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.
$\begingroup$
$\endgroup$
4
-
1$\begingroup$ I think if you take $g(x)=\left(\psi(x)-\psi\left(\frac x2\right)-\frac1x-\ln 2\right)\sin(\pi x)$, you should be able to show that $h(x):=g(x-1/2)-\pi/2$ is an odd function. From there you can take $\lim_{x\to\infty}h(x)$ easily by using the well known asymptotics for $\psi(x)$. $\endgroup$– K.defaoiteCommented Feb 8 at 4:09
-
$\begingroup$ @K.defaoite then re-writing as $-h(x-1/2)+\pi/2=g(-x)$ we get the answer. Nice $\endgroup$– Kamal SalehCommented Feb 8 at 4:34
-
$\begingroup$ how did you even get the polygamma function into desmos lmao $\endgroup$– Max0815Commented Feb 8 at 6:56
-
1$\begingroup$ @Max0815 It accepts the factorial function so I wrote digamma is the logarithmic derivative $\endgroup$– Kamal SalehCommented Feb 8 at 13:06
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
2
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}$
$\clubsuit$ 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 $\psi(x) + \frac{1}{x}$.
answered Feb 8 at 6:53
-
$\begingroup$ If you use big-$O$, you do not need $\sim$. The asymptotics of $\psi(1+x)$ can be obtained quickly from that of $\psi(x)$ via $\psi(1+x)=\psi(x)+1/x$. In some places you use $z$ in place of $x$. $\endgroup$– GaryCommented Feb 8 at 7:43
-
$\begingroup$ @Gary Thanks for pointing those things out. I was thinking about how I would obtain the expansion for $\psi(a+x)$. That's why I did it that way. $\endgroup$ Commented Feb 8 at 8:10