Need help in solving an integral

[Schulerbible]

Dear Physics Forums users,

My name is Tobias and I'm actually stucking in solving an integral with modified Bessel function of zeroth order. I haven't had that much math in my life to solve this horrible integral. This problem is concerning my private interest to build up a model predicting some weired behaviour in optical fibre cables. Basically function P has to be insert in F(θi,θ) and must be solved by this integral. I just want to know the outcome of the integral. Inserting the limits should be less problematic.

P(θi,θ) = 1/(bL) * exp(-(θ^2+θi^2)/(4DL))* Io ((θ*θi)/(2DL))

b = constant
L = length
D = constant
θ = output angle
θi = incident angle

F(θi,θ) = 2*pi ∫ P(θi,θ)*sin(θi) ∂θi

The integral limits are 0 → Δθ

Thanks a lot.

cheers Tobias

 

[jvicens]

Dear Tobias,

Thank you for reaching out to Physics Forums for help with your integral problem. The modified Bessel function of zeroth order can be quite tricky to work with, but with some careful manipulation and use of known identities, it can be solved.

First, let's rewrite the integral as follows:

F(θi,θ) = 2*pi ∫ 1/(bL) * exp(-(θ^2+θi^2)/(4DL))* Io ((θ*θi)/(2DL))*sin(θi) ∂θi

= 2*pi/(bL) * ∫ exp(-(θ^2+θi^2)/(4DL))* Io ((θ*θi)/(2DL))*sin(θi) ∂θi

Next, we can use the identity Io(x) = (1/π) ∫ cos(x*cos(t))*dt to rewrite the modified Bessel function within the integral:

F(θi,θ) = 2*pi/(bL*π) * ∫ exp(-(θ^2+θi^2)/(4DL))* ∫ cos((θ*θi)*cos(t)/(2DL))*sin(θi) ∂θi ∂t

= 2*pi/(bL*π) * ∫ ∫ exp(-(θ^2+θi^2)/(4DL))*cos((θ*θi)*cos(t)/(2DL))*sin(θi) ∂θi ∂t

Using the product-to-sum identity cos(A)*sin(B) = (1/2)*(sin(A+B)-sin(A-B)), we can further simplify the integral:

F(θi,θ) = pi/(bL*π) * ∫ ∫ exp(-(θ^2+θi^2)/(4DL))*[sin((θ+θi)*cos(t)/(2DL))-sin((θ-θi)*cos(t)/(2DL))] ∂θi ∂t

= pi/(bL) * ∫ [exp(-θ^2/(4DL))*sin(θ*cos(t)/(2DL))] ∫ [exp(-θi^2/(4DL))*sin(θi*cos(t)/(2DL))] ∂θi ∂t - pi/(bL) * ∫ [exp(-θ^2/(4DL))*sin(θ*cos(t)/(2DL