Inversion theorem for Dirichlet series

Question

Can someone come up with a proof for this little theorem?

Suppose that $F_a(s)$ is a Dirichlet series and $a(n)$ is its associated arithmetic function, that is:

$$F_a(s)=\sum_{n=1}^{\infty}\frac{a(n)}{n^s}$$

Then the $a(n)$ are given by:

\begin{equation} \label{eq:a(n)} \nonumber a(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}F_a(2j)}{(2i+1-2j)!} \end{equation}

The advantage of this formula is that if you know $F_a(s)$ at the even integers, you know the coefficients of its series expansion.

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Let me give an example, before questions rain down asking for clarification.

We know that: $$\frac{1}{\zeta(s)}=\sum_{n=1}^{\infty}\frac{\mu(n)}{n^s}$$ where $\mu(n)$ is the Mobius function. Therefore: $$\mu(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\zeta(2j)^{-1}}{(2i+1-2j)!}$$

Now, one good application of this formula is the generalization of the Mobius function. Whatever the $k$, real or complex, positive or negative, we have by definition that: $$\zeta(s)^{-k}=\sum_{n=1}^{\infty}\frac{\mu_k(n)}{n^s}$$ and you may ask, what is the $\mu_k(n)$ for each $n$ that satisfies the above equation?

Well, they are given by: $$\mu_k(n)=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\zeta(2j)^{-k}}{(2i+1-2j)!}$$

An example:

$\sqrt{\zeta(s)}=1+\frac{1}{2\cdot 2^s}+\frac{1}{2\cdot 3^s}+\frac{3}{8\cdot 4^s}+\frac{1}{2\cdot 5^s}+\frac{1}{4\cdot 6^s}+\frac{1}{2\cdot 7^s} +\frac{5}{16\cdot 8^s} +\frac{3}{8\cdot 9^s}+\frac{1}{4\cdot 10^s}+\frac{1}{2\cdot 11^s}+\cdots\\$

Another example, the Von Mangoldt function divided by the log is given by:

$$\frac{\Lambda(n)}{\log n}=-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}\log\zeta(2j)}{(2i+1-2j)!}\\$$

Answer 1

First let's prove this for $$a(n)=\delta_{m,n}=\begin{cases}1,&n=m\\0,&n\neq m\end{cases}$$ with a (fixed) positive integer $m$. Consider the RHS of the formula, $$A_{m,n}=-2\sum_{i=0}^{\infty}(-1)^i(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j(2\pi m)^{-2j}}{(2i+1-2j)!}.$$ Replacing $j$ by $i-j$ in the inner sum, we get $$A_{m,n}=-2\sum_{i=0}^{\infty}\Big(\frac{n}{m}\Big)^{2i}\sum_{j=0}^{i}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}.\tag{1}\label{eq1}$$ Since $\displaystyle\sum_{j=0}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}=\frac{\sin 2\pi m}{2\pi m}=0$, this implies $$A_{m,n}=2\sum_{i=0}^{\infty}\Big(\frac{n}{m}\Big)^{2i}\sum_{\color{blue}{j=i+1}}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}=2\sum_{i=0}^{\infty}\Big(\frac{n}{m}\Big)^{2i}\sum_{\color{blue}{j=i}}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}$$ which, after replacing $j$ with $i+j$, gives $$A_{m,n}=2\sum_{i,j=0}^{\infty}\frac{(-1)^{i+j}(2\pi)^{2i+2j}n^{2i}m^{2j}}{(2i+2j+1)!}.\tag{2}\label{eq2}$$ In particular, $A_{m,n}=A_{n,m}$. Now if $n<m$ then we can change the order of summation in $\eqref{eq1}$: $$A_{m,n}=-2\sum_{j=0}^{\infty}\frac{(-1)^j(2\pi m)^{2j}}{(2j+1)!}\sum_{i=j}^{\infty}\Big(\frac{n}{m}\Big)^{2i}=-2\Big(1-\frac{n^2}{m^2}\Big)^{-1}\frac{\sin 2\pi n}{2\pi n}=0,$$ and if $n=m$ then, collecting terms in $\eqref{eq2}$ with $i+j=k$, we obtain $$A_{n,n}=2\sum_{k=0}^{\infty}\frac{(-1)^k(2\pi n)^{2k}}{(2k+1)!}(k+1)=\cos 2\pi n+\frac{\sin 2\pi n}{2\pi n}=1.$$ Concluding, we have $A_{m,n}=\delta_{m,n}$, and this finishes the proof for $a(n)=\delta_{m,n}$.

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Essentially, we've proven the formula for functions $a(n)$ such that $\{n:a(n)\neq 0\}$ is finite. It's also easy to see that the formula holds when $\sum_{n=1}^{\infty}a(n)$ converges absolutely: the triple sum $$\sum_{i=0}^{\infty}(-1)^i(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j(2\pi)^{-2j}}{(2i+1-2j)!}\sum_{\color{blue}{m=n+1}}^{\infty}\frac{a(m)}{m^{2j}}$$ converges absolutely in this case, so we can change the order of summation, but each of the sums with a fixed value of $m$ equals $0$ as already proven; so the entire sum is $0$, and we're left with $\sum_{m=1}^{n}$, which again has been treated already.

Further results require analytic continuation. This can be done as usual (beginning with application of the above to $\bar{a}(n)=n^{-s}a(n)$ with $\Re s>\sigma$ the abscissa of absolute convergence of $F_a(s)$), but behaviour of the series in the RHS of the formula needs to be explored separately.

Answer 2

This proof is simple:

\begin{equation} \nonumber -2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}F_a(2j)}{(2i+1-2j)!} \Rightarrow \end{equation} \begin{equation} \nonumber -2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}}{(2i+1-2j)!}\sum_{k=1}^{\infty}\frac{a(k)}{k^{2j}} \Rightarrow \end{equation} \begin{equation} \nonumber \sum_{k=1}^{\infty}a(k)\left(-2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi)^{-2j}k^{-2j}}{(2i+1-2j)!}\right)\\ \end{equation} The theorem then follows from the following equation: \begin{equation} \nonumber -2\sum_{i=0}^{\infty} (-1)^{i}(2\pi n)^{2i}\sum_{j=0}^{i}\frac{(-1)^j (2\pi k)^{-2j}}{(2i+1-2j)!}=\begin{cases} 1, & \text{if}\ n=k\\ 0, & \text{otherwise}\\ \end{cases} \end{equation} And the above equation is justified for being the convolution of $\mu_0(n)$ (which is the same as $\delta_{n1}$) and the associated function of the series $k^{-s}$, $b(n)$, since from the convolution formula: \begin{equation} \nonumber c(n)=(\mu_0*b)(n)=\sum_{d|n}\mu_0(d) b\left(\frac{n}{d}\right)=b(n)=\begin{cases} 1, & \text{if}\ n=k\\ 0, & \text{otherwise} \end{cases} \end{equation}

Answer 3

This answer assumes the following definitions where $\tilde{f}_a(x)$ and $\tilde{f}_a'(x)$ are used to refer to analytic representations of $\sum\limits_{n}a(n)\,\theta(x-n)=\underset{\epsilon\to 0}{\text{lim}}\frac{f_a(x-\epsilon)+f_a(x+\epsilon)}{2}$ and it's first-order derivative $\sum\limits_{n}a(n)\,\delta(x-n)$ where $\theta(x)$ is the Heaviside step function and $\delta(x)$ is the Dirac delta function.

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$$f_a(x)=\sum\limits_{n=1}^x a(n)\tag{1}$$

$$F_a(s)=s\int\limits_0^\infty f_a(x)\,x^{-s-1}\,dx=\sum\limits_{n=1}^\infty\frac{a(n)}{n^s}\,,\quad\Re(s)\ge 2\tag{2}$$

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$$\tilde{f}_a'(x)=\underset{\substack{K,f\to\infty \\ K\gg f\,x}}{\text{lim}}\left(-4 f\sum\limits_{k=1}^K (-1)^k\ x^{2 k}\sum\limits_{j=1}^k \frac{(-1)^j\ (2 \pi f)^{2(k-j)}\ F_a(2 j)}{(2 k-2 j+1)!}\right)\tag{3}$$

$$\tilde{f}_a(x)=\underset{\substack{K,f\to\infty \\ K\gg f\,x}}{\text{lim}}\left(-4 f \sum\limits_{k=1}^K \frac{(-1)^k\ x^{2 k+1}}{2 k+1} \sum\limits_{j=1}^k\frac{(-1)^j\ (2 \pi f)^{2 (k-j)}\ F_a(2 j)}{(2 k-2 j+1)!}\right)\tag{4}$$

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Partial evaluation of formula (3) for $\tilde{f}_a'(x)$ at a finite valuation frequency $f\in\mathbb{Z}_{>0}$ leads to the following analytic formula for $a(n)$. The OP's formula uses the evaluation frequency $f=1$, but also includes some additional terms that affect the evaluation at $x=0$ but not at other integer values of $x$.

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$$\tilde{a}(x)=\frac{1}{2\,f}\tilde{f}_a'(x)\tag{5}$$

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For the cases $a(n)=1$, $a(n)=(-1)^{n-1}$ and $a(n)=\delta_{n-1}$ where $F_a(s)=\zeta(s)$, $F_a(s)=\eta(s)$, and $F_a(s)=1$, formulas (3) and (4) above are simply the power series for the functions defined in formulas (6) to (11) below. This proves the validity of formulas (3) and (4) above for these particular cases, but I believe formulas (3) and (4) are more generally applicable to any definition of $a(n)$ for which the Dirichlet series $F_a(s)=\sum\limits_n\frac{a(n)}{n^s}$ converges for $\Re(s)\ge 2$. All formulas below are for $x\ge 0$, but $\tilde{f}_a'(x)$ and $\tilde{f}_a'(x)$ are actually even and odd functions respectively.

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$\quad a(n)=1 \text{ where } F_a(s)=\zeta(s)$:

$$\tilde{f}_a'(x)=\sum\limits_n\delta(x-n)=\underset{f\to\infty}{\text{lim}}\left(-\frac{\sin (2 f \pi x)}{\pi x}+\sum\limits_{n=1}^f (\cos(2 n \pi x)+\cos(2 (n-1) \pi x))\right)\tag{6}$$

$$\tilde{f}_a(x)=\sum\limits_n\theta(x-n)=\underset{f\to\infty}{\text{lim}}\left(-\frac{\text{Si}(2 f \pi x)}{\pi}+\sum\limits_{n=1}^f \left(\frac{\sin(2 n \pi x)}{2 n \pi}+x\ \text{sinc}(2 (n-1) \pi x)\right)\right)\tag{7}$$

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$\quad a(n)=(-1)^{n-1} \text{ where } F_a(s)=\eta(s)$:

$$\tilde{f}_a'(x)=\sum\limits_n (-1)^{n-1}\delta(x-n)=\underset{f\to\infty}{\text{lim}}\left(\frac{\sin(2 f \pi x)}{\pi x}-2 \sum\limits_{n=1}^f \cos((2 n-1) \pi x)\right)\tag{8}$$

$$\tilde{f}_a(x)=\sum\limits_n (-1)^{n-1}\theta(x-n)=\underset{f\to\infty}{\text{lim}}\left(\frac{\text{Si}(2 f \pi x)}{\pi }-\frac{2}{\pi}\sum\limits _{n=1}^f \frac{\sin ((2 n-1) \pi x)}{2 n-1}\right)\tag{9}$$

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$\quad a(n)=\delta_{n-1} \text{ where } F_a(s)=1$:

$$\tilde{f}_a'(x)=\delta(x-1)=\underset{f\to\infty}{\text{lim}}\left(\frac{\sin(2 f \pi (x+1))}{\pi (x+1)}+\frac{\sin(2 f \pi (x-1))}{\pi (x-1)}\right)\tag{10}$$

$$\tilde{f}_a(x)=\theta(x-1)=\underset{f\to\infty}{\text{lim}}\left(\frac{\text{Si}(2 f \pi (x+1))+\text{Si}(2 f \pi (x-1))}{\pi }\right)\tag{11}$$

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Figures (1) to (3) below illustrated formulas (7), (9), and (11) above in orange overlaid on the corresponding blue reference function where all three formulas are evaluated at $f=10$. The red discrete portion of the three figures below represent the evaluation of the three formulas at integer values of $x$. Note that formulas (7) and (9) illustrated in Figures (1) and (2) below both evaluate to $0$ at $x=0$ which is atypical of the Fourier series related to these functions. This is because formulas (6) and (8) are both missing a tooth of a Dirac comb at $x=0$.

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Figure (1): Illustration of formula (7) for $\tilde{f}_a(x)=\sum\limits_n\theta(x-n)$

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Figure (2): Illustration of formula (9) for $\tilde{f}_a(x)=\sum\limits_n (-1)^{n-1}\theta(x-n)$

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Figure (3): Illustration of formula (11) for $\tilde{f}_a(x)=\theta(x-1)$

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Formulas (6) to (9) above lead to the following formulas for the Riemann zeta function $\zeta(s)$ and Dirichlet eta function $\eta(s)$. Formulas (12) and (13) for $\zeta(s)$ and $\eta(s)$ can also be used to derive formulas for $\zeta(s)$ and $\eta(s)$ which converge for $\Re(s)>-1$.

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$$\zeta(s)=\underset{f\to\infty}{\text{lim}}\left(2^s \pi^{s-1} \sin\left(\frac{\pi s}{2}\right) \Gamma(1-s) \left(-\frac{f^s}{s}+\frac{1}{2} \left(1+\sum\limits_{n=2}^f \left(n^{s-1}+(n-1)^{s-1}\right)\right)\right)\right),\ \Re(s)<2\tag{12}$$

$$\eta(s)=\underset{f\to\infty}{\text{lim}}\left(2 \pi^{s-1} \sin\left(\frac{\pi s}{2}\right)\ \Gamma(1-s) \left(\frac{2^{s-1} f^s}{s}-\sum\limits_{n=1}^f (2 n-1)^{s-1}\right)\right),\ \Re(s)<2\tag{13}$$

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Figure (4) below illustrates formula (13) for $\eta(s)$ evaluated at $f=100$ and $f=1000$ in orange and green overlaid on the blue reference function.

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Figure (4): Illustration of formula (13) for $\eta(s)$ evaluated at $f=100$ and $f=1000$ (orange and green)

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Figure (5) and (6) below illustrate the real and imaginary parts of formula (13) for $\eta(s)$ evaluated along the critical line $s=\frac{1}{2}+i\,t$ in orange overlaid on the blue reference function where formula (13) is evaluated at $f=1000$ for both figures. The red discrete portion of the two figures below represent the evaluation of formula (13) at the first ten non-trivial zeta zeros in the upper-half plane.

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Figure (5): Illustration of real part of formula (13) for $\eta(\frac{1}{2}+i\,t)$

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Figure (6): Illustration of imaginary part of formula (13) for $\eta(\frac{1}{2}+i\,t)$