- #1
schtruklyn
- 6
- 0
Hi. In Apostol's book "Introduction to analytic number theory", Teorem 3.2(b), Apostol proves
(1)
[itex]
\zeta (s) = \lim\limits_{x \to \infty} \left\{ \sum\limits_{n \leq x} \frac{1}{n^s} - \frac{x^{1-s}}{1-s} \right\}
[/itex]
for [itex]s[/itex] in critical strip. I know this translates to a Mellin transform integral representation
(2)
[itex]
\zeta(s) \Gamma(s)= \int\limits_0^\infty \left( \frac{1}{e^t -1}-\frac{1}{t} \right) t^{s-1} \rm{d} t
[/itex]
because it's stated as such by mathematician Petersen here, eq. (43). Petersen says (2) stems from the usual integral of zeta for [itex]\Re (s) > 1[/itex]
(3)
[itex]
\zeta(s) \Gamma(s)=\int\limits_0^\infty \frac{ t^{s-1}}{e^t -1} \rm{d} t
[/itex]
by analytic continuation, but I just can't find justification to it. It is probably trivial, but I fail to see it.
My question is:
Can someone please point me to a book-reference for Mellin transform integral representation (2), or can someone explain how one proves its validity, if proof turns trivial, possibly by use of analytic continuation or by use of (1).
Thanks
(1)
[itex]
\zeta (s) = \lim\limits_{x \to \infty} \left\{ \sum\limits_{n \leq x} \frac{1}{n^s} - \frac{x^{1-s}}{1-s} \right\}
[/itex]
for [itex]s[/itex] in critical strip. I know this translates to a Mellin transform integral representation
(2)
[itex]
\zeta(s) \Gamma(s)= \int\limits_0^\infty \left( \frac{1}{e^t -1}-\frac{1}{t} \right) t^{s-1} \rm{d} t
[/itex]
because it's stated as such by mathematician Petersen here, eq. (43). Petersen says (2) stems from the usual integral of zeta for [itex]\Re (s) > 1[/itex]
(3)
[itex]
\zeta(s) \Gamma(s)=\int\limits_0^\infty \frac{ t^{s-1}}{e^t -1} \rm{d} t
[/itex]
by analytic continuation, but I just can't find justification to it. It is probably trivial, but I fail to see it.
My question is:
Can someone please point me to a book-reference for Mellin transform integral representation (2), or can someone explain how one proves its validity, if proof turns trivial, possibly by use of analytic continuation or by use of (1).
Thanks