# An application of Mobius Inversion in a paper of Shintani

I've been reading about Shintani zeta functions and in particular with respect to finding the density of cubic discriminants as in the theorem of Davenport-Heilbronn. In Shintani's paper "On zeta-functions associated with the vector space of quadratic forms" [Tokyo Univ. J. Fac. Sci Sect. 1A Math 1975], in the proof of Theorem 4, Shintani writes

Hence, we have (by earlier results in this paper and in a previous paper): $$\sum_{nk^4 \leq x} h_r(n) = 2^{-1}\zeta(2)\zeta(4)x + O(x^{2/3 + \epsilon}) \qquad (x \to +\infty, \forall \epsilon > 0)$$ An application of the Mobius inversion formula now yields that
$$\sum_{n \leq x} h_r(n) = 2^{-1}\zeta(2)x + O(x^{2/3 + \epsilon})$$

When I think of Mobius Inversion, I think of two things: \begin{align*} g(n) &= \sum_{d \mid n} f(n/d) \iff f(n) = \sum_{d \mid n} \mu(n)g(n/d) \quad \text{or}\\\\ g(x) &= \sum_{n \leq x}f(n/x) \iff f(x) = \sum_{n \leq x} \mu(n)g(n/x) \end{align*}

But I don't see how I can use these here. Unfortunately, I also know that there are many things that might be called Mobius Inversion. This is one of those steps that taunts me. Qualitatively, we remove the fourth-power condition and end up losing a factor of $\zeta(4)$, and that feels very reasonable.

Further, in playing with it for a while, I re-stumbled upon the fact that $\displaystyle \dfrac{1}{\zeta(s)} = \sum_n \dfrac{\mu(n)}{n^s}$ (easy, but which I did not originally remember) and thus that $\displaystyle \dfrac{1}{\zeta(4)} = \sum \dfrac{\mu(n)}{n^4}$.

To make my question explicit - how can we arrive at the second equation from the first?

## 1 Answer

Let $g(n)$ denote the indicator function for the fourth powers. Then your sum equals

$$\sum_{n\leq x}\left(h_{r}*g\right)(n),$$

where $*$ denotes Dirichlet convolution. We may rewrite the given asymptotic as

$$\sum_{k\leq x}g(k)\sum_{n\leq\frac{x}{k}}h_{r}(n)=2^{-1}\zeta(2)\zeta(4)x+O\left(x^{2/3+\epsilon}\right),$$

noticing that this is of the form

$$G(x)=\sum_{n\leq x}\alpha(n)F\left(\frac{x}{n}\right).$$

Mobius inversion tells us that

$$F(x)=\sum_{n\leq x}\alpha^{-1}(n)G\left(\frac{x}{n}\right),$$

where $\alpha^{-1}$ is the multiplicative inverse of $\alpha$ with respect to Dirichlet convolution. Applying Mobius inversion to our sum, we have that

$$\sum_{n\leq x}h_{r}(n)=\sum_{j^{4}\leq x}\mu(j)\sum_{n\leq\frac{x}{j^{4}}}\left(h_{r}*g\right)(n)$$

which equals

$$2^{-1}\zeta(2)\zeta(4)x\sum_{j^{4}\leq x}\frac{\mu(j)}{j^{4}}+O\left(x^{2/3+\epsilon}\left(\sum_{j^{4}\leq x}\frac{1}{j^{4}}\right)\right)$$

$$=2^{-1}\zeta(2)x+O\left(x^{2/3+\epsilon}\right).$$

Notice that the Dirichlet inverse to the function $g(n)$, the indicator function for the fourth powers, is in some sense the mobius function on fourth powers.