I am studying Bessel function and found the good reference by G.N. Watson
At some point in page 58 he introduces the following expression due to Hankel:
\begin{eqnarray} \lim_{\nu \to n} \frac{J_{\nu}(x) - (-1)^{n} J_{-\nu}(x)}{\nu-n} \end{eqnarray}
This made perfect sense since $J_{\nu}$ and $J_{-\nu}$ are not linearly independent when $\nu = n \in \mathbb{Z}$. Then the author goes into great detail studying this equation. I was convinced that he would get to the equation:
\begin{eqnarray*} Y_{\nu}(x) = \frac{ \cos \nu \pi \; J_{\nu}(x) - J_{-\nu}(x)}{\sin \nu \pi}. \end{eqnarray*}
but.....no. He decided to introduce this equation as a new approach from Hankel with no motivation and let the other development there hanging on the air.
Any one can help me understand what could have motivated Hankel to introduce this equation?
I know it is good and satisfies the requirements....but....
Thanks.
Update:
It is easy to see (shown in Watson's book) that the first equation here implies
\begin{equation} \lim_{\nu \to n } \left [ \frac{\partial J_{\nu}}{\partial \nu} - (-1)^n \frac{\partial J_{-\nu}}{\partial \nu} \right ] \quad \quad (1). \end{equation}
On the other hand by applying L'Hôpital's rule we find from the second equation above the following chain:
(I avoid the argument $x$ on the functions to simplify notations)
\begin{eqnarray*} Y_{\nu}(x) &=& \frac{ \cos \nu \pi \; J_{\nu}(x) - J_{-\nu}(x)}{\sin \nu \pi} \\ &=& \lim_{\nu \to n} \frac{- \pi \sin \nu \pi J_{\nu} + \cos \nu \pi \frac{\partial}{\partial \nu} J_{\nu} - \frac{\partial}{\partial\nu}{J_{-\nu}} }{\pi \cos \nu \pi} \\ &=& \lim_{\nu \to n } \frac{1}{\pi} \left [ \frac{\partial J_{\nu}}{\partial \nu} - (-1)^n \frac{\partial J_{-\nu}}{\partial \nu} \right ] \end{eqnarray*}
So, up to a factor of $1/\pi$, we are back to equation (1).
I have no problem accepting equation (1) as a motivated linearly independent function of $J_{\nu}(x)$.
My question specifically is how to go back from equation (1) to the representation $Y_{\nu}(x)$. It is something like "inverse" L'Hôpital rule. Is there a way to do this? According to Watson, the formula above for $Y_{\nu}(x)$ was given by Weber after a small modification of a similar formula given by Hankel (which did not work for half integer numbers). Watson says that Weber got this formula as a limit from an integral representation. ??
Here is a possible argument but I would like to see a better one.
If $\nu=n\in\mathbb{Z}$ we could show that $J_{\nu}(x) = (-1)^n J_{-\nu}(x)$.
We know that $J_{\nu}(x)$ and $J_{-\nu}(x)$ are solutions, of the Bessel equation so any linear combination is a solution as well. We consider the difference:
\begin{eqnarray*} J_{-\nu}(x) - (-1)^{\nu} J_{\nu}(x) = 0. \end{eqnarray*} Since both, $J_{\nu}$, and $J_{-\nu}$ are solutions of the Bessel equation, their difference is solution as well. Of course this difference is not interesting since for $\nu=n\in \mathbb{Z}$ the difference is the trivial 0. Hankel's idea was to form a quotient so that the function is ``regularized''. That is Hankel thought about the function
\begin{eqnarray} \label{tolimit} \frac{J_{\nu}(x) - (-1)^{n} J_{-\nu}(x)}{\nu-n}. \end{eqnarray} which looks undefined but if, with the use of L'Hôpital rule rule, we can show that the limit as $\nu \to n$ exists and is well defined, we can adopt this new function as a linearly independent solution which works even for $\nu \in \mathbb{Z}$. Since for $\nu \ne n$, the above equation is a solution for the Bessel equation we could think that the limit as $\nu \to n$ is a solution as well.
We take the limit in the previous equation as $\nu$ approaches $n$. That is,
\begin{eqnarray*} \lim_{\nu \to n} \frac{J_{\nu}(x) - (-1)^{n} J_{-\nu}(x)}{\nu-n} &=&\lim_{\nu \to n} \frac{J_{\nu}(x) - J_{\nu}(x) + J_{\nu}(x)- (-1)^{n} J_{-\nu}(x)}{\nu-n} \\ &=& \lim_{\nu \to n} \frac{J_{\nu}(x) - J_{\nu}(x) - (-1)^n[ J_{-\nu}(x) - J_{-\nu}(x)]} {\nu-n} \\ &=& \left . \left [ \frac{\partial J_{\nu}(x)}{\partial \nu} - (-1)^n \frac{\partial J_{-\nu}(x)}{\partial \nu} \right ] \right |_{\nu=n}. \end{eqnarray*}
Hankel noted this function as
\begin{eqnarray} Y_{n}(x) = \left . \left [ \frac{\partial J_{\nu}(x)}{\partial \nu} - (-1)^n \frac{\partial J_{-\nu}(x)}{\partial \nu} \right ] \right |_{\nu=n}. \quad \quad (A) \end{eqnarray}
It can be shown (see Watson's book) that this function is a solution of Bessel equation.
Now for the motivation of the definition of the Bessel function of second kind.
let us go back to the linear combination
\begin{eqnarray} J_{\nu}(x) - (-1)^{\nu} J_{-\nu}(x). \label{jnux2} \end{eqnarray} We know that each of the two terms $J_{\pm \nu}(x)$ is a solution of Bessel equation and they are linearly independent as long as $\nu \notin \mathbb{Z}$. In the limit as $\nu \to n \in \mathbb{Z}$ we have
\begin{eqnarray*} J_{n}(x) - (-1)^{n} J_{-n}(x) = 0. \end{eqnarray*} Of course we do not want this trivial solution so we must ``regularize'' it. First, we observe that $(-1)^n= \cos n \pi$ so we might rewrite the expression on the left as
\begin{eqnarray*} \cos \nu \pi J_{\nu}(x) - J_{-\nu}(x). \end{eqnarray*} Then, we need to divide the equation by something that goes to 0 as $\nu \to n$. Why $\sin \nu \pi$ and not just $\nu$? The reason would be clear soon.
If we pick to divide by $\sin \nu \pi$, and use L'ôHpital rule to evaluate
\begin{eqnarray*} \lim_{\nu \to n} \; \frac{\cos \nu \pi \, J_{\nu}(x) - J_{-\nu}(x)}{\sin k \nu} \end{eqnarray*} we will inmediately find equation (A), which is what Hankel called $Y_n(x)$.
Well....there is a factor of $\pi$ missing here, but ...