DGFs for Dirichlet inverse functions

Recall that an arithmetic function is Dirichlet invertible, or has an inverse ${\displaystyle f^{-1}(n)}$ with respect to Dirichlet convolution such that ${\displaystyle (f\ast f^{-1})(n)=\delta _{n,1}}$, or equivalently ${\displaystyle f\ast f^{-1}=\mu \ast 1\equiv \varepsilon }$, if and only if ${\displaystyle f(1)\neq 0}$. It is not difficult to prove that is ${\displaystyle D_{f}(s)}$ is the DGF of f and is absolutely convergent for all complex s satisfying ${\displaystyle \Re (s)>\sigma _{0,f}}$, then the DGF of the Dirichlet inverse is given by ${\displaystyle D_{f^{-1}}(s)=1/D_{f}(s)}$ and is also absolutely convergent for all ${\displaystyle \Re (s)>\sigma _{0,f}}$. The positive real ${\displaystyle \sigma _{0,f}}$ associated with each invertible arithmetic function f is called the abscissa of convergence.

We also see the following identities related to the Dirichlet inverse of some function g that does not vanish at one:

${\displaystyle {\begin{aligned}(g^{-1}\ast \mu )(n)&=[n^{-s}]\left({\frac {1}{\zeta (s)D_{g}(s)}}\right)\\(g^{-1}\ast 1)(n)&=[n^{-s}]\left({\frac {\zeta (s)}{D_{g}(s)}}\right).\end{aligned}}}$

Summatory functions

Using the same convention in expressing the result of Perron's formula, we assume that the summatory function of a (Dirichlet invertible) arithmetic function ${\displaystyle f}$, is defined for all real ${\displaystyle x\geq 0}$ according to the formula

${\displaystyle S_{f}(x):={\sum _{n\leq x}}^{\prime }f(n)={\begin{cases}0,&0\leq x<1\\\sum \limits _{n<[x]}f(n),&x\in \mathbb {R} \setminus \mathbb {Z} ^{+}\wedge x\geq 1\\\sum \limits _{n\leq [x]}f(n)-{\frac {f(x)}{2}},&x\in \mathbb {Z} ^{+}.\end{cases}}}$

We know the following relation between the Mellin transform of the summatory function of f and the DGF of f whenever ${\displaystyle \Re (s)>\sigma _{0,f}}$:

${\displaystyle D_{f}(s)=s\cdot \int _{1}^{\infty }{\frac {S_{f}(x)}{x^{s+1}}}dx.}$

Some examples of this relation include the following identities involving the Mertens function, or summatory function of the Moebius function, the prime zeta function and the prime-counting function, and the Riemann prime-counting function:

${\displaystyle {\begin{aligned}{\frac {1}{\zeta (s)}}&=s\cdot \int _{1}^{\infty }{\frac {M(x)}{x^{s+1}}}dx\\P(s)&=s\cdot \int _{1}^{\infty }{\frac {\pi (x)}{x^{s+1}}}dx\\\log \zeta (s)&=s\cdot \int _{0}^{\infty }{\frac {\Pi _{0}(x)}{x^{s+1}}}dx.\end{aligned}}}$

Classical integral formula

For any s such that ${\displaystyle \sigma$ :=\Re (s)>\sigma _{0,f}} , we have that

${\displaystyle f(x)\equiv [x^{-s}]D_{f}(s)=\lim _{T\rightarrow \infty }{\frac {1}{2T}}\int _{-T}^{T}x^{\sigma +\imath t}D_{f}(\sigma +\imath t)\,dt.}$

If we write the DGF of f according to the Mellin transform formula of the summatory function of f, then the stated integral formula simply corresponds to a special case of Perron's formula. Another variant of the previous formula stated in Apostol's book provides an integral formula for an alternate sum in the following form for ${\displaystyle c,x>0}$ and any real ${\displaystyle \sigma >\sigma _{0,f}-c}$ where we denote ${\displaystyle \Re (s):=\sigma }$:

${\displaystyle {\sum _{n\leq x}}^{\prime }{\frac {f(n)}{n^{s}}}={\frac {1}{2\pi \imath }}\int _{c-\imath \infty }^{c+\imath \infty }D_{f}(s+z){\frac {x^{z}}{z}}dz.}$

Statements in the language of Mellin transformations

• The Dirichlet generating function of a sequence ${\displaystyle a(n)}$ is the Mellin transform of the sequence, evaluated at ${\displaystyle s=1}$: ${\textstyle \sum _{n=1}^{\infty }{\frac {a_{n}}{n^{s}}}=\int _{0}^{\infty }x^{s-1}\left(\sum _{n=1}^{\infty }a_{n}e^{-nx}\right)dx={\mathcal {M}}{a_{n}}(s)}$.
• The Dirichlet inverse of a sequence ${\displaystyle a(n)}$ is related to the inverse Mellin transform of its generating function: ${\textstyle a_{n}={\frac {1}{2\pi i}}\int _{c-i\infty }^{c+i\infty }{\frac {\sum _{m=1}^{\infty }{\frac {\mu (m)}{m^{s}}}{\mathcal {M}}{a_{n}}(s)}{n^{s}}}ds}$, where ${\displaystyle c}$ is a real number greater than the abscissa of convergence of the Dirichlet series ${\textstyle \sum _{n=1}^{\infty }a_{n}/n^{s}}$.
• The Mellin transform of a convolution of two sequences ${\displaystyle a(n)}$ and ${\displaystyle b(n)}$ is the product of their Mellin transforms: ${\textstyle {\mathcal {M}}{(a*b)_{n}}(s)={\mathcal {M}}{a_{n}}(s){\mathcal {M}}{b_{n}}(s)}$.
• If ${\displaystyle a(n)}$ is a sequence and ${\displaystyle f(x)}$ is a function such that the integral ${\textstyle \int _{0}^{\infty }{\frac {f(x)}{x^{s+1}}}dx}$ converges absolutely and uniformly for ${\displaystyle s}$ in some right half plane, then we can define a Dirichlet series by ${\textstyle a_{n}={\frac {1}{n^{s}}}\int _{0}^{\infty }f(x)e^{-nx}dx}$, and the Dirichlet series is the Mellin transform of ${\displaystyle f(x)}$.

A formal generating-function-like convolution lemma

Suppose that we wish to treat the integrand integral formula for Dirichlet coefficient inversion in powers of ${\displaystyle (\imath t)^{k}}$ where ${\displaystyle [(\imath t)^{k}]F(\sigma +\imath t)=F^{(k)})(\sigma )/k!}$, and then proceed as if we were evaluating a traditional integral on the real line. Then we have that

${\displaystyle {\begin{aligned}{\hat {D}}_{f}(x;\sigma ,T)&:=\int _{-T}^{T}x^{\sigma +\imath t}D_{f}(\sigma +\imath t)\,dt\\&=\sum _{m\geq 0}\int _{-T}^{T}x^{\sigma +\imath t}(\imath t)^{m}{\frac {D_{f}^{(m)}(\sigma )}{m!}}\\&=\sum _{m\geq 0}\int _{-T}^{T}t^{2m}x^{\sigma +\imath t}{\frac {(-1)^{m}D_{f}^{(2m)}(\sigma )}{(2m)!}}\,dt+\imath \times \sum _{m\geq 0}\int _{-T}^{T}t^{2m+1}x^{\sigma +\imath t}{\frac {(-1)^{m}A^{(2m+1)}(\sigma )}{(2m+1)!}}\,dt.\end{aligned}}}$

We require the result given by the following formula, which is proved rigorously by an application of integration by parts, for any non-negative integer ${\displaystyle m\geq 0}$:

${\displaystyle {\begin{aligned}{\hat {I}}_{m}(T)&:=\int _{-T}^{T}{\frac {(\imath t)^{m}}{m!}}x^{\imath t}\,dt\\&=\sum _{r=0}^{m}{\frac {\left(x^{\imath T}+(-1)^{r+1}x^{-\imath T}\right)(-\imath )^{r+1}}{\log ^{k+1-r}(x)}}\cdot {\frac {T^{r}}{r!}}\\&={\frac {\cos(T\log x)}{T\log x}}\times \sum _{r=0}^{\lfloor k/2\rfloor }{\frac {(-1)^{r+1}T^{2r}}{\log ^{k-2r}(x)(2r)!}}+{\frac {\sin(T\log x)}{T\log x}}\times \sum _{r=0}^{\lfloor k/2\rfloor }{\frac {(-1)^{r+1}T^{2r+1}}{\log ^{k-2r-1}(x)(2r+1)!}}.\end{aligned}}}$

So the respective real and imaginary parts of our arithmetic function coefficients f at positive integers x satisfy:

${\displaystyle {\begin{aligned}\operatorname {Re} (f(x))&=x^{\sigma }\times \lim _{T\rightarrow \infty }\sum _{m\geq 0}D_{f}^{(2m)}(\sigma )\cdot {\frac {{\hat {I}}_{2m}(T)}{2T}}\\\operatorname {Im} (f(x))&=x^{\sigma }\times \lim _{T\rightarrow \infty }\sum _{m\geq 0}D_{f}^{(2m+1)}(\sigma )\cdot {\frac {{\hat {I}}_{2m+1}(T)}{2T}}.\end{aligned}}}$

The last identities suggest an application of the Hadamard product formula for generating functions. In particular, we can work out the following identities which express the real and imaginary parts of our function f at x in the following forms:^[1]

${\displaystyle {\begin{aligned}\operatorname {Re} (f(x))&=\lim _{T\rightarrow \infty }\left[{\frac {x^{\sigma }}{2T}}\times {\frac {1}{2\pi }}\int _{-\pi }^{\pi }\left(D_{f}(\sigma +\imath T\cdot e^{\imath s})+D_{f}(\sigma -\imath Te^{\imath s})\right)\left(FUNC(e^{-\imath s})\right)\,ds\right]\\\operatorname {Im} (f(x))&=\lim _{T\rightarrow \infty }\left[{\frac {x^{\sigma }}{2T}}\times {\frac {1}{2\pi }}\int _{-\pi }^{\pi }\left(D_{f}(\sigma +\imath T\cdot e^{\imath s})-D_{f}(\sigma -\imath Te^{\imath s}\right)()\,ds\right].\end{aligned}}}$

Notice that in the special case where the arithmetic function f is strictly real-valued, we expect that the inner terms in the previous limit formula are always zero (i.e., for any T).