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If $\psi\left(x\right)$ is the Chebyshev psi function we know that $$\psi_{0}\left(x\right)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}-\frac{\zeta'}{\zeta}\left(0\right)-\frac{1}{2}\log\left(1-x^{-2}\right)$$ where $$\psi_{0}\left(x\right)=\begin{cases} \psi\left(x\right)-\frac{1}{2}\Lambda\left(x\right), & x\textrm{ is a prime power}\\ \psi\left(x\right), & \textrm{otherwise} \end{cases}$$ and $x>1$. Now if $0<x<1$ the identity does not hold due to the presence of log function. But the other terms are defined.

Question: Is it known an explicit formula for $\psi\left(x\right)$ if $0<x<1$? I mean $$\psi\left(x\right)=x-\sum_{\rho}\frac{x^{\rho}}{\rho}+\textrm{something}.$$

I ask this question since it would be very useful for my work. I have only found this identity $$\sum_{n\leq1/x}\frac{\Lambda\left(n\right)}{n}=\log\left(\frac{1}{x}\right)-\gamma+\sum_{\rho}\frac{x^{\rho}}{\rho}-x+\frac{1}{2}\log\left(\frac{1+x}{1-x}\right),\,0<x<1$$ but I don't see how to link that to the psi function.

Meaning of this question I have an integral of this type $$\int_{0}^{\infty}\psi\left(t\right)f\left(t,z\right)dt$$ where $z$ is a complex number with positive real part. I would to use the explicit formula for $\psi\left(t\right)$ and integrate termwise but I can only use it for $t>1$. And using another technique for evaluate the integral I get some terms that match with the integration termwise for $t>0$. Unfortunately the other technique is not useful for a complete evaluation. So I'm asking if it possible to extend in some way the formula.

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  • $\begingroup$ For $x>0$ there's always a formula for $\psi_0^{\phantom .}(x)$ as a contour integral with a factor $x^s \, ds$. When $x>1$ you make this smaller by moving the contour to the left, encountering poles that yield the contributions $x - \sum_\rho x^\rho/\rho$. When $x<1$, you make $x^s$ smaller by moving the contour to the right and encounter no singularities, so as expected $\psi_0^{\phantom .}(x) = 0$ in that case. $\endgroup$
    – Noam D. Elkies
    Commented Feb 24, 2017 at 3:02
  • $\begingroup$ @NoamD.Elkies Of course, but I'm interested in a formula involving the terms $x-\sum_{\rho}\frac{x^{\rho}}{\rho}$. In other words, a funny way to write $0$ using that terms. I repeat, studying my integral in two different ways I saw that the the first terms match with a calculation of the explicit formula for $t>0$, so I'm asking if it is possible to extend it in some way, keeping $x-\sum_{\rho}\frac{x^{\rho}}{\rho}$ (who have perfectly sense fot $0<x<1$). $\endgroup$
    – User
    Commented Feb 24, 2017 at 7:38

2 Answers 2

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The Chebyshev function $\psi(x)$ is constant zero for $0<x<1$. It does not get more explicit than that.

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  • $\begingroup$ Yes, but it is $0$ also for $1<x<2$. And in that case we have also the explicit formula. $\endgroup$
    – User
    Commented Feb 22, 2017 at 15:54
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    $\begingroup$ @User: Yes, but the explicit formula $\psi(x)=0$ is simpler. In general, the explicit formula you are talking about is useful for large $x$, not for small $x$. I don't see any reason why you would like to bring in the zeta zeros for expressing the constant zero function. $\endgroup$
    – GH from MO
    Commented Feb 22, 2017 at 15:57
  • $\begingroup$ I have an integral of this type $$\int_{0}^{\infty}\psi\left(t\right)f\left(t,z\right)dt$$ where $z$ is a complex number with positive real part. I would to use the explicit formula for $\psi\left(t\right)$ and integrate termwise but I can only use it for $t>1$. And using another technique for evaluate the integral I get some terms that match with the integration termwise for $t>0.$ Unfortunately the other technique is not useful for a complete evaluation. So I'm asking if it possible to extend in some way the formula. $\endgroup$
    – User
    Commented Feb 22, 2017 at 16:06
  • $\begingroup$ @User: I see. I think it would more useful if you asked a more specific question about your integral. Whatever formula we generate for $\psi(t)$ valid for $0<t<1$, it will not harmonize with the original formula valid for $t>1$, exactly because $\log(1-t^{-2})$ has a serious singularity at $t=1$ (and so does the sum over the zeros). $\endgroup$
    – GH from MO
    Commented Feb 22, 2017 at 16:14
  • $\begingroup$ Thank you. I proved that the integral converges if we use the complex log instead of the real log. If all terms converges can I use some type of analytic continuation? I essentially split the integral from $0$ to $1$ and $1$ to $\infty$ and all terms converges due to the particular form of $f(t,z).$ $\endgroup$
    – User
    Commented Feb 22, 2017 at 16:17
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Indeed the explicit formula is about the Laplace transform pairs :

$$\begin{array}{cc}\frac{1}{s}&\qquad & \frac{-\zeta'(s)}{s\zeta(s)} = \frac{1}{s} \left(\frac{1}{s-1}-\sum_\rho \frac{1}{s-\rho}- \sum_{n=1}^\infty (\frac{1}{s+2n}-\frac{1}{2n})+C\right) \\ \color{red}\uparrow & &\color{red}\uparrow \\ 1_{u > 0}& &\psi(e^u) = 1_{u > 0}(e^u-\sum_\rho \frac{e^{\rho u}}{\rho}- \sum_{n=1}^\infty \frac{e^{-2n u}}{-2n}- \ln 2\pi)\end{array}$$

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