What is the indefinite integral of Bessel function of 1 order (first k

In summary, the indefinite integral of the Bessel function of first order is given by ∫J₁(x)dx = xJ₀(x) + C, where J₀(x) is the Bessel function of zeroth order and C is the constant of integration. This integral is used to calculate the area under the curve of the Bessel function, and can also be used in solving differential equations involving Bessel functions. It is an essential tool in the study of special functions and has many applications in physics, engineering, and other fields.
  • #1
AhmedHesham
96
11
Hi
When we find integrals of Bessel function we use recurrence relations.
But this requires that we have the variable X raised to some power and multiplied with the function .
But how about when we have Bessel function of first order and without multiplication?
How should we integrate it ?
 
Physics news on Phys.org
  • #2
It is easier if you write out explicitly in correct notation what you are asking about. It seems to me you are asking about what the primitive function of ##J_n(x)## is. In the general case, like with many other functions, there is no closed form solution to this.
 
  • #3
Orodruin said:
It is easier if you write out explicitly in correct notation what you are asking about. It seems to me you are asking about what the primitive function of ##J_n(x)## is. In the general case, like with many other functions, there is no closed form solution to this.
The recurrence relations help writing the integrals of Bessel functions in terms of other Bessel functions of other orders.
 
  • #4
AhmedHesham said:
The recurrence relations help writing the integrals of Bessel functions in terms of other Bessel functions of other orders.
Again, these are special cases.
 
  • #5
AhmedHesham said:
But how about when we have Bessel function of first order and without multiplication?
How should we integrate it ?
Are you asking for ##\int J_1(x) \, dx##? If so, then it has a particularly simple answer. Do you know a function ##y## such that ##dy/dx = J_1(x)## ? Hint: ##y## is a Bessel function.

Anyway, in general for complicated integrals with special function the first place I look is
https://dlmf.nist.gov/
the section on Bessel function includes a fair number of integrals.
Jason
 
  • Like
Likes AhmedHesham
  • #6
jasonRF said:
Anyway, in general for complicated integrals with special function the first place I look is
https://dlmf.nist.gov/
the section on Bessel function includes a fair number of integrals.
Jason
Ok
Thanks
This really helps
I already solved it
 

Similar threads

Back
Top