Return to Answer

I am not sure if this result is explicitly mentioned in the literature, but it certainly is classical.

Let $$R(x) = \sum_{n \leq x}{\varphi(n)} - \frac{3x^2}{\pi^2}, \qquad H(x) = \sum_{n \leq x}{\frac{\varphi(n)}{n}} - \frac{6x}{\pi^2}.$$

Then by partial summation, $$\int^{x}_{0}{\frac{R(t)}{t^2} \: dt} = H(x) - \frac{R(x)}{x}.$$ A classical result of Chowla states that $$H(x) - \frac{R(x)}{x} = O\left((\log x)^{-4}\right).$$ See Lemma 13 of S. Chowla, "Contributions to the analytic theory of numbers", Mathematische Zeitschrift 35:1 (1932), 279-299. (If you have access to Springer Link then it is available here.)

From a cursory glance of Chowla's proof, the negative powers of a logarithm stem from the prime number theorem applied to the summatory function of the Möbius function, so it is likely that this bound could be improved with more modern estimates for this.

For what it's worth, I answered a question closely related to this here.

I am not sure if this result is explicitly mentioned in the literature, but it certainly is classical.

Let $$R(x) = \sum_{n \leq x}{\varphi(n)} - \frac{3x^2}{\pi^2}, \qquad H(x) = \sum_{n \leq x}{\frac{\varphi(n)}{n}} - \frac{6x}{\pi^2}.$$

Then by partial summation, $$\int^{x}_{0}{\frac{R(t)}{t^2} \: dt} = H(x) - \frac{R(x)}{x}.$$ A classical result of Chowla states that $$H(x) - \frac{R(x)}{x} = O\left((\log x)^{-4}\right).$$ See Lemma 13 of S. Chowla, "Contributions to the analytic theory of numbers", Mathematische Zeitschrift 35:1 (1932), 279-299. (If you have access to Springer Link then it is available here or here.)

From a cursory glance of Chowla's proof, the negative powers of a logarithm stem from the prime number theorem applied to the summatory function of the Möbius function, so it is likely that this bound could be improved with more modern estimates for this.

For what it's worth, I answered a question closely related to this here.

I am not sure if this result is explicitly mentioned in the literature, but it certainly is classical.

Let $$R(x) = \sum_{n \leq x}{\varphi(n)} - \frac{3x^2}{\pi^2}, \qquad H(x) = \sum_{n \leq x}{\frac{\varphi(n)}{n}} - \frac{6x}{\pi^2}.$$

Then by partial summation, $$\int^{x}_{0}{\frac{R(t)}{t^2} \: dt} = H(x) - \frac{R(x)}{x}.$$ A classical result of Chowla states that $$H(x) - \frac{R(x)}{x} = O\left((\log x)^{-4}\right).$$ See Lemma 13 of S. Chowla, "Contributions to the analytic theory of numbers", Mathematische Zeitschrift 35:1 (1932), 279-299. (If you have access to Springer Link then it is available here.)

From a cursory glance of Chowla's proof, the negative powers of a logarithm stem from the prime number theorem applied to the summatory function of the Möbius function, so it is likely that this bound could be improved with more modern estimates for this.

For what it's worth, I answered a question closely related to this here.

I am not sure if this result is explicitly mentioned in the literature, but it certainly is classical.

Let $$R(x) = \sum_{n \leq x}{\varphi(n)} - \frac{3x^2}{\pi^2}, \qquad H(x) = \sum_{n \leq x}{\frac{\varphi(n)}{n}} - \frac{6x}{\pi^2}.$$

Then by partial summation, $$\int^{x}_{0}{\frac{R(t)}{t^2} \: dt} = H(x) - \frac{R(x)}{x}.$$ A classical result of Chowla states that $$H(x) - \frac{R(x)}{x} = O\left((\log x)^{-4}\right).$$ See Lemma 13 of S. Chowla, "Contributions to the analytic theory of numbers", Mathematische Zeitschrift 35:1 (1932), 279-299. (If you have access to Springer Link then it is available here.)

From a cursory glance of Chowla's proof, the negative powers of a logarithm stem from the prime number theorem applied to the summatory function of the Möbius function, so it is likely that this bound could be improved with more modern estimates for this.

For what it's worth, I answered a question closely related to this here.

I am not sure if this result is explicitly mentioned in the literature, but it certainly is classical.

Let $$R(x) = \sum_{n \leq x}{\varphi(n)} - \frac{3x^2}{\pi^2}, \qquad H(x) = \sum_{n \leq x}{\frac{\varphi(n)}{n}} - \frac{6x}{\pi^2}.$$

Then by partial summation, $$\int^{x}_{0}{\frac{E(t)}{t^2} \: dt} = H(x) - \frac{R(x)}{x}.$$ A classical result of Chowla states that $$H(x) - \frac{R(x)}{x} = O\left((\log x)^{-4}\right).$$ See Lemma 13 of S. Chowla, "Contributions to the analytic theory of numbers", Mathematische Zeitschrift 35:1 (1932), 279-299. (If you have access to Springer Link then it is available here.)

From a cursory glance of Chowla's proof, the negative powers of a logarithm stem from the prime number theorem applied to the summatory function of the Möbius function, so it is likely that this bound could be improved with more modern estimates for this.

For what it's worth, I answered a question closely related to this here.

I am not sure if this result is explicitly mentioned in the literature, but it certainly is classical.

Let $$R(x) = \sum_{n \leq x}{\varphi(n)} - \frac{3x^2}{\pi^2}, \qquad H(x) = \sum_{n \leq x}{\frac{\varphi(n)}{n}} - \frac{6x}{\pi^2}.$$

Then by partial summation, $$\int^{x}_{0}{\frac{R(t)}{t^2} \: dt} = H(x) - \frac{R(x)}{x}.$$ A classical result of Chowla states that $$H(x) - \frac{R(x)}{x} = O\left((\log x)^{-4}\right).$$ See Lemma 13 of S. Chowla, "Contributions to the analytic theory of numbers", Mathematische Zeitschrift 35:1 (1932), 279-299. (If you have access to Springer Link then it is available here.)

From a cursory glance of Chowla's proof, the negative powers of a logarithm stem from the prime number theorem applied to the summatory function of the Möbius function, so it is likely that this bound could be improved with more modern estimates for this.

For what it's worth, I answered a question closely related to this here.