Applying Mellin Transformation

• Apr 3rd 2010, 07:40 AM
EinStone
Applying Mellin Transformation
Which Dirichlet Series is obtained as the Mellin transformation of
$\displaystyle \frac{1}{e^t+ 1}= \sum_{n=0}^\infty (-1)^n (e^{-t})^{n+1}$
• Apr 3rd 2010, 08:42 AM
chiph588@
Quote:

Originally Posted by EinStone
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.
• Apr 3rd 2010, 09:07 AM
EinStone
Ok thanks, what can you say about $\displaystyle \frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)}$ ? Deduce that

$\displaystyle \zeta(s) = \frac{1}{\Gamma(s)(1-2^{1-s})} \int_0^s \frac{t^{s-1}}{e^t+1}$
• Apr 3rd 2010, 09:48 AM
chiph588@
Quote:

Originally Posted by EinStone
Ok thanks, what can you say about $\displaystyle \frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)}$ ? Deduce that

$\displaystyle \zeta(s) = \frac{1}{\Gamma(s)(1-2^{1-s})} \int_0^s \frac{t^{s-1}}{e^t+1}$

Just use the equation $\displaystyle \zeta(s)(1-{2^{1-s}})= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}$.
• Apr 3rd 2010, 10:37 AM
chiph588@
Quote:

Originally Posted by chiph588@
Just use the equation $\displaystyle \zeta(s)(1-{2^{1-s}})= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}$.

Here's a proof of this.
• Apr 3rd 2010, 11:41 AM
EinStone
Ok good, can you still evaluate $\displaystyle \frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)}$ ?
• Apr 3rd 2010, 11:46 AM
chiph588@
Quote:

Originally Posted by EinStone
Ok good, can you still evaluate $\displaystyle \frac{\left\{\mathcal{M}f\right\}(s)}{\zeta(s)}$ ?

What do you mean?
• Apr 3rd 2010, 02:26 PM
EinStone
OK I will ask from beginning. I learned that the Mellin transformation transforms Power series in Dirichlet Series. If

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?
• Apr 3rd 2010, 03:05 PM
chiph588@
Quote:

Originally Posted by EinStone
OK I will ask from beginning. I learned that the Mellin transformation transforms Power series in Dirichlet Series. If

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 $\displaystyle g(s) = \frac{1}{\Gamma(s)}\int_0^\infty G(e^{-t})t^{s-1}dt$.
• Apr 4th 2010, 02:22 AM
EinStone
ah I found my mistake. My Power series had coefficient $\displaystyle e^{-t}$ and not just $\displaystyle t$. Thanks for the post :).