I would like to know the value of the following integral:

$$\displaystyle\int_2^\infty ({\theta(y)-y})\frac{\mathrm d}{\mathrm dy}\frac{\ln(y-1)}{\ln(y)} \mathrm dy$$

(Where $\theta(y)$ is Chebyshev's First Function)

It appears on (4.16) of J. Barkley and L. Schoenfeld "Approximated Formulas for some functions of Prime Numbers".

I do not know where to start from, and neither Mathematica nor Mathlab seem to help.

How can I find at least an approximated value for it?

  • $\begingroup$ By parts maybe? $\endgroup$ – Turing Jan 8 '17 at 23:25
  • $\begingroup$ Where does the $\sum\limits_{p\leqslant y} \frac{1}{p}$ come from? $\endgroup$ – Daniel Fischer Jan 9 '17 at 19:01
  • $\begingroup$ @Daniel Integrating by parts, it is the derivative of $\theta(y)$ $\endgroup$ – user3141592 Jan 9 '17 at 19:07
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    $\begingroup$ That's wrong. $\theta$ is constant between consecutive primes, and has a jump of height $\ln p$ at a prime $p$. Its distributional derivative is hence $\sum (\ln p)\delta_p$, where $\delta_a$ is the Dirac distribution at $a$. The classical derivative doesn't exist at primes, and is $0$ everywhere else, so that isn't useful here. If you integrate by parts, one term you get is $\int_2^k \frac{\ln (y-1)}{\ln y}\,d\theta(y)$, which is a Riemann-Stietjes integral. Due to the nature of $\theta$, you can also write that as $\sum\limits_{p\leqslant k}\frac{\ln (p-1)}{\ln p}\cdot \ln p$, which can $\endgroup$ – Daniel Fischer Jan 9 '17 at 19:16
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    $\begingroup$ be simplified to $\sum\limits_{p\leqslant k} \ln (p-1)$. Unfortunately, that doesn't look too helpful yet. $\endgroup$ – Daniel Fischer Jan 9 '17 at 19:16

Might be helpful. You have an integral of the form

$$\int f g'\ \text{d} y$$

Use by parts integration to get

$$\underbrace{\left(\theta(y) - y)\right)\frac{\ln(y-1)}{\ln(y)}\bigg|_2^{+\infty}}_{0} - \int_2^{+\infty} \left(\theta(y) - y\right)'\ \frac{\ln(y-1)}{\ln(y)}\ \text{d}y$$


$$\frac{\text{d}}{\text{d}y} \theta(y) = \sum_{p\leq y} \frac{1}{p}$$

$$-\left(\sum_{p\leq y} \frac{1}{p} - 1\right)\int_2^{+\infty} \frac{\ln(y-1)}{\ln(y)}\ \text{d}y$$

Your problem arises when you try to calculate the last integral, which does diverge.

We can approach it with a consideration like at infinity the integral approaches to the value $1$.

I think now it's your turn to make some approximation to that integral.

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  • $\begingroup$ Thank you! In your last step, why do you use $$-\left(\sum_{p\leq y} \frac{1}{p} - 1\right)\int_2^{+\infty} \frac{\ln(y-1)}{\ln(y)}\ \text{d}y$$ instead of $$-\left(\sum^\infty \frac{1}{p} - 1\right)\int_2^{+\infty} \frac{\ln(y-1)}{\ln(y)}\ \text{d}y$$ ? $\endgroup$ – user3141592 Jan 9 '17 at 6:42
  • $\begingroup$ Also, another problem I have is that, writting $$-\left(\sum_{p\leq y} \frac{1}{p} - 1\right)\int_2^{+\infty} \frac{\ln(y-1)}{\ln(y)}\ \text{d}y$$ as $$\left(\sum_{p\leq y} \frac{p-1}{p} \right)\int_2^{+\infty} \frac{\ln(y-1)}{\ln(y)}\ \text{d}y$$ , both terms diverge, and so its product, isn't it? $\endgroup$ – user3141592 Jan 9 '17 at 7:08
  • $\begingroup$ @user3141592 For the first question, I wrote the sum in that way because the definition I remember about the $\theta(y)$ function is that it's the sum of the logarithm of all the prime numbers less or equal than $y$. Hence if $y$ is a finite number the sum must be finite! For the second question, the $-1$ term IS NOT inside the sum. There would be less confusion if I had written it as $$\left(\sum_{p\leq y} \frac{1}{p}\right) - 1$$ $\endgroup$ – Turing Jan 9 '17 at 12:09
  • $\begingroup$ But since $$\sum_{p\le k} \frac{1}{p}$$ is arout $$ln(ln(k))$$ and the integral $$\int_2^{k} \frac{\ln(y-1)}{\ln(y)}\ \text{d}y is around $k$, it does not converge, doest it? $\endgroup$ – user3141592 Jan 9 '17 at 12:48
  • $\begingroup$ And is it sure that $$\theta(\infty)-\infty = 0$$?? $\endgroup$ – user3141592 Jan 9 '17 at 14:03

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