# A question related to Riemann zeta function

Does anyone know why the following statement is correct?

Let $f(x)$ be the function whose value on the interval $m\pi<x<(m+1)\pi, m=0,1,2,\cdots$, is $(-1)^m\frac{\pi}{4}$. Let $0<s<1$. Then $$\int_0^\infty x^{s-1}f(x)\,dx$$ represents an analytic function for $\Re s<1$ and is equal to $$2(1-2^{s+1})\zeta(1-s)$$ for $\Re s<0$. Here $\zeta(\cdot)$ is the Riemann zeta function.

Edit. I think now I partially understand the first part. Let $\alpha(x)$ be the function defined by $$\int_0^x u(t)\,dt$$ where $u(t)=(-1)^m$ on $[m\pi,(m+1)\pi], m=0,1,2,\cdots.$ Then the integral $\int_0^\infty x^{s-1}f(x)\,dx$ is equal to the Riemann-Stieltjes integral $\frac{\pi}{4}\int_0^\infty x^{s-1}\,d\alpha$. For $0<s<1$, this can be written as $$\frac{\pi}{4}\left(\int_0^\pi x^{s-1}\,dx-\pi^s-(s-1)\int_{\pi}^\infty\frac{\alpha(x)}{x^{2-s}}\,dx\right),$$ which is analytic in $s$ for $0<\Re s<1$.

Now I need to know how to do it for $\Re s\le 0$ and how to make it equal to $2(1-2^{s+1})\zeta(1-s)$ for $\Re s<0$.

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Note: Pb. 21 in Andrews-Askey-Roy contains a few misprints and is formulated inaccurately.

• The sum over $n$ should start from $n=0$ and not from $1$.

• As you might have noticed, the functional relation for $\zeta$ in the 1st line of the problem (correct) does not coincide with the relation (incorrect) implicitly stated in question (b). Namely, instead of $\zeta(1-s)$ one should have $\zeta(-s)$ - this is precisely what is obtained below.

• as for the prefactor $\frac{\pi^{1+s}}{2s}$ in my formula, it should also be there (check this by computing the lhs in AAR problem) if one want correct functional relation for $\zeta$.

By definition of $f$, \begin{align}\int_0^{\infty}x^{s-1}f(x)dx&=\frac{\pi}{4}\sum_{m=0}^{\infty}(-1)^m\int_{m\pi}^{(m+1)\pi}x^{s-1}dx. \end{align} Introduce instead of this $$F(a)=\frac{\pi}{4}\sum_{m=0}^{\infty}(-1)^m\int_{ma}^{(m+1)a}x^{s-1}dx.$$ Differentiating w.r.t. parameter $a$, we have \begin{align}\frac{\partial F}{\partial a}&=\frac{\pi a^{s-1}}{4}\left(1+\sum_{m=1}^{\infty}(-1)^m\left[(m+1)^{s}-m^{s}\right]\right)=-\frac{\pi a^{s-1}}{2}\sum_{m=1}^{\infty}(-1)^m m^s.\tag{1} \end{align} Recall that the "odd" part of the Riemann $\zeta$-function can be determined from $$\zeta(s)=\underbrace{\sum_{m=0}^{\infty}(2m+1)^{-s}}_{\zeta_{\mathrm{odd}}(s)}+\underbrace{2^{-s}\zeta(s)}_{\zeta_{\mathrm{even}}(s)}.$$ But our sum in (1) is then just $$\zeta_{\mathrm{even}}(-s)-\zeta_{\mathrm{odd}}(-s)=\left(2^{1+s}-1\right)\zeta(-s).$$ Using this, integrating (1) back with respect to $a$, taking into account that $F(0)=0$, and setting $a=\pi$, we get $$F(\pi)=\frac{\pi^{1+s}}{2s}\left(1-2^{1+s}\right)\zeta(-s).$$

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Sorry, it should be $x^{s-1}$, not $x^{1-s}$, edited. Maybe this time your argument works? The statement is on p.51 of Andrews-Askey-Roy's book, Problem 21. –  TCL May 27 at 16:04
@TCL: I have made the corresponding changes and pointed out what are the misprints in Andrews-Askey-Roy. If you have problems with the lhs, don't hesitate to ask me. –  O.L. May 27 at 20:35
For $0<s<1$, the sum in your equation (1) is divergent. I don't see why we could proceed as though it was convergent. –  TCL May 27 at 21:10
@TCL I do not claim it is convergent for $s\in(0,1)$. Once the $m=0$ term in the first two formulas is separated, I can (and I do) choose $\mathrm{Re}\,s$ sufficiently negative for all subsequent series to converge. –  O.L. May 27 at 21:19