# Need help with this integral!

I am having some difficulty figuring out how to do this integral analytically.

∫r1-x*I0(alpha*r)dr

I have attempted to do integration by parts, but am unable to find any recursion. This would be easy if x was not variable. Also, I have attempted just expanding this term into a Taylor series, but this is not resulting in satisfying results. Any thoughts on how I might approach this one?

tried: u= I0(alpha*r) and dv=r1-x
also tried: dv=r*I0(alpha*r) and u= r-x

LCKurtz
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What is I0? What difference does it make what x is if the integral is a dr integral?

Thanks for the response. I0 is the modified Bessel function of the 1st kind. If r was raised to an integer power, there is a well-known recursion formula. I cannot take advantage of that recursion, because there are only special instances where x will give me an integer value for the power of r.

arildno
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Dearly Missed
Hmm..
Why is not "x" to be regarded as some fixed, arbitrary parameter??
Try to regard it as just that, and see if you, by means of integration by parts, can simplify your integral by the defining differential equation of the bessel function

LCKurtz
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Gold Member
Thanks for the response. I0 is the modified Bessel function of the 1st kind. If r was raised to an integer power, there is a well-known recursion formula. I cannot take advantage of that recursion, because there are only special instances where x will give me an integer value for the power of r.

If I had known what your problem really was I wouldn't have replied. Sorry but I don't have the time or inclination to delve into this one. Perhaps someone else will pick it up.

 After posting, I see Arildno already has.

Hmm..
Why is not "x" to be regarded as some fixed, arbitrary parameter??
Try to regard it as just that, and see if you, by means of integration by parts, can simplify your integral by the defining differential equation of the bessel function

I think maybe that is the problem. This is but one step in a Taylor Aris dispersion problem. I am unsure if I can even go back and change my differential equation into a more friendly form. Essentially, what I have presented here is a multiplication of two functions which each were obtained independently earlier in the analysis which I now need to integrate. Only one of these terms contained a Bessel function.

step 1:

u=r-x , du=-x*r-(x+1)dr, dv=r *I0(alpha*r) dr, v=r*I1(alpha*r)/alpha

r1-xI1(alpha*r)/alpha+x/alpha*int(r-xI1(alpha*r)dr)

step2:

u=r-x, du= -x*r-(x+1) dr, dv= I1(alpha*r)dr, v=I0(alpha*r)/alpha

r-xI0(alpha*r)/alpha+x/alpha*int(r-(x+1)I0(alpha*r)dr)

step3:
u=r-(x+2), du= -(x+2)r-(x+3)dr, dv=r*I0(alpha*r)dr, v=r*I1(alpha*r)

I'm not really getting anywhere here. This looks like an infinite series solution.