I would like to resolve this integral numerically . However, I'm not sure about the best way to do it because it is an improper integral:
$$ \Phi(\rho)=\frac{1}{2\pi}\int_0^\infty\phi_1(s)sJ_0(s\rho)\mathrm{d}s$$ $$ \phi_1(s)=\frac{1}{D_1\alpha_1}\left[\sinh(\alpha_1z_0)\cdot\frac{D_1\alpha_1\cosh(\alpha_1l)+D_2\alpha_2\sinh(\alpha_1l)}{D_1\alpha_1\cosh(\alpha_1(l+z_b))+D_2\alpha_2\sinh(\alpha_1(l+z_b))}-\sinh(\alpha_1z_0)\right], $$ $$ \alpha_i=\sqrt{\frac{D_is^2+\mu a}{D_i}}. $$
I have studied the function $\phi_1$.
The function $\phi_1$ converges when s approaches infinity.
I have already simplified $\phi_1$'s expression with Taylor's formula. $$\phi_1(s) = \frac{1}{2 D_1 \alpha_1 }\cdot [(e^{\alpha_1 (z_b+z_0)}-e^{-\alpha_1 (z_b+z_0)})\times \frac{(1+k)e^{\alpha_1 l} +(1-k)e^{-\alpha_1 l}}{(1+k)e^{\alpha_1 (l+z_b)} +(1-k)e^{-\alpha_1 (l+z_b)}}]$$
with $k = \frac{\alpha_2 D_2}{\alpha_1 D_1}$ Then, $$\phi_1 (s)= \frac{1}{2 D_1 \alpha_1} \cdot [(1+\alpha_1(z_b+z_0) - 1 + \alpha_1(z_b+z_0))\times \frac{(1+\alpha_1l)(1+k)+(1-\alpha_1l)(1-k)}{(1+\alpha_1(l+z_b))(1+k)+(1-\alpha_1(l+z_b))(1-k)}]$$ $$\lim_{\alpha_1\to 0} \frac{z_b + z_0}{D_1}$$
I have also tried the following variable substitution: $$ u=s\rho, $$ $$ \frac{1}{2\pi}\int_0^x\phi_1 \left(\frac{u}{\rho}\right)\frac{u}{\rho}J_0(u)\frac{\mathrm{d}u}{\rho}. $$
I would appreciate some advice how to proceed.
Thank you.