How to show that the Laurent series of the Riemann Zeta function has $\gamma$ as its constant term?

I mean the Laurent series at $s=1$.

I want to do it by proving $\displaystyle \int_0^\infty \frac{2t}{(t^2+1)(e^{\pi t}+1)} dt = \ln 2 - \gamma$,

based on the integral formula given in Wikipedia. But I cannot solve this integral except by using Mathematica. Tried complex analytic ways but no luck. Any suggestions? Thanks for your attention!

Solution 1:

We can show that $$ \zeta(s)=\frac1{1 - 2^{1-s}}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\tag{1} $$ converges for $\mathrm{Re}(s)>0$ by applying the Generalized Dirichlet Test. To apply the test, we need to show that the partial sum of $(-1)^{n-1}$ is bounded, $|n^{-s}|\to0$, and $$ \sum_{n=1}^\infty\left|n^{-s}-(n+1)^{-s}\right|\tag{2} $$ converges.

When $s\in\mathbb{R}$, $n^{-s}$ head straight from $1$ to $0$, so the sum in $(2)$ is $1$. However, if $s\not\in\mathbb{R}$, then $n^{-s}$ spirals into $0$, and it is not immediately obvious that that spiral has finite length.

Let's look at how $n^{-s}$ spirals into $0$:

$\hspace{6pt}$(a)$\hspace{6pt}$ $\arg(n^{-s}) = -\log(n)\mathrm{Im}(s)$

$\hspace{6pt}$(b)$\hspace{6pt}$ $|n^{-s}| = n^{-\mathrm{Re}(s)} = e^{-\log(n)\mathrm{Re}(s)}$

Thus, $n^{-s}$ lies on the spiral $r = e^{t\theta}$ where the constant $t = \mathrm{Re}(s)/\mathrm{Im}(s)$. The length of this curve from $r=1$ to $r=0$ is easily computed to be $|s|/\mathrm{Re}(s)$. Thus, the total variation of $n^{-s}$, as given in $(2)$, is bounded by $|s|/\mathrm{Re}(s)$, and therefore, the sum in $(1)$ converges.

Consider $(1)$ to first order in $s-1$. $$ \frac1{1 - 2^{1-s}}=\frac1{s-1}\frac1{\log(2)}+\frac12+O(s-1)\tag{3} $$ and $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}=\log(2)+(s-1)\sum_{n=1}^\infty(-1)^n\frac{\log(n)}{n}+O\left((s-1)^2\right)\tag{4} $$ Therefore, $$ \zeta(s)=\frac1{s-1}+\frac{\log(2)}{2}+\frac1{\log(2)}\sum_{n=1}^\infty(-1)^n\frac{\log(n)}{n}+O(s-1)\tag{5} $$ Next, we can use the Euler-Maclaurin Sum Formula to compute $$ \sum_{k=1}^n\frac{\log(k)}{k}=C+\frac{\log(n)^2}{2}+O\left(\frac{\log(n)}{n}\right)\tag{6} $$ and $$ \sum_{k=1}^n\frac1{k}=\log(n)+\gamma+O\left(\frac1n\right)\tag{7} $$ Then, we can use $(6)$ and $(7)$ to get $$ \begin{align} \sum_{n=1}^\infty(-1)^n\frac{\log(n)}{n} &=-\lim_{n\to\infty}\left(\sum_{k=1}^{2n}\frac{\log(k)}{k}-2\sum_{k=1}^n\frac{\log(2k)}{2k}\right)\\ &=\small-\lim_{n\to\infty}\left(\left(C+\frac{\log(2n)^2}{2}\right)-\left(C+\frac{\log(n)^2}{2}\right)-\log(2)(\log(n)+\gamma)\right)\\ &=\gamma\log(2)-\frac{\log(2)^2}{2}\tag{8} \end{align} $$ Combining $(5)$ and $(8)$ yields $$ \zeta(s)=\frac1{s-1}+\gamma+O(s-1)\tag{9} $$

Solution 2:

In my previous answer, it is shown that $$ \zeta(s)=\frac1{s-1}+\gamma+O(s-1)\tag{1} $$ We can compute more terms of the Laurent expansion using the formula $$ \begin{align} (-1)^n\frac{\mathrm{d}^n}{\mathrm{d}s^n}\left[(s-1)\sum_{k=1}^m\frac1{k^s}\right] &=(s-1)\sum_{k=1}^m\frac{\log(k)^n}{k^s}-n\sum_{k=1}^m\frac{\log(k)^{n-1}}{k^s}\tag{2} \end{align} $$ Euler-Maclaurin gives $$ \begin{align} \sum_{k=1}^m\frac{\log(k)^n}{k^s} &=\int_1^m\frac{\log(x)^n}{x^s}\,\mathrm{d}x+c_n(s)+O\left(\frac{\log(m)^n}{m^s}\right)\\ &=\int_0^{\log(m)}x^ne^{(1-s)x}\,\mathrm{d}x+c_n(s)+O\left(\frac{\log(m)^n}{m^s}\right)\\ &=\frac1{(s-1)^{n+1}}\int_0^{(s-1)\log(m)}x^ne^{-x}\,\mathrm{d}x+c_n(s)+O\left(\frac{\log(m)^n}{m^s}\right)\tag{3} \end{align} $$ Since $c_n(s)=\sum\limits_{k=1}^\infty\frac{\log(k)^n}{k^s}-\int_1^\infty\frac{\log(x)^n}{x^s}\,\mathrm{d}x$, we have $$ c_n'(s)=-c_{n+1}(s)\tag{4} $$ and since $\frac{\log(x)^n}{x^s}$ increases then decreases for $x\ge1$, $$ \left|c_n(s)\right|\le2\sup\limits_{x\ge1}\frac{\log(x)^n}{x^s}=2\left(\frac n{es}\right)^n\tag{5} $$

The left term of $(2)$ is $(s-1)$ times $(3)$ $$ \begin{align} (s-1)\sum_{k=1}^m\frac{\log(k)^n}{k^s} &=\frac1{(s-1)^n}\int_0^{(s-1)\log(m)}x^ne^{-x}\,\mathrm{d}x+(s-1)c_n(s)\\ &+(s-1)O\left(\frac{\log(m)^n}{m^s}\right)\tag{6} \end{align} $$ The right term of $(2)$ is $n$ times $(3)$ at $n-1$ $$ \begin{align} n\sum_{k=1}^m\frac{\log(k)^{n-1}}{k^s} &=\frac{n}{(s-1)^n}\int_0^{(s-1)\log(m)}x^{n-1}e^{-x}\,\mathrm{d}x+nc_{n-1}(s)+O\left(\frac{\log(m)^{n-1}}{m^s}\right)\\ &=\log(m)^nm^{1-s}+\frac1{(s-1)^n}\int_0^{(s-1)\log(m)}x^ne^{-x}\,\mathrm{d}x+nc_{n-1}(s)\\ &+O\left(\frac{\log(m)^{n-1}}{m^s}\right)\tag{7} \end{align} $$ Putting $(2)$, $(6)$, and $(7)$ together, we get $$ \begin{align} (-1)^n\frac{\mathrm{d}^n}{\mathrm{d}s^n}(s-1)\zeta(s) &=\lim_{m\to\infty}\left[(s-1)\sum_{k=1}^m\frac{\log(k)^n}{k^s}-n\sum_{k=1}^m\frac{\log(k)^{n-1}}{k^s}\right]\\[6pt] &=(s-1)c_n(s)-nc_{n-1}(s)\tag{8} \end{align} $$ Evaluating $(8)$ at $s=1$ gives $$ \begin{align} \left.(-1)^n\frac{\mathrm{d}^n}{\mathrm{d}s^n}(s-1)\zeta(s)\,\right|_{\,s=1} &=-nc_{n-1}(1)\\ &=\lim_{m\to\infty}\left[\log(m)^n-n\sum_{k=1}^m\frac{\log(k)^{n-1}}k\right]\tag{9} \end{align} $$ Applying $(9)$, we get the Laurent series for $\zeta(s)$ at $s=1$ to be $$ \bbox[5px,border:2px solid #C0A000]{\zeta(s)=\frac1{s-1}+\sum_{n=0}^\infty\frac{(1-s)^n}{n!}\,\gamma_n}\tag{10} $$ where $$ \gamma_n=\lim_{m\to\infty}\left[\sum_{k=1}^m\frac{\log(k)^n}k-\frac{\log(m)^{n+1}}{n+1}\right]\tag{11} $$

Solution 3:

There is a nice proof in Titchmarsh's "The Theory of the Riemann Zeta Function": putting $\,[x]=\,$ the greatest integer not greater than $\,x\in\mathbb R\,$, we get: $$\lim_{x\to 1^+}\left[\zeta(s)-\frac{1}{s-1}\right]\stackrel{\mathbf{(7)}}=\int_1^\infty\frac{[x]-x+\frac{1}{2}}{x^2}\,dx+\frac{1}{2}=\\=\int_1^\infty\frac{[x]-x}{x^2}+\frac{1}{2}\int_1^\infty\frac{dx}{x^2}+\frac{1}{2}=\int_1^\infty\frac{[x]-x}{x^2}\,dx+1=$$$$=\lim_{n\to\infty}\left[\sum_{m=1}^{n-1}\left(\int_m^{m+1}\frac{[x]dx}{x^2}-\int_m^{m+1}\frac{dx}{x}\right)+1\right]=$$$$\lim_{n\to\infty}\left[\sum_{m=1}^{n-1}m\left(\int_m^{m+1}\frac{dx}{x^2}\right)-\log n+1\right]=$$$$=\lim_{n\to\infty}\left[\left(1-\frac{1}{2}+1-\frac{2}{3}+...+1-\frac{m-1}{m}\right)+1-\log n\right]=$$$$=\lim_{n\to\infty}\left(\sum_{m=1}^{n-1}\frac{1}{m+1}+1-\log n\right)=\lim_{n\to\infty}\left(\sum_{m=1}^n\frac{1}{m}-\log n\right)=:\gamma$$

Finally, since we know $\,\displaystyle{\lim_{s\to 1^+}(s-1)\zeta(s)=1}\,$, we have that $\,s=1\,$ is a simple pole of $\,\zeta(s)\,$ with residue $\,1\,$, so the above gives the free coefficient of the Laurent expansion of $\,\zeta(s)\,$ around $\,1$

Proof of (7): We use the next form of Abel's summation by parts formula (all the time, $\,n\in\mathbb N\,$):

Lemma: Let $\,\phi(x)\,$ be any function with continuous derivative in $\,[a,b]\,$, then $$\sum_{a< n\leq b}\phi(n)=\int_a^b\phi(x) dx+\int_a^b\left(x-[x]-\frac{1}{2}\right)\phi'(x)dx+\left(a-[a]-\frac{1}{2}\right)\phi(a)-\left(b-[b]-\frac{1}{2}\right)\phi(b)\,\,\,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\color{blue}{(1)}$$

$\color{red}{\text{Proof}}$: Taking $\,a=n\,,\,b=n+1\,$ and integrating by parts we get at once: $$\int_n^{n+1}\left(x-n-\frac{1}{2}\right)\phi'(x)dx=\left.\left(x-n-\frac{1}{2}\right)\phi(x)\right|_n^{n+1}-\int_n^{n+1}\phi(x)dx=$$ $$=\frac{1}{2}\left(\phi(n)+\phi(n+1)\right)-\int_n^{n+1}\phi(x)dx\Longrightarrow$$ $$\Longrightarrow \int_n^{n+1}\phi(x)dx+\int^{n+1}_n\left(x-[x]-\frac{1}{2}\right)\phi'(x)dx+\left(n-[n]-\frac{1}{2}\right)\phi(n)-\left(n+1-[n+1]-\frac{1}{2}\right)\phi(n+1)=\phi(n+1)=\sum_{n<m\leq n+1}\phi(m)\,\,,\,\,m\in\mathbb N$$

which shows both that the formula works for the above particular case and that it's enough to check for the case $\,n\leq a<b\leq n+1\,$ , so again integrating by parts: $$\int_a^b\left(x-n-\frac{1}{2}\right)\phi'(x)dx=\left(b-n-\frac{1}{2}\right)\phi(b)-\left(a-n-\frac{1}{2}\right)\phi(a)-\int_a^b\phi(x)dx$$ Comparing with the equality promised by the lemma, we see the RHS of $\,(1)\,$ above reduces to $$-\left(b-[b]-\frac{1}{2}\right)\phi(b)+\left(b-n-\frac{1}{2}\right)\phi(b)=\left([b]-n\right)\phi(b)$$ and this equals zero unless $\,b=n+1\,$, but then the last expression above equals $\,\phi(b)=\phi(n+1)\,$, which is the LHS in $\,(1)\,\;\;\;\;\;\;\;\square$

Now $\,(7)\,$ follows from the above with $$a_n=1\,\,,\,\forall n\in\mathbb N\,\,,\,\,\phi(n):=n^{-s}\quad\text{ and }\quad\,\,A(x):=\sum_{0\leq n\leq x}a_n=[x]$$ and we get

$$\zeta(s):=\sum_{n=1}^\infty\frac{1}{n^s}=\sum_{n=1}^\infty a_n\phi(n)=s\int_1^\infty\frac{[x]dx}{x^{1+s}}$$

Finally, we just write $$\frac{1}{s-1}=\int_1^\infty\frac{dx}{x^s}$$

Solution 4:

This is my third answer to this question, but I just came up with this much simpler derivation.

A Simple Derivation of the Laurent Series for Zeta $$ \begin{align} &\frac1{s-1}+\sum_{k=1}^mk^{-s}-\frac{m^{1-s}-1}{1-s}\tag1\\ &=\frac1{s-1}+\sum_{k=1}^m\frac1ke^{(1-s)\log(k)}-\frac{e^{(1-s)\log(m)}-1}{1-s}\tag2\\ &=\frac1{s-1}+\sum_{n=0}^\infty\left[\sum_{k=1}^m\frac1k\frac{(1-s)^n\log(k)^n}{n!}-\frac{(1-s)^n\log(m)^{n+1}}{(n+1)!}\right]\tag3\\ &=\frac1{s-1}+\sum_{n=0}^\infty\frac{(1-s)^n}{n!}\left[\sum_{k=1}^m\frac{\log(k)^n}k-\frac{\log(m)^{n+1}}{n+1}\right]\tag4 \end{align} $$ Explanation:
$(2)$: convert powers to exponentials
$(3)$: expand exponentials about $s=1$
$(4)$: pull out a common factor

Taking the limit as $m\to\infty$, for $s\gt1$, $$ \bbox[5px,border:2px solid #C0A000]{\zeta(s)=\frac1{s-1}+\sum_{n=0}^\infty\frac{(1-s)^n}{n!}\,\gamma_n}\tag5 $$ where $$ \bbox[5px,border:2px solid #C0A000]{\gamma_n=\lim_{m\to\infty}\left[\sum_{k=1}^m\frac{\log(k)^n}k-\frac{\log(m)^{n+1}}{n+1}\right]}\tag6 $$ $\gamma_n$ is the $n^\text{th}$ Stieltjes constant; $\gamma_0$ is the Euler-Mascheroni constant.

A Rough Bound on the Coefficients

We can get the following bound on $\gamma_n$ using a truncated version of the Euler-Maclaurin Sum Formula with $f_n(x)=\frac{\log(x)^n}{x}$ $$ \begin{align} |\gamma_n| &=\lim_{m\to\infty}\left|\,\sum_{k=1}^mf_n(k)-\int_1^mf_n(x)\,\mathrm{d}x\,\right|\\ &=\lim_{m\to\infty}\left|\,\frac12(f_n(m)+f_n(1))+\int_1^mf_n'(x)\left(\{x\}-\tfrac12\right)\mathrm{d}x\,\right|\\ &\le\frac12[n=0]+\frac12\operatorname*{Var}_{(1,\infty)}(f_n)\\[3pt] &=\frac{n^n}{e^n}\tag7 \end{align} $$ $(7)$ guarantees a radius of convergence in $(5)$ of no less than $1$. Since $(s-1)\zeta(s)$ is entire, the radius of convergence of $(5)$ is actually $\infty$.