4
$\begingroup$

I'm trying to do the integral

Integrate[ B^2*BesselK[0, ko*ρ]^2*2 π*ρ, {ρ, a, ∞}]

which I figured should be relatively simple as the integral of a Bessel Function outside of a certain circle (of radius a) from a to infinity. Mathematica throws me the answer

ConditionalExpression[(B^2 π^(3/2) MeijerG[{{}, {3/2}}, {{0, 1, 1}, {}}, a^2 ko^2])/(2 ko^2), Re[a] > 0 && Im[a] == 0 && Re[ko] > 0]

The conditional expression is expected, but I was wondering if it's possible to expand the G function in terms of other Bessel Functions, especially the modified one $K_o$. Integrating across both regions and subtracting them gives a similar answer; while the answer for the integration from $0\to \inf$ is a constant, the integral from 0 to a involves the G function again, like

Integrate[B^2*BesselK[0, ko*ρ]^2*2 π*ρ, {ρ, 0, a}] 

ConditionalExpression[( B^2 π^(3/2) MeijerG[{{1}, {3/2}}, {{1, 1, 1},{0}},a^2 ko^2])/(2 ko^2), Re[a ko] > 0]

I'm hoping to get a wholly symbolic answer or at least an expansion of one, I was wondering if it's possible to have Mathematica get rid of the G function stuff in these results?

$\endgroup$
  • $\begingroup$ I don't know if your integrand is continuous between a and infinity but, if it is, you could use the Fundamental Theorem of Calculus: Ask Mathematica to instead determine the indefinite integral, substitute the limits of integration, and subtract: int = Integrate[B^2*BesselK[0, ko*\[Rho]]^2*2 \[Pi]*\[Rho], \[Rho]]; Limit[int, \[Rho] -> Infinity, Direction -> "FromBelow"] - (int /. \[Rho] -> a) $\endgroup$ – theorist Nov 20 '18 at 4:23
4
+50
$\begingroup$

Based on the comment by theorist

$Assumptions = a > 0 && ko > 0 && B ∈ Reals

int = 2*Pi*B^2*Integrate[BesselK[0, ko*ρ]^2*ρ, ρ]
(*π B^2 ρ^2 (BesselK[0, ko ρ]^2 - BesselK[1, ko ρ]^2)*)

Check answer

D[int, ρ] // FullSimplify
(* 2 π B^2 ρ BesselK[0, ko ρ]^2*)

ok

intinf = Limit[int, ρ -> ∞]
(*0*)

inta = int /. ρ -> a
(*π a^2 B^2 (BesselK[0, a ko]^2 - BesselK[1, a ko]^2)*)

int0 = Limit[int, ρ -> 0]
(*-((π B^2)/ko^2)*)

int0a = inta - int0
(*π a^2 B^2 (BesselK[0, a ko]^2 - BesselK[1, a ko]^2) + (π B^2)/ko^2*)

intainf = intinf - inta
(*π (-a^2) B^2 (BesselK[0, a ko]^2 - BesselK[1, a ko]^2)*)

A quick spot check numerically with B = 1, ko = 1, a = 1 both the above solutions matche the MeijerG answers.

$\endgroup$
1
$\begingroup$

I do not know the answer, however I would like to share what I have observed.

Basically, what we are trying to do is simplifying the expression (1/x) MeijerG[{{}, {3/2}}, {{0, 1, 1}, {}}, x] (upto an overall factor) into some Bessel functions.

I do not now the generally true answer. However, I do know the approximate answer for very small $x$ values, such that the error percentage is less than 1% for approximately $x<10^{-3.2}$.

The key point is to analyze the Mellin transformation. This is suggested by the definition of MeijerG, so we check its MellinTransform:

In[294]:= MellinTransform[MeijerG[{{},{3/2}},{{0,1,1},{}},z]/z,z,x]

$\frac{\Gamma (x-1) \Gamma (x)^2}{\Gamma \left(x+\frac{1}{2}\right)}$

We will now try to match this by Mellin transform of Bessel functions. In general, we see that

In[296]:= MellinTransform[{BesselK[n,z^(1/2)]/z^k,BesselJ[n,z^(1/2)]/z^k},z,x]

$\left\{2^{2 (x-k)-1} \Gamma \left(-k-\frac{n}{2}+x\right) \Gamma \left(-k+\frac{n}{2}+x\right),\frac{2^{2 (x-k)} \Gamma \left(-k+\frac{n}{2}+x\right)}{\Gamma \left(k+\frac{n}{2}-x+1\right)}\right\}$

Now, by trial and error, the closest I managed to get was

MellinTransform[2/Sqrt[\[Pi]] BesselK[0, z^(1/2)]^2, z, x]

$\frac{\Gamma (x)^3}{\Gamma \left(x+\frac{1}{2}\right)}$

One can check that $\frac{2 K_0\left(\sqrt{z}\right){}^2}{\sqrt{\pi }}$ indeed matches MeijerG in small $z$ in leading order.

Even though it is far from being a nice approximation in general, it still works well for small x.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.