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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 $\zeta(s)$?

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Please feel free to edit if this isn't what you had in mind (you need to put {} around things you want LaTeX to treat as a single group). –  Zev Chonoles Dec 12 '11 at 6:15
    
$\psi(x)$ is the Chebyshev function here, no? –  J. M. Dec 12 '11 at 6:20
    
Yes, $\psi$ is the Chebyshev function –  Rob Dec 12 '11 at 6:20

1 Answer 1

up vote 3 down vote accepted

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 $$ \Psi(x)=\sum_{n<x}\Lambda(n). $$

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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And a very nice book it is! –  lhf Dec 12 '11 at 18:18

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