How prove this limit $\lim_{\alpha\to n}\frac{J_{\alpha}(x)\cos{(\alpha \pi)}-J_{-\alpha}(x)}{\sin{\alpha\pi}}$? Let $$J_{\alpha}(x)=\sum_{m=0}^{\infty}\dfrac{(-1)^m}{m!\Gamma{(m+\alpha+1)}}\left(\dfrac{x}{2}\right)^{2m+\alpha}$$
show that：
\begin{align*}&\lim_{\alpha\to n}\dfrac{J_{\alpha}(x)\cos{(\alpha \pi)}-J_{-\alpha}(x)}{\sin{\alpha\pi}}\\
&=\dfrac{2}{\pi}J_{n}(x)\left(\ln{\dfrac{x}{2}}+\gamma\right)-\dfrac{1}{\pi}\sum_{m=0}^{n-1}\dfrac{(n-m-1)!}{m!}\left(\dfrac{x}{2}\right)^{2m-n}-\dfrac{1}{\pi}\sum_{m=0}^{\infty}\dfrac{(-1)^m\left(\dfrac{x}{2}\right)^{n+2m}}{m!(n+m)!}\left(\sum_{k=0}^{n+m-1}\dfrac{1}{k+1}+\sum_{k=0}^{m-1}\dfrac{1}{k+1}\right)
\end{align*}
where $$\gamma=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{1}{n}-\ln{n}\right)=0.5772\cdots$$ call Eluer constant.
and where $J_{\alpha}(x)$ is Bessel function:http://en.wikipedia.org/wiki/Bessel_function
This problem is from a book,My book say this proof  use L'Hôpital's rule and it is very very  trouble,so we can't post all solution, only post reslut,so I hope see solution,Thank you someone can help
 A: Let $J_{\nu}(z)$ the Bessel function of first kind defined as
$$
J_\nu(z) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\nu+1)} {\left(\frac{z}{2}\right)}^{2m+\nu} \tag 1
$$
and $Y_{\nu}(z)$ the Bessel function of second kind (or Weber function or Neumann function) defined as
$$
Y_{\nu}(z)=\frac{J_{\nu}(z)\cos(\nu\pi)-J_{-\nu}(z)}{\sin(\nu\pi)} \tag 2
$$
or the limit for integral $\nu$, $\nu=n$. For integral $\nu$, using the l'Hopital's rule we have
$$\begin{align}
Y_{n}(z)
&=\lim_{\nu\to n}\frac{J_{\nu}(z)\cos(\nu\pi)-J_{-\nu}(z)}{\sin(\nu\pi)}\\
&=\lim_{\nu\to n}\frac{\frac{\partial}{\partial \nu}\left(J_{\nu}(z)\cos(\nu\pi)-J_{-\nu}(z)\right)}{\frac{\partial}{\partial \nu}\sin(\nu\pi)}\\
&=\lim_{\nu\to n}\frac{\frac{\partial J_{\nu}(z)}{\partial \nu}\cos(\nu\pi)-\pi J_{\nu}(z)\sin(\nu\pi)-\frac{\partial J_{-\nu}(z)}{\partial \nu}}{\pi\cos(\nu\pi)}\\
&=\frac{1}{\pi}\left[\left.\frac{\partial J_{\nu}(z)}{\partial \nu}\right|_{\nu=n}-(-1)^n\left.\frac{\partial J_{-\nu}(z)}{\partial \nu}\right|_{\nu=n}\right]\tag 3
\end{align}
$$
where we used $\cos(n\pi)=(-1)^n$ and $\sin(n\pi)=0$.
We shall now obtain Hankel's expansion of function $Y_n(z)$, where $n$ is any positive integer. It is clear that
$$
\begin{align}
\frac{\partial J_{\nu}(z)}{\partial \nu}
&= \sum_{m=0}^\infty \frac{(-1)^m}{m!}\frac{\partial }{\partial \nu}\left\{\frac{1}{\Gamma(m+\nu+1)} {\left(\frac{z}{2}\right)}^{2m+\nu} \right\}\\
&= \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\nu+1)} {\left(\frac{z}{2}\right)}^{2m+\nu} \left\{\log\left(\tfrac{z}{2}\right)-\psi(\nu+m+1)\right\}\tag 4
\end{align}
$$
observing that 
$$
\frac{\partial }{\partial \nu} {\left(\frac{z}{2}\right)}^{2m+\nu}={\left(\frac{z}{2}\right)}^{2m+\nu}\log\left(\tfrac{z}{2}\right)
$$
and
$$
\frac{\partial }{\partial \nu}\frac{1}{\Gamma(m+\nu+1)}=-\frac{\frac{\partial }{\partial \nu}\Gamma(m+\nu+1)}{[\Gamma(m+\nu+1)]^2}=-\frac{\psi(m+\nu+1)}{\Gamma(m+\nu+1)}
$$
and $\psi(z)=\frac{\operatorname{d}\log \Gamma(z)}{\operatorname{d}z}=\frac{\Gamma'(z)}{\Gamma(z)}$ is the digamma function.
For $\nu\to n$ the (4) becomes
$$
\begin{align}
\left.\frac{\partial J_{\nu}(z)}{\partial \nu}\right|_{\nu=n}
&= \sum_{m=0}^\infty \frac{(-1)^m}{m! \, (m+n)!} {\left(\frac{z}{2}\right)}^{2m+n} \left\{\log\left(\frac{z}{2}\right)-\psi(n+m+1)\right\}. \tag 5
\end{align}
$$
The evaluation of $\left.\frac{\partial J_{-\nu}(z)}{\partial \nu}\right|_{\nu=n}$ is a little more tedious because of the pole of $\psi(-\nu+m+1)$ at $\nu=n$ in the terms of which $m=0,1, \ldots,n-1$. We brack the series for $J_{-\nu}(z)$ into two parts, thus 
$$
J_{-\nu}(z) =  \sum_{m=0}^{n-1} \frac{(-1)^m}{m! \, \Gamma(m-\nu+1)} {\left(\frac{z}{2}\right)}^{2m-\nu}+\sum_{m=n}^{\infty} \frac{(-1)^m}{m! \, \Gamma(m-\nu+1)} {\left(\frac{z}{2}\right)}^{2m-\nu}
$$
and in the former part we replace 
$$
\frac{1}{\Gamma(m-\nu+1)}=\frac{\Gamma(\nu-m)\sin((\nu-m)\pi)}{\pi}
$$
using the reflection formula for the Gamma function $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$.
Now, when $0\le m\le n$,
$$
\begin{align}
&
\left.\frac{\partial }{\partial \nu} \left\{\frac{\Gamma(\nu-m)\sin((\nu-m)\pi)}{\pi}{\left(\frac{z}{2}\right)}^{2m-\nu}\right\} \right|_{\nu=n}\\
&\quad=
{\left(\tfrac{z}{2}\right)}^{2m-\nu}\Gamma(\nu-m)\frac{1}{\pi}\times\\
&\qquad\times\left.
\left\{
\psi(\nu-m)\sin((\nu-m)\pi)+
\pi\cos((\nu-m)\pi)-\log\left(\tfrac{z}{2}\right)\sin((\nu-m)\pi)
\right\}\right|_{\nu=n}\\
&\quad=
{\left(\tfrac{z}{2}\right)}^{2m-n}\Gamma(n-m)\cos((n-m)\pi).
\end{align}
$$
Hence
$$
\begin{align}
\left.\frac{\partial J_{-\nu}(z)}{\partial \nu}\right|_{\nu=n}
&= \sum_{m=0}^{n-1} \frac{(-1)^n\, \Gamma(m-n)}{m!} {\left(\frac{z}{2}\right)}^{2m-n}+\\
&\quad+\sum_{m=n}^{\infty} \frac{(-1)^m}{m!(m-n)!} {\left(\frac{z}{2}\right)}^{2m-n}\left\{-\log\left(\frac{z}{2}\right)+\psi(-n+m+1)\right\}
\end{align}
$$
that is
$$
\begin{align}
\left.\frac{\partial J_{-\nu}(z)}{\partial \nu}\right|_{\nu=n}
&= (-1)^n\, \sum_{m=0}^{n-1} \frac{(n-m-1)!}{m!} {\left(\frac{z}{2}\right)}^{2m-n}+\\
&\quad+(-1)^{n-1}\sum_{m=0}^{\infty} \frac{(-1)^m}{m!(m+n)!} {\left(\frac{z}{2}\right)}^{2m+n}\left\{\log\left(\frac{z}{2}\right)-\psi(m+1)\right\}
\end{align}\tag 6
$$
when we replace $m$ by $n+m$ in the second series. 
Combining (5) and (6), the (3) becomes
$$
\begin{align}
\pi Y_n(z)
&= -\sum_{m=0}^{n-1} \frac{(n-m-1)!}{m!} {\left(\frac{z}{2}\right)}^{2m-n}+\\
&\quad+\sum_{m=0}^{\infty} \frac{(-1)^m}{m!(m+n)!} {\left(\frac{z}{2}\right)}^{2m+n}\left\{2\log\left(\frac{z}{2}\right)-\psi(m+1)-\psi(n+m+1)\right\}\\
&= 2\left[\gamma+ \log\left(\tfrac{z}{2}\right)\right]J_n(z)
-\sum_{m=0}^{n-1} \frac{(n-m-1)!}{m!} {\left(\frac{z}{2}\right)}^{2m-n}+\\
&\quad-\sum_{m=0}^{\infty} \frac{(-1)^m}{m!(m+n)!} {\left(\frac{z}{2}\right)}^{2m+n}\left\{\sum_{k=1}^{m}\frac{1}{k}+
\sum_{k=1}^{n+m}\frac{1}{k}
\right\}\\
\end{align}
$$
using the relation $\psi(n+1)=\sum_{k=1}^n\frac{1}{k}-\gamma$ where $\gamma$ is the Euler’s Constant.
Thus we have
$$
Y_n(z)=\frac{2}{\pi}\left[\gamma+ \log\left(\tfrac{z}{2}\right)\right]J_n(z)
-\frac{1}{\pi}\sum_{m=0}^{n-1} \frac{(n-m-1)!}{m!} {\left(\frac{z}{2}\right)}^{2m-n}+\\
\qquad\qquad\quad-\frac{1}{\pi}\sum_{m=0}^{\infty} \frac{(-1)^m}{m!(m+n)!} {\left(\frac{z}{2}\right)}^{2m+n}\left\{\sum_{k=0}^{m-1}\frac{1}{k+1}+
\sum_{k=0}^{n+m-1}\frac{1}{k+1}
\right\}\tag 7
$$
known also as Hankel's formula.
A: Hint: This DLMF link may help to start if you complement it with this expression for the derivative and an expansion of the digamma function.
ADDITION:
If we suppose $x$ out of $(-\infty,0]$ then l'hospital rules (applied to an entire function of $\alpha$) returns $$\lim_{\alpha\to n}\dfrac{J_{\alpha}(x)\cos{(\alpha \pi)}-J_{-\alpha}(x)}{\sin{\alpha\pi}}=\lim_{\alpha\to n}\;\frac{\frac{\partial{J_{\alpha}(x)}}{\partial\,\alpha}\cos{(\alpha \pi)}-\pi J_{\alpha}(x)\sin{(\alpha \pi)}-\frac{\partial{J_{-\alpha}(x)}}{\partial\,\alpha}}{\pi\,\cos(\alpha\,\pi)}=\frac 1{\pi}\left.\frac{\partial{J_{\alpha}(x)}}{\partial\,\alpha}\right|_{\alpha=n}+\left.\frac{(-1)^n}{\pi}\frac{\partial{J_{-\alpha}(x)}}{\partial\,(-\alpha)}\right|_{\alpha=n}$$
i.e. the first link.  
After that you may compute the derivative of the series for $J$ using $\;\displaystyle\frac d{dx}\frac 1{\Gamma(x+a)}=-\frac{\psi(x+a)}{\Gamma(x+a)}\;$ to obtain the second result and so on...
A: Since @china math "Looking for an answer drawing from credible and/or official sources" I give it.
B.G. Korenev, Bessel Functions and their Applications (Taylor & Francis, 2002) page 10, (2.1), (2.2) and (2.3). I quote "It is sufficient to obtain this formula for the special case $n=0$ (then using the recurrence relations considered below in Section 6, one can obtain the result for any $n$)." The proof for $n=0$ is in page 11. The recurrence relations can be found also on Recurrence Relations and Derivatives.
