# A Perron-like formula

From basic Dirichlet series ,if $f(s)=\sum_{n=1}^{\infty}\frac{a_n}{n^s},s=\sigma+it , \sigma>\sigma_0$ where $\sigma_0$ is the abscissa of convergence, we know that : $$\sum_{n<x}\frac{a_n}{n^s}=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i \infty}f(s+z) \frac{x^z}{z}dz, c>0 ,c>\sigma-\sigma_0$$. I tried using some other integrand to create a similar formula and i came to this: $$\frac{1}{\pi}\sum_{n=1}^{\infty}\frac{a_n}{n^s}\log\left(1+\frac{1}{n^2}\right)=$$ $$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}-\frac{f(s+2z)}{z\sin(\pi z)} dz$$ , for $\Re{z}>0$ . In order to get the result , one has to compute the integral on the right side. I used as a contour the semicircle from the right side of the vertical line $(c-iT,c+iT)$ which contains the poles $1,2,3,...,n,...$ of the integrand function. I think my result is right , but can somebody confirm it?

• It suffices to check the formula when $a_n=0$ for all $n$ except $a_k=1$ for some positive integer $k$. Have you done that calculation? Commented Nov 24, 2016 at 0:23
• @GregMartin yes, but your argument is valid only when assuming everything converges absolutely Commented Nov 24, 2016 at 5:38
• Yes this is an assumption in my formula . I took it into account. Thanks Commented Nov 24, 2016 at 7:19

There is a harmonic sum hiding here which we now evaluate.

Put $$L(s) = \sum_{n\ge 1} \frac{a_n}{n^s}$$ with abscissa of convergence $\sigma_0$ and introduce

$$S(x; z)= \sum_{n\ge 1} \frac{a_n}{n^z} \log\left(1+\frac{1}{x^2 n^2}\right)$$

The sum term is harmonic and may be evaluated by inverting its Mellin transform.

Recall the harmonic sum identity $$\mathfrak{M}\left(\sum_{k\ge 1} \lambda_k g(\mu_k x);s\right) = \left(\sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} \right) g^*(s)$$ where $g^*(s)$ is the Mellin transform of $g(x).$

In the present case we have $$\lambda_k = \frac{a_k}{k^z}, \quad \mu_k = k \quad \text{and} \quad g(x) = \log\left(1+\frac{1}{x^2}\right).$$

We need the Mellin transform $g^*(s)$ of $g(x)$ which is $$\int_0^\infty \log\left(1+\frac{1}{x^2}\right) x^{s-1} dx \\ = \left[\log\left(1+\frac{1}{x^2}\right) \frac{x^s}{s}\right]_0^\infty + 2 \int_0^\infty \frac{1}{x(1 + x^2)} \frac{x^s}{s} dx \\ = \frac{2}{s} \int_0^\infty \frac{1}{1 + x^2} x^{s-1} dx.$$

The fundamental strip of the term from the integration by parts is $\langle 0, 2\rangle,$ which is also the fundamental strip of the second Mellin transform, call it $h^*(s)$ of $h(x).$

The latter transform is simple -- use a keyhole contour with the branch of the logarithm producing arguments between $0$ and $2\pi$ (branch cut on the positive real axis) and two poles at $x=\pm i.$ Above the cut we obtain $h^*(s)$ and below we put $x^{s-1} = \exp(\log(x)(s-1))$ to obtain $-h^*(s) \exp(2\pi i(s-1))$ so that

$$h^*(s) (1-\exp(2\pi i s)) = 2\pi i \times \frac{1}{2i} (\exp(i\pi (s-1)/2) - \exp(3 i\pi(s-1)/2)) \\ = 2\pi i \times \frac{1}{2i} (-i\exp(i\pi s/2) - i\exp(3 i\pi s/2))$$

and hence

$$h^*(s) = -i\pi\frac{\exp(i\pi s/2) + \exp(3 i\pi s/2)} {1-\exp(2\pi i s)} = -i\pi\frac{\exp(-i\pi s/2) + \exp(i\pi s/2)} {\exp(-\pi i s)-\exp(\pi i s)} \\ = i\pi \frac{2\cos(\pi s/2)}{2i\sin(\pi s)} = \pi \frac{\cos(\pi s/2)}{\sin(\pi s)} = \frac{\pi}{2} \frac{1}{\sin(\pi s/2)}.$$

Therefore $$g^*(s) = \frac{\pi}{s\sin(\pi s/2)}.$$

It follows that the Mellin transform $Q(s)$ of the harmonic sum $S(x;z)$ is given by

$$Q(s) = \frac{\pi}{s\sin(\pi s/2)} L(z+s) \quad\text{because}\quad \sum_{k\ge 1} \frac{\lambda_k}{\mu_k^s} = \sum_{k\ge 1} \frac{a_k}{k^z} \frac{1}{k^s} = L(z+s)$$ for $\Re(z+s) > \sigma_0.$

The Mellin inversion integral here is $$\frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} Q(s)/x^s ds$$

where we choose $c$ in the intersection of $\langle 0,2\rangle$ and $\Re(z+s) \gt \sigma_0.$ Putting $x=1$ we obtain the formula

$$\bbox[5px,border:2px solid #00A000]{ \sum_{n\ge 1} \frac{a_n}{n^z} \log\left(1+\frac{1}{n^2}\right) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{\pi}{s\sin(\pi s/2)} L(z+s) \; ds.}$$

Now of course we need to check the usefulness of this result.

For example, fixing first $z=4$ and $a_n = 1$ we get $L(s) = \zeta(s)$ and for $c$ we need $\Re(4+s) \gt 1$ so we may take $c=3/2$ to get

$$\sum_{n\ge 1} \frac{1}{n^4} \log\left(1+\frac{1}{n^2}\right) = \frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \frac{\pi}{s\sin(\pi s/2)} \zeta(s+4) \; ds.$$

Shifting the integral to the left for an expansion about zero yields

$$\frac{1}{4} + \frac{\pi^2}{6} - \frac{\pi}{3} + 2\zeta'(4) \approx 0.709913984$$

when the correct value is $0.7087317181.$ This simply says that the remainder integral does not vanish. Here we observe cancelation of the poles from the sine term by the trivial zeroes of the zeta function. With $z$ an odd integer numerics indicate an expansion that does not converge.

As another example we get

$$\sum_{n\ge 1} \frac{\tau(n)}{n^4} \log\left(1+\frac{1}{n^2}\right) = \frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \frac{\pi}{s\sin(\pi s/2)} \zeta^2(s+4) \; ds$$

with the divisor function. The expansion abouz zero produces

$$-\frac{1}{8} + \frac{\pi^4}{36} + \frac{2}{45} \pi^4 \zeta'(4) - \frac{2}{3}\gamma\pi - \frac{\pi}{9} \approx 0.7244875166$$

when the correct value is $0.7246090404.$

Finally consider the sum-of-divisors function to obtain

$$\sum_{n\ge 1} \frac{\sigma(n)}{n^4} \log\left(1+\frac{1}{n^2}\right) = \frac{1}{2\pi i} \int_{3/2-i\infty}^{3/2+i\infty} \frac{\pi}{s\sin(\pi s/2)} \zeta(s+4) \zeta(s+3) \; ds$$

We get

$$-\frac{1}{48} + \frac{1}{6}\gamma\pi^2 + \zeta'(2) + \frac{\pi^2}{12} + \frac{1}{45} \pi^4 \zeta'(3) + 2\zeta'(4) \zeta(3) + \frac{\pi}{6} \approx 0.7426221350$$

with the true value being $0.74267514.$

Using a function like $2^{v_2(n)}$ with $v_2(n)$ the largest power of two that divides $n$ we get

$$L(s) = \zeta(s) \frac{1-1/2^{s}}{1-2/2^{s}}$$

and a Fourier series due to the poles at $\rho_k = 1+2\pi i k/\log(2)$ appears. We have deliberately refrained from using arithmetic functions whose Dirichlet series contains inverse zeta function terms, where the non-trivial zeroes of the zeta function enter into the game.

Summary. An interesting formula whose use for numeric approximations is limited and which does not appear to produce novel closed forms, possibly meriting additional investigation e.g. for smaller $x$ that yield better convergence (when we take $x=1/10$ in the first example we obtain $4.8618872988137896537$ which has a remarkable $20$ good digits, a phenomenon that was observed with the other two examples as well) or cases where $L(s)$ in conjunction with the sine term might produce a functional equation.

• sorry but I don't understand why you need such a loing answer Commented Nov 24, 2016 at 5:32
• Wow great answer and very useful results. Good summary also. Thank you for showing so much interest in my question . Your insight was very useful. I am undergraduate and i would like to study more about Dirichlet Series and L functions. Do you have a book to suggest? Commented Nov 24, 2016 at 7:13
• This is very kind of you considering I did not answer your question exactly! I recommend Flajolet & Sedgewick, Mellin Transform Asymptotics, Technical Report RR 2956, INRIA, Roquencourt, 1996. Commented Nov 24, 2016 at 14:43

If $F(s) = \sum_{k=1}^\infty a_k k^{-s}$ converge abslutely for $Re(s) > 0$ then on $Re(s) > \epsilon$ : $F(s)$ is analytic and $|F(s)| = \mathcal{O}(1)$ so that

• For $c \in (0,1)$ and $Re(s) > 0$ : $$\int_{c - i\infty}^{c+i \infty} \frac{F(s+2z)}{z \sin(\pi z)}dz$$ converges absolutely.
• With $R_{c,d,T}$ the boundary of the rectangle $c\pm i T,d \pm i T$ ($d > c,d \not \in \mathbb{Z}$) you have by the residue theorem $$\int_{R_{c,d,T}}\frac{F(s+2z)}{z \sin(\pi z)}dz = 2i\pi\sum_{n \in (c,d)} Res(\frac{F(s+2z)}{z \sin(\pi z)}, n) = 2i\pi\sum_{n \in (c,d)} (-1)^n\frac{F(s+ 2n)}{n}$$

• and since $\frac{F(s+2z)}{z \sin(\pi z)}$ decreases exponentially as $Im(z) \to \infty$ and $\int_{d-iT}^{d+iT} \frac{F(s+z)}{z \sin(\pi z)}dz \to 0$ as $d \to \infty$ and that everything converges absolutely, you have

$$\int_{c - i\infty}^{c+i \infty} \frac{F(s+2z)}{z \sin(\pi z)}dz= \lim_{d,T \to \infty}\int_{R_{c,d,T}}\frac{F(s+2z)}{z \sin(\pi z)}dz = 2i\pi\sum_{n=1}^\infty (-1)^n\frac{F(s+2 n)}{ n}$$ $$= 2i\pi\sum_{k=1}^\infty a_k k^{-s}\sum_{n=1}^\infty(-1)^n \frac{k^{- 2n}}{ n}= 2i\pi\sum_{k=1}^\infty a_k k^{-s} \log(1+k^{-2})$$

The conclusion is that you need $F$ to be analytic, and everything to converges (if it doesn't converge absolutely, you have to be very careful when taking the $\lim_{d, T \to\infty}$ and when inverting $\sum_k \sum_n$)

• O suppose f is a convergent dirichlet series in the hal plane of absolute convergence. In the proof of the formula i consider the series absolutely convergent in order to change the integration-summation order Commented Nov 24, 2016 at 7:08