I was messing around with the Zeta-Function and I got what I thought was an interesting limit:

$$\lim_{x \to 0} \Gamma(x)(\gamma+\psi(1+x)) = \frac{\pi^2}{6} $$ Where $\Gamma$ is the gamma function, $\gamma $ is the Euler-Mascheroni constant, and $\psi$ is the digamma function. I got it by writing the Zeta Function as:

$$\sum_{n=1}^{\infty} \frac{1}{n^2} = -\int_{0}^1 \frac{\ln(x)}{1-x}dx$$ Then using the beta function and differentiating with respect to x you get the limit.

What other ways can be used to evaluate the limit?

Here's how I got to my answer:

We know $$\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6} $$ To convert the sum into an integral consider: $$\int_{0}^1 x^{n-1}dx=\frac{1}{n}$$ Differentiating once: $$\int_{0}^1 \ln(x)x^{n-1}dx=\frac{-1}{n^2}$$ Plugging in the integral into the sum you get: $$-\sum_{n=1}^\infty \int_0^1 \ln(x)x^{n-1}dx=-\int_0^1\ln(x)\sum_{n=1}^\infty x^{n-1}dx=-\int_0^1 \frac{\ln(x)}{1-x}dx$$

Euler's Beta Function is defined as:

$$\operatorname{B(x,y)}=\int_0^1 t^{x-1}(1-t)^{y-1} = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$

Differentiation with respect to x: $$\operatorname{B_x(x,y)}=\int_0^1 \ln(t)t^{x-1}(1-t)^{y-1} = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}(\psi(x)-\psi(x+y))$$

$$-\int_0^1 \frac{\ln(x)}{1-x}dx = -\operatorname{B_x(1,0)}=\frac{\pi^2}{6} $$

Taking the limit as (x,y)->(1,0)

$$-\lim_{(x,y)\to(1,0)} \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}(\psi(x)-\psi(x+y))=-\lim_{y \to 0} \Gamma(y)(\psi(1)-\psi(1+y)) = \lim_{y \to 0} \Gamma(y)(\gamma+\psi(1+y))$$

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    $\begingroup$ What are $\psi$ and $\gamma$ though? Are these standard notations? $\endgroup$ – Suzet Aug 2 '18 at 14:48
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    $\begingroup$ They're the digamma function and the euler-mascheroni constant. I'll edit the post so it's more clear. $\endgroup$ – Tom Himler Aug 2 '18 at 14:49
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    $\begingroup$ I am just interested. How exactly did you derived the limit from the given integral representation with your steps. I tried to do it by myself and failed. Could you maybe explain it to me or add it to the question? $\endgroup$ – mrtaurho Aug 2 '18 at 15:27
  • $\begingroup$ @mrtaurho I put how I got my answer if that helps. It's not super rigorous since I don't really know that sort of shtuff yet. $\endgroup$ – Tom Himler Aug 2 '18 at 18:25
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    $\begingroup$ Thank you for adding it nevertheless. I am not that familiar as I want to with these kinds of reshapings and this stuff. It is great for me to learn more about it. $\endgroup$ – mrtaurho Aug 2 '18 at 18:27

One may use $\Gamma(x)=\Gamma(x+1)/x$ to get


and since $\psi(1)=-\gamma$, we may use the definition of the derivative to get $\psi'(1)$, which has many different forms, of which the result may be deduced from.

  • $\begingroup$ Thanks! That's very clever. $\endgroup$ – Tom Himler Aug 2 '18 at 14:58

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