Evaluate Integral of Bessel K Function

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SUMMARY

The integral of the form ∫x^(2/3) * cos(2ax) * [Kv(x)]^2 dx, where Kv(x) is the modified Bessel function of the second kind of order v=1/3, evaluates to a hypergeometric function 2F1. Using Mathematica, the result is given by (π² * Hypergeometric2F1[5/6, 7/6, 4/3, -a²])/(4 * Gamma[1/3]) under the condition that |Im[a]| ≤ 1. This confirms the expected form of the result, which should be verified through multiple methods for accuracy.

PREREQUISITES
  • Understanding of modified Bessel functions, specifically BesselK
  • Familiarity with hypergeometric functions, particularly Hypergeometric2F1
  • Knowledge of integral calculus, especially improper integrals
  • Experience with Mathematica for symbolic computation
NEXT STEPS
  • Explore the properties and applications of modified Bessel functions in mathematical physics
  • Learn about the derivation and applications of hypergeometric functions
  • Investigate advanced integral techniques in Mathematica, focusing on conditional expressions
  • Review the Gamma function and its role in complex analysis and integrals
USEFUL FOR

Mathematicians, physicists, and engineers working with special functions, particularly those involved in integral calculus and mathematical modeling using Bessel functions and hypergeometric functions.

kschau
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Hey All

Got a tough one and I'm just not seeing the path here. I need to find the close form expression of:

The integral from zero to infinity:

∫xλ * cos(2ax) * [Kv(x)]2 dx

where Kv(x) is the modified Bessel function of the second kind of order v and argument x. If it helps, the value of v=1/3 and the value of λ=2/3

The result will have the form of a hypergeometric function 2F1

I've just been racking my brain for too long with this one. If anyone has some experience with Bessel functions, any help would be appreciated.
 
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Have you tried plugging it into Mathematica ?
 
In[1]:= Integrate[x^(2/3)*Cos[2 a x]*BesselK[1/3, x]^2, {x, 0, Infinity}]

Out[1]= ConditionalExpression[(Pi^2*Hypergeometric2F1[5/6, 7/6, 4/3, -a^2])/(4*Gamma[1/3]), Abs[Im[a]] <= 1]

so if the absolute value of the imaginary component of a is less than or equal to 1 then you have your hypergeometric as expected.

Verify this independently several different ways before you depend on it.

http://reference.wolfram.com/mathematica/ref/Hypergeometric2F1.html
http://reference.wolfram.com/mathematica/ref/Gamma.html
http://reference.wolfram.com/mathematica/ref/ConditionalExpression.html
 
Last edited:

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