I am wondering, if it is possible to find the limit of


with $\mu_1\,k_1^2=\mu_2\,k_2^2$ and $k_1=\sqrt{k_1^2}$ and $k_2=\sqrt{k_2^2}$ and $\mathrm{Im}[k_1]<0$ and $\mathrm{Im}[k_2]>0$.

I am pretty sure that there should be a limit different from $0$. (From a physical point of view)

I tried the limiting form of the modified Bessel functions see here $$\mathrm{K}_0(z)\approx -\ln(z)$$

but I did not get any useful results. It needs to be mentioned, that $k_1$ and $k_2$ are not idenpendent from each other. So I am also happy with a solution having a restriction on $k_1$ and $k_2$.


I would have the following idea. Replacing $\mathrm{K}_0(z)$ with $-\ln(z)$ for small $z$ leads to


Using some logarithmic identities leads to

$$\begin{align}\lim_{r\to0}\left(\ln\left((\mathrm{j}\,k_1\,r)^{-\frac{1}{\mu_1-1}}\right)+\ln\left((-\mathrm{j}\,k_2\,r)^{\frac{1}{\mu_2-1}}\right)\right)=\\ \lim_{r\to0}\ln\left((\mathrm{j}\,k_1\,r)^{-\frac{1}{\mu_1-1}}\,(-\mathrm{j}\,k_2\,r)^{\frac{1}{\mu_2-1}}\right)=\\ \lim_{r\to0}\ln\left((\mathrm{j}\,k_1)^{-\frac{1}{\mu_1-1}}\,(-\mathrm{j}\,k_2)^{\frac{1}{\mu_2-1}}\,r^{-\frac{1}{\mu_1-1}+\frac{1}{\mu_2-1}}\right)=\\ \ln\left((\mathrm{j}\,k_1)^{-\frac{1}{\mu_1-1}}\right)+\ln\left((-\mathrm{j}\,k_2)^{\frac{1}{\mu_2-1}}\right)+\lim_{r\to0}\ln\left(r^{-\frac{1}{\mu_1-1}+\frac{1}{\mu_2-1}}\right) \end{align}$$

Therefore, it seems to me, that a limit only exists if


which results in $\mu_1=\mu_2$. Are there any errors in this approach? Because I accutally hoped for a different result.

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