How prove this $f(x,y)=\frac{1}{2\pi}\ln{\frac{1}{|x-y|}}+\frac{i}{4}-\frac{1}{2\pi}\ln{\frac{k}{2}}-\frac{C}{2\pi}$?

Question

let $$J_{0}(x)=\sum_{p=0}^{\infty}\dfrac{(-1)^p}{(p!)^2}\left(\dfrac{x}{2}\right)^{2p}$$

and $$Y_{0}(x)=\dfrac{2}{\pi}J_{0}(x)\left(\ln{\dfrac{x}{2}}+C\right)-\dfrac{2}{\pi}\sum_{p=0}^{\infty}\dfrac{(-1)^p(x/2)^{2p}}{(p!)^2}\left(\sum_{k=1}^{p}\dfrac{1}{k}\right)$$ and $$H^{(1)}_{n}=J_{n}+iY_{n}$$ where $i=\sqrt{-1}$ define $$f(x,y)=\dfrac{i}{4}H^{(1)}_{0}(k|x-y|),x\neq y$$

show that:

$$f(x,y)=\dfrac{1}{2\pi}\ln{\dfrac{1}{|x-y|}}+\dfrac{i}{4}-\dfrac{1}{2\pi}\ln{\dfrac{k}{2}}-\dfrac{C}{2\pi}+O\left(|x-y|^2\ln{\dfrac{1}{|x-y|}}\right)$$ For $|x-y|\to 0$,where $C$ is Euler's constant,

My try: since $$J_{0}(x)=\sum_{p=0}^{\infty}\dfrac{(-1)^p}{(p!)^2}\left(\dfrac{x}{2}\right)^{2p}$$ $$Y_{0}(x)=\dfrac{2}{\pi}J_{0}(x)\left(\ln{\dfrac{x}{2}}+C\right)-\dfrac{2}{\pi}\sum_{p=0}^{\infty}\dfrac{(-1)^p(x/2)^{2p}}{(p!)^2}\left(\sum_{k=1}^{p}\dfrac{1}{k}\right)$$ then \begin{align*} &H^{(1)}_{0}(x)=J_{0}(x)+iY_{0}(x)\\ &=\sum_{p=0}^{\infty}\dfrac{(-1)^p}{(p!)^2}(\dfrac{x}{2})^{2p}\left(\dfrac{2i}{\pi}\ln{\dfrac{x}{2}}+1\right)+\dfrac{2i}{\pi}\left(C\sum_{p=0}^{\infty}\dfrac{(-1)^p}{(p!)^2}(\dfrac{x}{2})^{2p}-\sum_{p=0}^{\infty}\left(\dfrac{(-1)^p}{(p!)^2}(\dfrac{x}{2})^{2p}\sum_{k=1}^{p}\dfrac{1}{k}\right)\right) \end{align*}

then I find this sum,Thank you

Answer 1

(Lemma) $H_0^{(1)}(x) = 1+ \frac{2i}{\pi}\left(\ln\frac{x}{2}+C\right)+O(x^2\ln{|x|})$.

The proof of this is just collecting constant terms and natural logarithms, everything else can be consumed in the error term.

The rest of the argument is just multiplying $i/4$ and plugging in $k|x-y|$ in the above lemma.