Euler Maclaurin Formula to prove $\zeta(0)=-\frac12$ and $\zeta(-1)=-\frac{1}{12}$

Question

This is a quick question regarding the analytic continuation of the Riemman Zeta function by application of the Euler Maclaurin Formula and the evaluation of $\zeta(0)=-\frac12$ and $\zeta(-1)=-\frac{1}{12}$ .

Choosing $f(x)=\frac{1}{x^s}$ in the second order Euler Maclaurin Formula we obtain

$$\zeta(s)=\frac{1}{s-1}+\frac12+\frac{s}{12}-\frac{s(s+1)}{2}\int_1^\infty\frac{P_2(x)}{x^{s+2}}\,dx \tag{1}$$

Where $P_2(x)$ is the second Periodic Bernoulli polynomial. $(1)$ extends analytically the Riemann zeta function for $-1<\Re(s)$. Obviously, there is a simple pole @ $1$.

If we now set $s \to 0$ in$(1)$ we immediately get that

$$\zeta(0)=-\frac12 \tag{2}$$

Similarly, if we integrate by parts the integral in $(1)$, we get the third order Euler Maclaurin formula that extends analytically the Riemann zeta function for $-2<\Re(s)$.

$$\zeta(s)=\frac{1}{s-1}+\frac12+\frac{s}{12}+\frac{s(s+1)(s+2)}{2}\int_1^\infty\frac{P_3(x)}{x^{s+3}}\,dx \tag{3}$$

Letting $s \to -1$ in $(3)$ we immediately obtain

$$\zeta(-1)=-\frac{1}{12} \tag{4}$$

Apparently this procedure is all fine because we have extended the domain of the Zeta function to the right of the line $-2$ in the S-complex plane, excluding only the point $s=1$, and now it makes sense to ask what is $\zeta(0)=\, ?$ for instance. Note also that by letting $s \to 0$ in $(3)$ we again recover that $\zeta(0)=-\frac12$.

My question is: Is this procedure correct and valid or there are any caveat that should be taken into account before we perform it, or some additional condition that should be added?

Answer 1

As mentioned in the comments by @WADon if we manage to prove that the integrals in $(1)$ and $(3)$ above are finite, the analytic continuation is valid. So, I found an article (see below) where the author proves that the Bernoulli Polynomials on the interval $[0,1]$ satisfy the following inequalities for $k \geq 1$:

$$|B_{2k}(x)|\leq |B_{2k}| \,\,\text{and}\,\,|B_{2k+1}(x)|\leq(2k+1) |B_{2k}|.$$

Therefore we can prove that the two integrals are indeed finite:

$$ \left|\int_1^\infty \frac{P_2(x)}{x^{s+2}}\,dx\right| \leq \frac12 \int_1^\infty \frac{1}{x^{\sigma+2}}\,dx=\frac12 \frac{1}{\sigma+1}<\infty\\ $$

For $\Re(s)=\sigma$ and $\sigma>-1$

and

$$ \left|\int_1^\infty \frac{P_3(x)}{x^{s+3}}\,dx\right| \leq \frac12 \int_1^\infty \frac{1}{x^{\sigma+3}}\,dx=\frac12 \frac{1}{\sigma+2}<\infty\\ $$

For $\Re(s)=\sigma$ and $\sigma>-2$

• D. H. Lehmer, On the maxima and minima of Bernoulli polynomials, Amer. Math. Monthly, v. 47 (1940), pp. 533-538.

Answer 2

As described in this answer, and computed in this answer, for $\mathrm{Re}(z)\gt-5$, the Euler-Maclaurin Sum Formula says that $$ \sum_{k=1}^n\frac1{k^z}=\zeta_n(z)+\frac1{1-z}n^{1-z}+\frac12n^{-z}-\frac{z}{12}n^{-1-z}+\frac{z(z+1)(z+2)}{720}n^{-3-z}\tag1 $$ where $\lim\limits_{n\to\infty}\zeta_n(z)=\zeta(z)$.

If we set $z=0$, we get $$ \sum_{k=1}^n1=\zeta_n(0)+n+\frac12\tag2 $$ If we let $n\to\infty$, we get $\zeta(0)=-\frac12$.

If we set $z=-1$, we get $$ \sum_{k=1}^nk=\zeta_n(-1)+\frac12n^2+\frac12n+\frac1{12}\tag3 $$ If we let $n\to\infty$, we get $\zeta(-1)=-\frac1{12}$.

If we set $z=-2$, we get $$ \sum_{k=1}^nk^2=\zeta_n(-2)+\frac13n^3+\frac12n^2+\frac16n\tag4 $$ If we let $n\to\infty$, we get $\zeta(-2)=0$.

If we set $z=-3$, we get $$ \sum_{k=1}^nk^3=\zeta_n(-3)+\frac14n^4+\frac12n^3+\frac14n^2-\frac1{120}\tag5 $$ If we let $n\to\infty$, we get $\zeta(-3)=\frac1{120}$.

If we set $z=-4$, we get $$ \sum_{k=1}^nk^4=\zeta_n(-4)+\frac15n^5+\frac12n^4+\frac13n^3-\frac1{30}n\tag6 $$ If we let $n\to\infty$, we get $\zeta(-4)=0$.