Help-Integration of Bessel functions
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Nikolaos Xenidis
el 7 de Abr. de 2016
Comentada: Roger Stafford
el 8 de Abr. de 2016
Hello. I want to calculate the symbolic definite integrals including Bessel function of first kind and modified Bessel function of second kind:

and

I've tried with the int() function, but matlab returns: int(z*besselj(0, z)^4, z, 0, a) (for the first integral). Similarly, for the second integral returns int(z*besselk(0, z)^4, z, a, Inf). Is there a way to calculate these integrals? I'm new to matlab, and I need some advice
Thank you!
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Roger Stafford
el 7 de Abr. de 2016
Editada: Walter Roberson
el 7 de Abr. de 2016
It is likely that an explicit expression for either of those integrals as functions of 'a' is not known in mathematics and therefore not known to the symbolic toolbox or mupad. However, in the case of your integrals, cumulative numerical integration such as given by the matlab function 'cumtrapz' can give you a numerical result for an entire vector of increasing values of 'a' for any desired size of increment and do this with only one call on the function.
[Note: I wrote a cubic cumulative integration function for the File Exchange called 'cumint3' which can be accessed at:
For a smooth integrand such as you have, 'cumint3' might well be more accurate than 'cumtrapz' for the same size steps.]
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Roger Stafford
el 8 de Abr. de 2016
Actually it's the other way around. The 'integral' function is more recent and has more capabilities than 'quad'. For example the limits of integration for 'integral' may be infinite, whereas that is not permitted for 'quad'. Read about 'integral' at:
http://www.mathworks.com/help/matlab/ref/integral.html
Más respuestas (2)
Roger Stafford
el 7 de Abr. de 2016
Use numerical integration with matlab's 'integral' function or one of the other quadrature functions.
1 comentario
Walter Roberson
el 7 de Abr. de 2016
MuPad does not know how to do these integrals.
I find, though, a table of integrals that is relevant: http://www.eah-jena.de/~rsh/Forschung/Stoer/besint.pdf . If you look on page 391 of the pdf, the indefinite int(x*J[sub 0](x)^4) integral is given as x*J[sub 0](x)^3 + 3 * X[sub 1][super (22)](x) where earlier on that page it shows X[sub 1][super (22)](x) = int( x * J[sub 0](x)^2 * J[sub 1](x)^2
That X[sub 1][super (22)](x) seems to be listed as a basic integral. I do not see any applicable formula that would allow that to be further broken down, but I might have overlooked it.
2 comentarios
Walter Roberson
el 7 de Abr. de 2016
Not all integrals have closed forms. (There is a body of theory that shows that there are integrals that cannot have closed forms over a body of standard operations. The proof is well beyond my skills, though.)
And sometimes it happens that the expanded terms cancel or combine in interesting ways that make the expansion worth while.
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