Question: How to deal with improper integrals from zero to infinity when the integrand has not a finite limit at zero?

Dear all,

I would like to compute numerically using Maple the following improper integral


Integrand := (1/4)*(((((6*I)*beta-3-6*C+(6*I)*C*beta)*s^4+((24*I)*C*beta-24*C-12)*s^2+(24*I)*(1+C)*beta)*BesselK(0, s)+12*BesselK(1, s)*(C+1/2)*s^3)*BesselI(1, s)^3+6*BesselI(0, s)*(-(2*(I*beta*C*s^2+(2*I)*beta*C+(2*I)*beta+4*C+2))*s*BesselK(0, s)+((I*beta*C+I*beta-C-1/2)*s^4+((4*I)*C*beta+4*C+2)*s^2+(4*I)*(1+C)*beta)*BesselK(1, s))*BesselI(1, s)^2-(12*(-(1/2*((C+1/2)*s^2+I*beta*C+I*beta+8*C+4))*s*BesselK(0, s)+((I*beta*C+2*C+1)*s^2+(2*I)*(1+C)*beta)*BesselK(1, s)))*s*BesselI(0, s)^2*BesselI(1, s)+6*s^2*((-2*C-1)*s*BesselK(0, s)+BesselK(1, s)*((C+1/2)*s^2+I*(1+C)*beta))*BesselI(0, s)^3)/((BesselI(0, s)^2*s-BesselI(1, s)^2*s-2*BesselI(1, s)*BesselI(0, s))^2*(C+1/2)*s*Pi):


However, Maple does seem to give a result for this integral. I have tried to compute from e.g. 0.001 as an approximation but it turns out that the integrand diverges as s goes to zero. I have also tried some options such as method = _d01amc but I get Error, (in evalf/int) powering may produce overflow.


I would appreciate it if someone here could provide with some help with regards to the computation of such improper integrals. Thank you.



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