Which Dirichlet Series is obtained as the Mellin transformation of
$\displaystyle \frac{1}{e^t+ 1}= \sum_{n=0}^\infty (-1)^n (e^{-t})^{n+1}$
$\displaystyle \left\{\mathcal{M}f\right\}(s) = \int_0^\infty \frac{x^{s-1}}{e^x+1}dx = \sum_{n=0}^\infty (-1)^n\int_0^\infty x^{s-1}e^{-x(n+1)} dx $
Now let $\displaystyle t=x(n+1) $ which means $\displaystyle dt=(n+1)dx $.
We then get $\displaystyle \left\{\mathcal{M}f\right\}(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^s}\int_0^\infty t^{s-1}e^{-t} dt $
Therefore $\displaystyle \left\{\mathcal{M}f\right\}(s) = \Gamma(s)\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^s} = \Gamma(s)\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} $
This Dirichlet series can then be written in terms of $\displaystyle \zeta(s) $.
Note: We require $\displaystyle x>0 $ for the original sum to converge. It also converges absolutely for $\displaystyle x>0 $ and that's why we can move the integral sign into the summation.
OK I will ask from beginning. I learned that the Mellin transformation transforms Power series in Dirichlet Series. If
and
then .
Now in my case $\displaystyle F(s) = \sum_{n=0}^\infty (-1)^n (e^{-s})^{n+1}$
Then we should have $\displaystyle f(s) = \frac{1}{\Gamma(s)} \int_0^\infty \sum_{n=0}^\infty ((-1)^n (e^{-e^{-t}})^{n+1}) * t^{s-1}dt$. How do I get your result from here?
The definition of the Mellin transformation is $\displaystyle \left\{\mathcal{M}f\right\}(s) = \int_0^\infty x^{s-1}f(x)dx $
We have $\displaystyle f(x) = \sum_{n=0}^\infty (-1)^n e^{-x(n+1)} $, so $\displaystyle \left\{\mathcal{M}f\right\}(s) = \int_0^\infty x^{s-1}\sum_{n=0}^\infty (-1)^n e^{-x(n+1)}dx $.
Now simplify a bit to get $\displaystyle \left\{\mathcal{M}f\right\}(s) = \sum_{n=0}^\infty (-1)^n \int_0^\infty x^{s-1}e^{-x(n+1)}dx $.
Now let $\displaystyle t=x(n+1) $ which means $\displaystyle dt=(n+1)dx $.
We then get $\displaystyle \left\{\mathcal{M}f\right\}(s) = \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^s}\int_0^\infty t^{s-1}e^{-t} dt $
Therefore $\displaystyle \left\{\mathcal{M}f\right\}(s) = \Gamma(s)\sum_{n=0}^\infty \frac{(-1)^n}{(n+1)^s} = \Gamma(s)\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} $
So now, let $\displaystyle g(s) = \sum_{n=0}^\infty \frac{a_n}{n^s} $ and $\displaystyle G(z) = \sum_{n=1}^\infty a_n z^n $, where $\displaystyle a_n=(-1)^{n+1} $.
Note $\displaystyle G(z) = \sum_{n=0}^\infty a_{n+1} z^{n+1} $, thus $\displaystyle G(e^{-t}) = \sum_{n=0}^\infty a_{n+1} (e^{-t})^{n+1} $.
So as you can see (hopefully), we do get your identity for this case, namely $\displaystyle g(s) = \frac{1}{\Gamma(s)}\int_0^\infty G(e^{-t})t^{s-1}dt $.