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Take $p$ as prime, $\text{li}(x)$ as logarithmic integral and

$$ R(x)=\sum_{p\leq x}\frac{\ln p}p-\ln x $$

Without using Mertens' theorem find

$$ \int_0^x\frac{tR'(t)}{\ln t}dt $$

I tried using integration by parts and got stuck here:

$$ \frac{xR(x)}{\ln x}-\int_0^x\sum_{p\leq t}\frac{\ln p}p\left(\frac1{\ln t}-\frac1{\ln^2 t}\right)dt+x-\text{li}(x) $$

where that integral is the term I can't solve!

An approximate solution is also welcome!

Gerry Myerson
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Shivanshu
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  • What ARE you allowed to use? – Igor Rivin Jan 03 '14 at 00:52
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    If you are going to post here regularly, I suggest learning enough to do some simple formatting. See, e.g., http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Gerry Myerson Jan 03 '14 at 00:56

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Note that $$ \sum_{p\leq t}\frac{\ln p}{p} $$ remains constant for $t$ between two successive primes. So if we write all the primes smaller than $x$: $$ p_1 \leq p_2 \leq\ldots\leq p_n\leq x, $$ then we can split up the integral into intervals: $$ \begin{multline} \frac{xR(x)}{\ln x}- \sum_{k=1}^{n-1}\sum_{j=1}^k\frac{\ln p_j}{p_j}\int_{p_k}^{p_{k+1}}\left(\frac1{\ln t}-\frac1{\ln^2 t}\right)dt\\ - \sum_{j=1}^n\frac{\ln p_j}{p_j}\int_{p_n}^{x}\left(\frac1{\ln t}-\frac1{\ln^2 t}\right)dt + x-\text{li}(x), \end{multline} $$ which is $$ \begin{multline} \frac{xR(x)}{\ln x}- \sum_{k=1}^{n-1}\sum_{j=1}^k\frac{\ln p_j}{p_j}\left(\frac{p_{k+1}}{\ln p_{k+1}} - \frac{p_{k}}{\ln p_{k}}\right)\\ - \sum_{j=1}^n\frac{\ln p_j}{p_j}\left(\frac{x}{\ln x} - \frac{p_{n}}{\ln p_{n}}\right) + x -\text{li}(x), \end{multline} $$ and it reduces to $$ \begin{multline} - \sum_{k=1}^{n-1}\sum_{j=1}^k\frac{\ln p_j}{p_j}\left(\frac{p_{k+1}}{\ln p_{k+1}} - \frac{p_{k}}{\ln p_{k}}\right) + \sum_{j=1}^n\frac{\ln p_j}{p_j}\frac{p_{n}}{\ln p_{n}} -\text{li}(x). \end{multline} $$

Pulsar
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