Dirichlet series inversion

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:

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

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 Riemmann prime counting function:

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

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 ${\displaystyle \Re (s):=\sigma }$:

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

If we are interested in expressing formulas for the Dirichlet inverse of f, denoted by ${\displaystyle f^{-1}(n)}$ whenever ${\displaystyle f(1)\neq 0}$, we write ${\displaystyle D_{f}(s)=1+s\cdot A_{f}(s)}$. Then we have by absolute convergence of the DGF for any ${\displaystyle \Re (s)>\sigma _{0,f}}$ that

${\displaystyle {\begin{aligned}\int {\frac {x^{\imath t}}{D_{f}(\sigma +\imath t)}}\,dt&=\int \left(\sum _{j\geq 0}(\sigma +\imath t)^{j}\times \sum _{k=0}^{j}(-1)^{k}D_{f}(\sigma +\imath t)^{k}{\frac {\log ^{j-k}(x)}{(j-k)!}}\right)\,dt.\end{aligned}}}$

Now we can call on integration by parts to see that if we denote by ${\displaystyle F^{(-m)}(x)=\sum _{n\geq 0}{\frac {F^{(n)}(0)}{n!}}x^{n+m}\times {\frac {n!}{(n+m)!}}}$ denotes the ${\displaystyle m^{th}}$ antiderivative of F, for any fixed non-negative integers ${\displaystyle k\geq 0}$, we have

${\displaystyle \int (ax+b)^{k}\cdot F(ax+b)\,dx=\sum _{j=0}^{k}{\frac {k!}{a\cdot j!}}(-1)^{k-j}t^{j}F^{(j+1-k)}(ax+b).}$

Thus we obtain that

${\displaystyle {\begin{aligned}\int _{-T}^{T}{\frac {x^{\imath t}}{D_{f}(\sigma +\imath t)}}\,dt&={\frac {1}{\imath }}\left(\sum _{j\geq 0}\sum _{k=0}^{j}\sum _{m=0}^{k}{\frac {k!}{m!}}(-1)^{m}(\sigma +\imath t)^{m}\left[D_{f}^{k}\right]^{(j+1-k)}(\sigma +\imath t){\frac {\log ^{j-k}(x)}{(j-k)!}}\right){\Biggr |}_{t=-T}^{t=+T}.\end{aligned}}}$

We also can relate the iterated integrals for the ${\displaystyle k^{th}}$ antiderivatives of F by a finite sum of k single integrals of power-scaled versions of F:

${\displaystyle F^{(-k)}(x)=\sum _{i=0}^{k-1}{\binom {k-1}{i}}{\frac {(-1)^{i}}{(k-1)!}}\int {\frac {F(x)}{x^{i}}}\,dx.}$

In light of this expansion, we can then write the partially limiting T-truncated Dirichlet series inversion integrals at hand in the form of

${\displaystyle {\begin{aligned}{\frac {1}{2T}}\int _{-T}^{T}{\frac {x^{\imath t}}{D_{f}(\sigma +\imath t)}}\,dt&={\frac {1}{2T\cdot \imath }}\left(\sum _{j\geq 0}\sum _{k=0}^{j}\sum _{m=0}^{k}\sum _{n=0}^{j-k}{\binom {j-k}{n}}{\frac {(-1)^{k+n+m}\cdot k!}{(j-k)!\cdot m!}}(\sigma +\imath t)^{m}{\frac {\log ^{j-k}(x)}{(j-k)!}}\int _{0}^{\sigma +\imath t}\left[A_{f}^{k}\right](v){\frac {dv}{v^{n}}}\right){\Biggr |}_{t=-T}^{t=+T}\\&={\frac {1}{2T\cdot \imath }}\left(\sum _{j\geq 0}\sum _{k=0}^{j}\sum _{m=0}^{k}{\frac {(-1)^{j-k}\cdot (-s)^{m}}{m!}}{\frac {\log ^{k}(x)}{k!}}\int _{0}^{1}s\cdot D_{f}^{j-k}(rs)\left(1-{\frac {1}{rs}}\right)^{k}\,dv\right){\Biggr |}_{s=\sigma -\imath T}^{s=\sigma +\imath T}\\&={\frac {s\left(e^{-s}+O_{s}(1)\right)}{2T\cdot \imath }}\int _{0}^{1}{\frac {x^{1-{\frac {1}{rs}}}}{1+D_{f}(rs)}}dr{\Biggr |}_{s=\sigma -\imath T}^{s=\sigma +\imath T}\\&={\frac {\left(e^{-s}+O_{s}(1)\right)}{2T\cdot \imath }}\int _{0}^{s}{\frac {x^{1-{\frac {1}{v}}}}{1+D_{f}(v)}}\,dv{\Biggr |}_{s=\sigma -\imath T}^{s=\sigma +\imath T}.\end{aligned}}}$

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).