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I have two functions of the radius $r$ in the polar plane $(r,\theta)$ as

$$ f_{1}(kr)=H_{n}^{(1)}[kr]$$ $$ f_{2}(\gamma r)=K_{n}[\gamma r]$$ where the former is the Hankel function of the first kind, and the latter is the modified Bessel function of second kind. Here $k,\gamma$ are constants that can be assumed positive and real.

Both functions are candidate solutions to a physical problem. However, I need to check that these solutions are "physical" by checking if the an integral of the following type is finite

$$\int\limits_{r=a}^{\infty}\ f_{j}\ f_{j}^{' *}\ r\ dr = \ \text{finite?}$$ where the index $j$ is equal to either 1 or 2 (i.e. we work with either $f_{1}$ and its derivative's conjugate, or $f_{2}$ and its derivative's conjugate).

If the integral is finite, we accept it, otherwise we reject it as non-physical. Here $a>0$ to avoid a potential singularity at $r=0$, the prime indicated derivative with respect to argument, and $^{*}$ indicated complex conjugation.

When I try to evaluate this integral by taking the asymptotic approximation (e.g. see here and here) of $H_{n}^{(1)}$ and $K_{n}$, I see that they generally have expansions (written here up to two terms) like

$$ K_{n}(z) \sim \sqrt{\frac{\pi}{2 z}}e^{-z}\left( 1+ \frac{4n^{2}-1}{8z}\right) $$ $$ K'_{n}(z) \sim -\sqrt{\frac{\pi}{2 z}}e^{-z}\left( 1+ \frac{(4n^{2}-1)(4n^{2}+3)}{8z}\right) $$ $$ H^{(1)}_{n}(z) \sim \sqrt{\frac{\pi}{2 z}}e^{i(z-n\pi/2-\pi/4)}\left( 1+ \frac{i(4n^{2}-1)}{8z}\right) $$ $$ H^{(1) \prime}_{n}(z) \sim i\sqrt{\frac{\pi}{2 z}}e^{i(z-n\pi/2-\pi/4)}\left( 1+ \frac{i(4n^{2}-1)(4n^{2}+3)}{8z}\right) $$ where we can substitute $z$ later by $k r$ or $\gamma r$, as needed.

For large $z$, some approximations further drop the second term in the brackets, leaving only the factor 1 in the bracket. I tried to do the integration using the above expansions or the further simplified ones.

If I take $f_{2}$ (the modified Bessel) option, the integrand turns out to be basically $e^{-\gamma r}$ and gives a finite result. So, this solution seems to be an acceptable (physical) solution.

However, for the $f_{1}$ (the Hankel) option, the exponents cancel upon conjugation, and the integrand is basically (if I keep both two terms) $$ \frac{1}{2\pi z^{2}}-i\frac{2}{\pi z} $$ or (if I keep only the larger term) $$ -i\frac{2}{\pi z} $$ which, upon integration, seems to give $\infty$ and thus should be rejected (non-physical).

Questions:

(1) Is this calculation correct?

(2) Generally, when we have a small real term we can approximately neglect it compared to a larger real term that is added to it. However, can the same be done when one is real and the other is imaginary (i.e. wouldn't removing one of them alter the nature of the number fundamentally from complex to purely real or purely imaginary)? This question is related to the last two expressions above and which one should be used. I am assuming that we will still get $\infty$ in both cases. Does that make sense - am I missing something?

user135626
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  • I do not understand your problem. You mulytiply $f_1$ with $f'_2$ conjugate. First, $f'_2$ is real valued for $\gamma, r>0$. Second, it is exponentially decaying making the integral convergent. – Gary Jun 06 '21 at 19:32
  • @Gary Oh, I am very sorry, I had made a mistake in the post. I have now corrected/updated it. The integration should be over the same function (either $f_{1}$ or $f_{2}$) with its own derivative's conjugate. I hope this clarifies the question now. Yes, for the modified Bessel (when using $f_{2}$) function, I think I can see how it converges upon integration (as stated in the post). The other case is a bit more tricky, which is why I ask about it. I hope it makes more sense now. – user135626 Jun 07 '21 at 00:33
  • Then it seems correct to me. You can integrate the terms two terms coming from the of the Hankel asymptotics separately up to a given $R$. You will get something of the form $c \log R +d$ where $c$ and $d$ depend on $a$ but not on $R$. Letting $R\to +\infty$ yields the divergence. – Gary Jun 07 '21 at 05:19

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