3} \left( \frac{6}{\alpha^2 s^2} -2e^{-\alpha s} \left( 1 + \frac{3}{\alpha s} + \frac{3}{\alpha^2 s^2} \right) \right) \, . $$ Here, we have defined for the sake of convenience the abbreviations $K = \sqrt{k^2+\alpha^2}$, $Q = \sqrt{q^2+\alpha^2}$, and $s=\sqrt{1+r^2}$. Here, $\alpha$ is a positive real number.

Please note that $t \in [0,R]$ and that $k,q \in [0,\infty)$.

A numerical evaluation shows that the result is surprisingly accurate up to a very small additive constant that I do not succeed to evaluate exactly yet.

As $R\to\infty$, it can be checked that the identity is exact.

How I proceeded is to use inverse Hankel transform wrt the variable $q$ and inverse sine Fourier transform wrt the variable $k$. Even though I know that $$ \int_0^\infty r g(r) J_1(qr) \, \mathrm{d}r = \frac{2q}{\alpha^2} \left( e^{-q} - e^{-Q}\right) \, , $$ I was unable to rigorously prove that.

In particular, in the limit $\alpha \to 0$, this can easily be shown upon using the identity $$ \int_0^\infty \sin(qt) J_1(qr) \, \mathrm{d}q = \frac{t H(r-t)}{r \left( r^2-t^2\right)^\frac{1}{2}} \, . $$

Any insight is highly appreciated. Thank you.

Answer 1

Partial solution

First of all we note that in the distribution sense $\int_{-\infty}^\infty e^{ikx}dx=2\pi\delta (k)$

Therefore, $$\int_0^\infty\sin(qt)\sin(kt)dt=\frac12\int_{-\infty}^\infty\sin(qt)\sin(kt)dt=\frac\pi2\left(\delta(q-k)-\delta(q+k)\right)$$ Integrating with respect to $q$ $$I(R)=\int_0^\infty \int_0^\infty \int_0^R f(k,q,t,r) \,\mathrm{d}t \,\mathrm{d}q \,\mathrm{d}k=\frac2{\alpha^2}\int_0^Rk^2J_1(kr)\left(e^{-k}-e^{-\sqrt{k^2+\alpha^2}}\right)dk$$ $$=\frac2{\alpha^2}(I_1-I_2)\tag{1}$$ We will find $I(R=\infty)$.

Using the relation and taking derivatives with respect to $\alpha$,$$\displaystyle \int_0^\infty e^{-\alpha x}J_\nu(\beta x)dx=\frac{\beta^{-\nu}\left(\sqrt{\alpha^2+\beta^2}-\alpha\right)^\nu}{\sqrt{\alpha^2+\beta^2}}$$ $$I_1(\infty)=\int_0^\infty k^2J_1(kr)e^{-k}dk=\frac{3r}{s^5}\tag{2}$$ There are different way to evaluate $I_2(\infty)$. Using the relation $J_1(x)=-J'_0(x)$ and integrating by part $$I_2=\frac2r\int_0^\infty J_0(kr)e^{-\sqrt{k^2+\alpha^2}}k\,dk-\frac1r\int_0^\infty J_0(kr)\frac{e^{-\sqrt{k^2+\alpha^2}}}{\sqrt{k^2+\alpha^2}}k^3\,dk=I_a-I_b\tag{3}$$ Using the integral form $\displaystyle J_0(x)=\frac1{2\pi}\int_0^{2\pi}e^{ix\cos t}dt$ $$I_a=\frac1{\pi r}\int_0^\infty e^{-\sqrt{k^2+\alpha^2}} kdk\int_0^{2\pi}e^{ikr\cos t}dt$$ $$=\frac1{\pi r}\int_{-\infty}^\infty\int_{-\infty}^\infty dk_xdk_ye^{-\sqrt{k_x^2+k_y^2+\alpha^2}}e^{i(k_xx+k_yy)}\tag{4}$$ where we denoted $k=\sqrt{k_x^2+k_y^2}$ and $r=\sqrt{x^2+y^2}$.

Using the relation $\displaystyle\int_{-\infty}^\infty e^{-iz\alpha}e^{ik_zz}dz=2\pi\delta(z-\alpha)$ and changing the order of integration, we can write (4) in the form $$I_a=\frac1{\pi r}\frac1{2\pi}\int_{-\infty}^\infty e^{-iz\alpha}dz\int_{-\infty}^\infty\int_{-\infty}^\infty \int_{-\infty}^\infty dk_xdk_ydk_ze^{-\sqrt{k_x^2+k_y^2+k_z^2}}e^{i(k_xx+k_yy+k_zz)}$$ Switching to the polar system of coordinates and denoting $p=\sqrt{k_x^2+k_y^2+k_z^2};\,d=\sqrt{x^2+y^2+z^2}=\sqrt{r^2+z^2}$ $$I_a=\frac{2\pi}{2\pi^2r}\int_{-\infty}^\infty e^{-iz\alpha}dz\int_0^\infty e^{-p}p^2dp\int_0^{\pi}e^{ipd\cos\theta}\sin\theta d\theta$$ $$=\frac2{\pi r}\int_{-\infty}^\infty e^{-iz\alpha}dz\int_0^\infty e^{-p}\frac{\sin(pd)}dp\,dp=\frac{4}{\pi r}\int_{-\infty}^\infty \frac{e^{-iz\alpha}}{(1+r^2+z^2)^2}dz$$ Closing the contour in the lower half of the complex plane, we get $$I_a=\frac{2e^{-s\alpha}}{rs^2}\left(\frac1s+\alpha\right)\tag{5}$$ where $s=\sqrt{1+r^2}$.

In the same way we evaluate $I_b$ $$I_b=\frac1{2\pi r}\frac1{2\pi}\int_{-\infty}^\infty e^{-iz\alpha}dz\int_{-\infty}^\infty\int_{-\infty}^\infty \int_{-\infty}^\infty \frac{k_x^2+k_y^2+k_z^2-\alpha^2}{\sqrt{k_x^2+k_y^2+k_z^2}}e^{-p}e^{i(k_xx+k_yy+k_zz)}dk_xdk_ydk_z$$ $$=\frac1{\pi r}\int_{-\infty}^\infty e^{-iz\alpha}dz\int_0^\infty e^{-p}(p^2-\alpha^2)\frac{\sin(pd)}{d}dp$$ $$=\frac1{\pi r}\int_{-\infty}^\infty e^{-iz\alpha}\left(\frac8{(z^2+s^2)^3}-\frac2{(z^2+s^2)^2}-\frac{\alpha^2}{z^2+s^2}\right)dz$$ $$I_b=e^{-\alpha s}r\left(\frac3{s^5}+\frac{3\alpha}{s^4}+\frac{\alpha^2}{s^3}\right)\tag{6}$$ Putting (6) and (5) into (3), and (3) and (2) into (1), we get $$I(R)= \frac2{\alpha^2}\int_0^Rk^2J_1(kr)\left(e^{-k}-e^{-\sqrt{k^2+\alpha^2}}\right)dk$$ $$I(\infty)=\frac{r}{s^3}\bigg[\frac6{\alpha^2s^2}-2e^{-\alpha s}\left(1+\frac3{\alpha s}+\frac3{\alpha^2s^2}\right)\bigg]=g(r)$$ Asymptotics of $I(R)$ can also be obtained - depending on the relationship between the parameters: $\,R, \, r,\,\alpha$