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 $\zeta(s)$
0
$\begingroup$
analytic-number-theory
-
0Please 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). – 2011-12-12
-
0$\psi(x)$ is the Chebyshev function here, no? – 2011-12-12
-
0Yes, $\psi$ is the Chebyshev function – 2011-12-12
1 Answers
3
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.)
-
0And a very nice book it is! – 2011-12-12