Prove that $\lim_{x\to-\infty}\left(\psi(x)-\psi\left(\frac x2\right)-\frac1x-\ln 2\right)\sin(\pi x)=\pi$

Question

I was messing around on Desmos when I saw this interesting limit: $$\lim_{x\to-\infty}\left(\psi(x)-\psi\left(\frac x2\right)-\frac1x-\ln 2\right)\sin(\pi x)=\pi$$This is of course what I think is the limit, but it seems to be true. Here $\psi$ denotes the digamma function. I do know that $$\lim_{x\rightarrow0}\left(\psi(x)-\psi\left(\frac x2\right)-\frac1x\right)=0$$But I don't know how this can help. Also, removing the $\ln2$ term and replacing it with some constant $a\ne\ln2$ makes it [seem] non-convergent.

Answer 1

We can use the reflection formula $$\psi(x) = \psi(1-x) - \pi \cot(\pi x), $$

along with the asymptotic expansion $$\psi(x) = \ln(x) - \frac{1}{2x} +O\left( \frac{1}{x^{2}}\right)$$ as $x \to + \infty. $

$ \begin{align} &\lim_{x \to - \infty} \left(\psi(x) - \psi \left(\frac{x}{2} \right)- \frac{1}{x} - \ln(2) \right)\sin(\pi x) \\ &= \lim_{x \to \infty} \left( \psi(-x) - \psi \left(-\frac{x}{2} \right)+ \frac{1}{x} - \ln(2)\right) \sin (- \pi x) \\ &= -\lim_{x \to \infty} \left(\psi(1+x) + \pi \cot(\pi x)- \psi\left(1+ \frac{x}{2} \right) - \pi \cot\left(\frac{\pi x}{2} \right) + \frac{1}{x}- \ln(2)\right) \sin(\pi x) \\ &\overset{\clubsuit}{=} \small -\lim_{x \to \infty} \left(\ln(x) + \frac{1}{2x} + O(x^{-2})+ \pi \cot(\pi x)- \ln \left(\frac{x}{2} \right) - \frac{1}{x} +O(x^{-2}) - \pi \cot \left(\frac{\pi x}{2} \right) + \frac{1}{x} - \ln(2) \right) \sin(\pi x) \\&= -\lim_{x \to \infty} \left(\frac{1}{2x} + \pi \cot(\pi x)- \pi \cot \left(\frac{\pi x}{2} \right) +O(x^{-2}) \right) \sin(\pi x) \\ &= -\lim_{x \to \infty} \left(\pi \cot(\pi x)- \pi \cot \left(\frac{\pi x}{2} \right) \right) \sin(\pi x) \\&=- \lim_{x \to \infty} \left(\frac{\pi \cos(\pi x)}{\sin(\pi x)} - \pi \, \frac{1+ \cos(\pi x)}{\sin(\pi x )} \right) \sin(\pi x) \\ &= - \lim_{x \to \infty} \left(- \frac{\pi}{\sin(\pi x)}\right) \sin(\pi x)\\ &= \pi \end{align}$

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$\clubsuit$ As $x \to +\infty$, $$\begin{align} \psi(1+x) &\sim \ln(1+x) - \frac{1}{2(1+x)} \\ &= \ln(x) + \ln \left(1+ \frac{1}{x} \right) - \frac{1}{2x} \frac{1}{1+ \frac{1}{x}} \\ &= \ln(x)+ \frac{1}{x} - \frac{1}{2x}+O \left(\frac{1}{x^{2}} \right) \\ &= \ln(x) + \frac{1}{2x} + O \left(\frac{1}{x^{2}} \right). \end{align}$$

UPDATE:

As Gary pointed out in the comments, the asymptotic expansion for $\psi(1+x)$ can be obtained more easily by using the recurrence relation to write $\psi(1+x)$ as $\psi(x) + \frac{1}{x}$.