# NIntegrate failed to converge to prescribed accuracy for integral contain Bessel function of first kind

I am new in Mathematica, I try to find the following numerical integral. but I get the error message ( NIntegrate failed to converge to prescribed accuracy ...), I know that there is a divergence since there is a rho in the denominator of the fraction but how can I solve that?I think I should transform the integral any help? can any one transform the integral for me

NIntegrate[ρ*Replace[f[5],
f[l_] :> {0, ((155*2)/ρ)*
Integrate[
BesselJ[l, 23*y]^2*y, {y, 0, ρ}], (6.75*2*l*
Integrate[BesselJ[l, 23*y]^2/y, {y, ρ, Infinity}])/
10^2} .
{0, -D[(6.75*2*l*
Integrate[BesselJ[l, 23*y]^2/y, {y, ρ, Infinity}])/
10^2, ρ], (1/ρ)*
D[ρ*(((155*2)/ρ)*
Integrate[
BesselJ[l, 23*y]^2*y, {y,
0, ρ}]), ρ]}], {ρ, 0, Infinity},
WorkingPrecision -> 10]

• Please edit your question: prior to copy and paste, convert your cell to InputForm (or Raw InputForm) Commented Aug 17, 2022 at 23:08
• How I edit that?
– sara
Commented Aug 19, 2022 at 11:03
• Click on the word Edit below the question Commented Aug 19, 2022 at 13:40

Step 1. The expression provided by OP can be cleaned up a bit. For example, this is a subexpression of the expression provided by OP, with the exception that I put an Inactive in to get Mathematica to exploit the fact that there is derivative D just outside:

D[ρ*(((155*2)/ρ)*Inactive[Integrate][BesselJ[l,23*y]^2*y,{y,0,ρ}]),ρ]


This simplifies to

310*ρ*BesselJ[l,23*ρ]^2


Step 2. Doing Step 1 for the entire integrand that OP has provided, I get the following simpler formula for the integrand, where I also replaced 41.85 by 4185/100 to avoid problems with floating point numbers:

integrand[l_]:=integrand[l]=Assuming[ρ>0, 4185/100 l BesselJ[l, 23 ρ]^2 (
ρ^2 Integrate[BesselJ[l, 23 y]^2/y, {y, ρ, \[Infinity]}]
+ Integrate[y BesselJ[l, 23 y]^2, {y, 0, ρ}])/ρ//Simplify];


This can be evaluated. For example, taking the value l==5 in the question:

integrand[5]
(* (1/(1566219705620 ρ^7))837 (529 ρ^2 (-4608 +
101568 ρ^2 - 5596820 ρ^4 +
740179445 ρ^6) BesselJ[0, 23 ρ]^2 -
23 ρ (-18432 + 1625088 ρ^2 - 22387280 ρ^4 +
740179445 ρ^6) BesselJ[0, 23 ρ] BesselJ[1,
23 ρ] + (-18432 + 2843904 ρ^2 - 102981488 ρ^4 -
2960717780 ρ^6 + 391554926405 ρ^8) BesselJ[1,
23 ρ]^2) BesselJ[5, 23 ρ]^2 *)


Step 3. Let us plot this:

Plot[integrand[5],{ρ,0,5},PlotRange->All]


This gives

Looks nice, but decay seems slow. In fact, let us also plot ρ times the integrand:

Plot[integrand[5]*ρ,{ρ,0,5},PlotRange->All]


This gives

It seems likely that this oscillates forever with some fixed amplitude (see remark below). Thus integrand[5] is 1/ρ times this oscillation. That is too slow for convergence of the integral over {ρ,0,\[Infinity]}. Hence the integral does in fact not converge, it is $$+\infty$$.

The divergence is not due to the behavior near ρ==0, in fact the integrand decays quickly towards ρ==0. Rather, the divergence is due to slow decay as ρ goes to infinity.

Remark. The 1/ρ decay is due to the following terms in integrand[5]//Expand:

  ρ BesselJ[0, 23 ρ]^2 BesselJ[5, 23 ρ]^2
ρ BesselJ[1, 23 ρ]^2 BesselJ[5, 23 ρ]^2

• following your step, I get this result in step 2 (Assuming[[Rho] > 0, {0, (310/[Rho])* Integrate[ BesselJ[l, 23*y]^2*y, {y, 0, [Rho]}], ((675/100)*2*l* Integrate[BesselJ[l, 23*y]^2/y, {y, [Rho], Infinity}])/10^2} . {0, (0.135*l*BesselJ[l, 23*[Rho]]^2)/[Rho], 310*BesselJ[l, 23*[Rho]]^2}]).
– sara
Commented Aug 23, 2022 at 11:42