# Von Mangoldt function estimate

In many treatments of Vinogradov's three prime theorem, one considers the summation $S(\alpha) = \sum_{k \leq N}\Lambda(k)e^{2\pi i\alpha k}$ in place of $T(\alpha) = \sum_{p \leq N}e^{2\pi i\alpha p}$ (where this sum runs over primes). I assume these two expressions are roughly the same, but I can't seem to be able to estimate $|S(\alpha) - T(\alpha)|$. Can anyone give an estimate?

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Do you mean $T(\alpha) = \sum_{p \leq N} \log p e^{2 \pi i \alpha p}$? –  user27126 Dec 10 '12 at 5:39
I do not actually, I do know that $|\sum_{p \leq N}\log p e^{2\pi i\alpha p} - S(\alpha)| \ll N^{1/2}$. –  quant10 Dec 10 '12 at 7:18
I see, I am then just curious about why $S$ and $T$ should be roughly the same. –  user27126 Dec 10 '12 at 7:36

The reason we consider $\Lambda(n)$ rather that just the sum over primes is because this new sum is slightly nicer to work with. The idea here is that we may freely weight by a smooth function without affecting the end result significantly.
We should expect that $|S(\alpha,N)-\log N T(\alpha,N)|$ to be small, and one can try to manipulate it with partial summation. However that misses the point, and is in some sense the wrong "time" to compare the two sums.
In Vinogradov's theorem, we are looking at the primes which satisfy $k_1+k_2+k_3=N$, and we look at the slightly weighted version $$\sum_{k_1+k_2+k_3=N}\Lambda(k_1)\Lambda(k_2)\Lambda(k_3).$$ Let $\chi_{\mathcal{P}}$ be the characteristic function for the primes. Then lets it is not hard to see how the above relates to $$\sum_{k_1+k_2+k_3=N}\chi_{\mathcal{P}}(k_1)\chi_{\mathcal{P}}(k_2)\chi_{\mathcal{P}}(k_3),$$ at least in terms of the order of magnitude. This connection is easier to prove things about, rather than comparing the exponential sums.
Can you say a word on why $S(\alpha, N) \sim \log N T(\alpha, N)$, and what is the general statement behind? –  user27126 Dec 13 '12 at 11:14