1. The problem statement, all variables and given/known data By integrating (2-55), over a small volume containing the origin, substituting ψ = Ce-jβr/r, and letting r approach zero, show that C = 1/4π, thus proving (2-58). 2. Relevant equations (2-55): ∇2ψ + β2ψ = -δ(x)δ(y)δ(z) (2-58): ψ = e-jβr/(4πr) 3. The attempt at a solution I am SO lost on this. I integrated the RHS first. I would assume I have to do this in spherical coordinates. I know from the internet that to convert the delta function between coordinate systems you need to incorporate the jacobian, but because it's at the origin (i.e., x,y,z = 0) I just assumed that: ∫∫∫(-δ(x)δ(y)δ(z))r2sin(θ)dr dθ d∅ = -∫∫∫r2sin(θ)dr dθ d∅ = -(4π/3)r03 Where I've integrated from r = 0 to r0, θ from 0 to pi, and ∅ from 0 to 2pi. Is this even correct? For the next part, the LHS, I'm totally lost. I'm trying to follow the instructions chronologically, thus ∫∫∫(∇2 + β2)ψ r2sin(θ)dr dθ d∅ Where I'm using the same integration limits as before. I'm really, really unsure how to even do this, but I assumed that psi is regarded as scalar (or otherwise does not have dependency on r, theta or phi). Thus, starting with the r component: ∫∫∫(1/r2)(∂/∂r)[r2(∂/∂r)] (ψ r2sin(θ)dr dθ d∅) = ∫∫∫(1/r2)(∂/∂r)[r2 (ψ 2r sin(θ)dr dθ d∅)] = ∫∫∫(1/r2)(∂/∂r)[2r3 (ψ sin(θ)dr dθ d∅)] = ∫∫∫(1/r2)[6r2 (ψ sin(θ)dr dθ d∅)] = ∫∫∫[6 (ψ sin(θ)dr dθ d∅)] = 4ψ∏r0 For the next part (θ component): ∫∫∫(1/[r2 sin(θ)])(∂/∂θ)[sin(θ)(∂/∂θ)] (ψ r2sin(θ)dr dθ d∅) =∫∫∫(1/[r2 sin(θ)])(∂/∂θ)[sin(θ)cos(θ)] (ψ r2dr dθ d∅) =∫∫∫(1/[r2 sin(θ)])(∂/∂θ)[(1/2)sin(2θ)] (ψ r2dr dθ d∅) =∫∫∫(1/[r2 sin(θ)])[cos(2θ)] (ψ r2dr dθ d∅) =∫∫∫(cos(2θ)/ sin(θ)]) (ψ dr dθ d∅) But I can't integrate this because (cos(2θ)/ sin(θ)]) doesn't converge. I'm really lost on where to go from here and don't fully understand what I'm being asked to do. ANY help is appreciated. Thank you. Please note that I'm an electrical eng so our notation may be different, and I may have some difficulty with more of the hardcore physics notation. I have posted this on the intro physics board but nothing doing, so I'm not sure if maybe this is more advanced. If it isn't, my apologies.