How to express the following integral: $s\int_0^{\infty}e^{-st} \psi(e^t) dt$ where $\psi(x)$ represents the second Chebyshev function, in terms of $\zeta(s)$?
Express this integral in terms of $\zeta(s)$
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analytic-number-theory
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0Yes, $\psi$ is the Chebyshev function – 2011-12-12
1 Answers
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First change the variables via $x=e^t$, $\log(x)=t$, $dx/x=dt$ and the Laplace Transform becomes the Mellin Transform. (The integral is then from $1$ to $\infty$, but since $\Psi(x)=0$ for $0\le x<1$, you can extend the integral to be from $0$ to $\infty$.)
Then substitute $ \Psi(x)=\sum_{n
Now interchange the sum and integral to get
$ s\sum_{n=1}^\infty \Lambda(n)\int_n^\infty x^{-s-1}dx. $
Finally, compute the integral and see that you have the series expansion of $ -\frac{\zeta^\prime(s)}{\zeta(s)}. $
(For details, see my book A Primer of Analytic Number Theory.)
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0And a very nice book it is! – 2011-12-12