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4

Every integral over a function behaving asymptotically (when $x$ goes to infinity) as $\frac{1}{x^\alpha}$ where $\alpha \leq1$ is divergent, it's a mathematical theorem which could be found in every reasonable handbook of calculus. Since Tanh[ π Sqrt[x]] goes to one rapidly we find that the integral is indeed divergent. We can demonstrate this fact with ...

3

You can do the whole thing analytically: (*ϵ=78.36; λ=0.548881;*) Clear[ϵ, λ]; funcin = 1/(ϵ * π) * Exp[-(2*r + λ * Sqrt[r^2 + ρ^2 - 2*r*ρ * Cos[θ]])] * r^2 * Sin[θ] * 1 / Sqrt[ρ^2 + r^2 - 2*r*ρ * Cos[θ]]; intfuncinang = Assuming[r > 0 && ρ > 0 && ϵ > 0 && λ > 0, Integrate[funcin, {θ, 0, π}, {ϕ, 0, 2 π}] ] (* ...

2

If you avoid the intermediate integration of Ef1, you can do it all at once: eqnBo = D[Eb1[r, t], t] - (Ef1[r, t]) * (((p[r, t])/(p[r, t] + kmn)) * ((kme)/(kme + p[r, t])) + (1 - (p[r, t])/(kmn + p[r, t]))*j); x = ParametricNDSolve[{eqnBo == 0, Eb1[r, 0] == 0, eqnDe == 0, Ef1[r, 0] == 0, Derivative[1, 0][Ef1][micron, t] == 0, Ef1[ro, ...

1

You can do the integrals over the angular variables analytically; the result will be a function of r, ρ : intth = 2 Pi Integrate[funcin, θ]; integ[ρ_] = (intth /. θ -> Pi) - (intth /. θ -> 0) ; output = Table[{ρ, NIntegrate[integ[ρ], {r, 0, Infinity}]}, {ρ, 0.1, 3,0.05}] ; ListLinePlot[output]

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