5
$\begingroup$

Consider the following improper integral : $ I = \int_1^\infty \left(\{t\}-\frac{1}{2}\right)\frac{dt}{t}. $

Comparing with Stirling's formula, we can see that $I = \ln(\sqrt{2\pi}) - 1$. Is there a more direct way to compute this ?

Edit : I have split my two questions : the other part is here.

  • 0
    @Marvis: Perhaps in some sense you could view what I did as deriving the constant in Stirlings formula in a different way.2012-12-23

1 Answers 1

4

The Riemann zeta function is defined as $\zeta(s)=\sum_{n=1}^{\infty}n^{-s}.$ Writing this as a Riemann Stieltjies integral and applying integration by parts we have that $\zeta(s)=\int_{1}^{\infty}x^{-s}d\left[x\right]=\int_{1}^{\infty}x^{-s}dx-\int_{1}^{\infty}x^{-s}d\left\{ x\right\}$

$=\frac{s}{s-1}-s\int_{1}^{\infty}\left\{ x\right\} x^{-s-1}dx,$ and this holds for all $\text{Re}(s)>0.$ Notice that taking the limit as $s\rightarrow0$ yields $\zeta(0)=-\frac{1}{2}.$ Now, since $\frac{s}{2}\int_{1}^{\infty}x^{-s-1}=\frac{1}{2},$ it follows that $\zeta(s)+\frac{1}{2}=\frac{s}{s-1}-s\int_{1}^{\infty}\left(\left\{ x\right\} -\frac{1}{2}\right)x^{-s-1}dx,$ and so for $\text{Re}(s)>0,$ we have that $\int_{1}^{\infty}\left(\left\{ t\right\} -\frac{1}{2}\right)t^{-s-1}dt=\frac{1}{s-1}+\frac{-\zeta(s)-\frac{1}{2}}{s}.$ Taking the limit as $s\rightarrow0,$ and using the fact that $\zeta(0)=-\frac{1}{2},$ we find that $\int_{1}^{\infty}\left(\left\{ t\right\} -\frac{1}{2}\right)\frac{dt}{t}=-\zeta^{'}(0)-1.$ Since $\zeta^{'}(0)=-\frac{1}{2}\log(2\pi)$, this yields $\int_{1}^{\infty}\left(\left\{ t\right\} -\frac{1}{2}\right)\frac{dt}{t}=\frac{1}{2}\log(2\pi)-1,$ as desired.

Proving that $\zeta^{'}(0)=-\frac{1}{2}\log(2\pi).$

Recall the functional equation for the zeta function, $\zeta(z)=2^{z}\pi^{z-1}\sin\left(\frac{\pi z}{2}\right)\Gamma\left(1-z\right)\zeta\left(1-z\right).$ Taking the logarithmic derivative, we have that $\frac{\zeta^{'}(z)}{\zeta(z)}=\log2+\log\pi-\frac{\Gamma^{'}\left(1-z\right)}{\Gamma\left(1-z\right)}+\frac{d}{dz}\log\left(\sin\left(\frac{\pi z}{2}\right)\zeta\left(1-z\right)\right).$ Since $\zeta(1-z)=-\frac{1}{z}+\gamma+O(z),$ and $\sin\left(\frac{\pi z}{2}\right)=\frac{\pi z}{2}+O\left(z^{3}\right),$ we have that $\sin\left(\frac{\pi z}{2}\right)\zeta\left(1-z\right)=\frac{\pi}{2}-\frac{\pi\gamma}{2}z+O\left(z^{2}\right),$

and so $\frac{d}{dz}\log\left(\sin\left(\frac{\pi z}{2}\right)\zeta\left(1-z\right)\right)=\frac{-\frac{\pi\gamma}{2}+O\left(z\right)}{\frac{\pi}{2}+O\left(z\right)}=-\gamma+O(z).$ Thus $\frac{\zeta^{'}(0)}{\zeta(0)}=\log2\pi-\Gamma^{'}\left(1\right)-\gamma.$ As $\Gamma^{'}(1)=-\gamma,$ and $\zeta(0)=-\frac{1}{2},$ we conclude that $\zeta^{'}(0)=-\frac{1}{2}\log2\pi.$

Remark: This reminded me of my answer regarding the evaluation of $\Gamma\left(\frac{1}{2}\right).$

  • 0
    @Marvis: At first I thought the same thing, and I was thinking "ooo no, there are too many terms all multiplied together. I don't want to take the product rule that many times" but then I remembered the logarithmic derivative is a shortcut out of multi product rule, and it turned out rather nice.2012-12-23