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Here I have a limit to which I arrived while working on a seperate integral through Mellin Transforms. $$\lim\limits_{s\to -1^{-}}\Big[\psi_{(0)}(s)-\frac{\pi}{2}\tan\left(\frac{\pi s}{2}\right)\Big]$$ Here, we have $\psi_{(0)}(s)$ which represents the digamma function. I graphed the whole thing on Desmos to see what it looked like approaching $-1$, and it seems very likely that the limit approaches $$1-\gamma$$ Here, $\gamma$ is the Euler-Mascheroni constant. I would like to know if there is a concrete way of evaluating this limit. I tried doing some work with it: $$=\lim\limits_{x\to 0}\Big[\psi_{(0)}(x-1)-\frac{\pi}{2}\tan\left(\frac{\pi}{2}(x-1)\right)\Big]$$ $$=\lim\limits_{x\to 0}\Big[\frac{1}{1-x}+\psi_{(0)}(x)-\frac{\pi}{2}\tan\left(\frac{\pi}{2}(x-1)\right)\Big]$$ I don't really know where to go from here. I figure maybe a Taylor expansion could do the trick. However, the expansions for both the digamma and tangent functions are largely unrelated, it seems. I'm curious to see a solution to this problem, and wish you all good luck!

A natural extension of this question would be to find: $$\lim\limits_{s\to (-1-2k)^{-}}\Big[\psi_{(0)}(s)-\frac{\pi}{2}\tan\left(\frac{\pi s}{2}\right)\Big]\,\,\forall\,\,k\in Z^{+}$$ This generalized limit might be the bane of my existance.

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You're correct about the limiting value and correct that a series expansion is the way to go.

Starting with the series expansion at $s=0$, which is $$ \Gamma(s) = \frac1s + \gamma + O(|s|), $$ and repeatedly using the functional equation $\Gamma(s+1)=s\Gamma(s)$, one can prove by induction that the series expansion at $s=-k$ ($k$ a nonnegative integer) is $$ \Gamma(s) = \frac{(-1)^k}{k!} \bigg( \frac1{s+k} + H_k - \gamma + O(|s+k|) \bigg), $$ where $H_k = \sum_{j=1}^k \frac1j$ is the $k$th harmonic number. Therefore $$ \Gamma'(s) = \frac{(-1)^k}{k!} \bigg( {-}\frac1{(s+k)^2} + O(1) \bigg), $$ from which we calculate (by long division) that $$ \psi_{(0)}(s) = \frac{\Gamma'(s)}{\Gamma(s)} = -\frac1{s+k} + H_n - \gamma + O(|s+k|). $$ Since $\tan$ is an odd function (and is itself a logarithmic derivative of $\sin$), its series expansion at odd negative integers $-k$ is going to be simply $-1/(s+k) + O(|s+k|)$. Therefore the series expansion of their difference is $$ \psi_{(0)}(s) - \frac\pi2 \tan\bigg( \frac{\pi s}2 \bigg) = H_n - \gamma + O(|s+k|), $$ which gives you the desired limits.

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  • $\begingroup$ Thanks for the detailed explanation Mr. Martin, I could follow along perfectly. I haven't worked a lot with the digamma function before so I was at a loss for an expansion with negative arguments. This is awesome...again, thanks for sharing! $\endgroup$ – Brian Constantinescu Jul 20 at 8:30
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Let $t=s+1\to 0^-$, then $$\psi_{(0)}(s)-\frac{\pi}{2}\tan\left(\frac{\pi s}{2}\right)=\psi_{(0)}(t-1)+\frac{\pi/2}{\tan\left(\frac{\pi t}{2}\right)}=\psi_{(0)}(t-1)+\frac{1}{t}+o(1).$$ Now recall that $\psi_0(1+x)-\psi_0(x)=\frac{1}{x}$ which implies $$\psi_{(0)}(t-1)=\psi_{(0)}(t)-\frac{1}{t-1}=\psi_{(0)}(t+1)-\frac{1}{t}-\frac{1}{t-1}.$$ Therefore, as $s\to 1^-$, we have that $t\to 0^-$, and
$$\psi_{(0)}(s)-\frac{\pi}{2}\tan\left(\frac{\pi s}{2}\right)=\psi_{(0)}(t+1)-\frac{1}{t-1}+o(1)\to \psi_{(0)}(1)+1=1-\gamma.$$

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  • $\begingroup$ Thanks Robert! Funny enough, right as you made this post I edited my question to show I had thought of the same manipulation that you provided in your hint. Alas, I didn't know where to go from there (and it appears I arrived to something wrong too). I appreciate the quick response! $\endgroup$ – Brian Constantinescu Jul 20 at 8:23
  • $\begingroup$ @BrianConstantinescu I edited my answer. $\endgroup$ – Robert Z Jul 20 at 9:01
  • $\begingroup$ Cheers, it's an elegant and concise solution! Thanks for taking the time to help me out. $\endgroup$ – Brian Constantinescu Jul 20 at 9:49

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