I am stuck on this computation of the Fourier coefficients of Eisenstein series. For $\Gamma = SL(2, \mathbb{Z})$ and $\Gamma\_\infty = \left\{ \left( \begin{array}{cc} 1 & m \\ 0 & 1 \end{array}\right): m \in \mathbb{Z} \right\}$ we can define an Eisenstein series as the sum over the cosets:


$$ E\_\Gamma(z,s) = \sum\_{\gamma \in \Gamma\_\infty \backslash \Gamma} (y(\gamma z))^s $$


The Fourier coefficients of this Eisenstein series are written in terms of an L-function and Bessel functions:


$$ \int\_0^1 E\_\Gamma(z,s) e^{2\pi i m x} \, dx= \begin{cases} 
y^s + \left( \sum\_{c \geq 0} \frac{r\_{0, \Gamma}(c)}{c^{2s}} \right) \frac{\pi^{1/2}\Gamma(s - \frac{1}{2})}{\Gamma(s)} y^{1-s} & m = 0 \\
 \left( \sum\_{c \geq 0} \frac{r\_{n, \Gamma}(c)}{c^{2s}} \right) \frac{2\pi^s|m|^{s - \frac{1}{2}}K\_{s- \frac{1}{2}}(2\pi|m|y)}{\Gamma(s)}& m \neq 0
\end{cases} $$


Not begin an expert on Automorphic forms (in fact a total newbie), I consulted Anton Dietmar's text and found not-quite the same formula:


$$ E(z,s) = \pi^{-s} \Gamma(s) \zeta(2s) \sum\_{\Gamma\_\infty \backslash \Gamma} \mathrm{Im}(\gamma z)^s $$


This formula looks the same up to some factors of gamma and zeta functions. Then 


$$ \int\_0^1 E(z,s) e^{2\pi i m x} \, dx= \begin{cases} 
\pi^{-s}\Gamma(s) \zeta(2s)y^s + \pi^{s-1} \Gamma(1-s)\zeta(2-2s)y^{1-s} & m = 0 \\
2|m|^{2 - \frac{1}{2}}\sigma\_{1-2s}(|m|)\sqrt{y}K\_{s - \frac{1}{2}}(2\pi|m|y)& m \neq 0
\end{cases} $$


I don't understand why in the first case I get [Ramanujan sums](https://en.wikipedia.org/wiki/Ramanujan%27s_sum) and in the second case, I get the [sum-of-divisors](https://en.wikipedia.org/wiki/Divisor_function) function:


$$ r\_m(c) = \sum\_{(c,d)=1} e^{2\pi i m \frac{d}{c}}$$


Is this consistent? I know the two version of $E$ are written slightly differently, but I have not yet verified that they agree.



Fourier Transform of Eisenstein Series - Sum of Divisors or Ramanujan Sums?
I think all your questions are answered by the following calculation (assume $m\geq 1$ and $\Re(s)>1$):
$$ \sum\_{c=1}^\infty\frac{r\_m(c)}{c^{2s}} = \sum\_{c=1}^\infty\frac{1}{c^{2s}}\sum\_{\substack{\text{$d$ mod $c$}\\{(d,c)=1}}}e\left(m\frac{d}{c}\right) = \sum\_{c=1}^\infty\frac{1}{c^{2s}}\sum\_{\text{$d$ mod $c$}}e\left(m\frac{d}{c}\right)\sum\_{n\mid(c,d)}\mu(n) $$
$$ = \sum\_{n=1}^\infty\mu(n)\sum\_{\substack{c=1\\{n\mid c}}}^\infty\frac{1}{c^{2s}}\sum\_{\substack{\text{$d$ mod $c$}\\{n\mid d}}}e\left(m\frac{d}{c}\right) = \sum\_{n=1}^\infty\mu(n)\sum\_{c=1}^\infty\frac{1}{(nc)^{2s}}\sum\_{\text{$nd$ mod $nc$}}e\left(m\frac{nd}{nc}\right)$$
$$ = \sum\_{n=1}^\infty\frac{\mu(n)}{n^{2s}}\sum\_{c=1}^\infty\frac{1}{c^{2s}}\sum\_{\text{$d$ mod $c$}}e\left(m\frac{d}{c}\right) = \sum\_{n=1}^\infty\frac{\mu(n)}{n^{2s}}\sum\_{c\mid m}\frac{c}{c^{2s}} = \frac{\sigma\_{1-2s}(m)}{\zeta(2s)}.$$


**P.S.** The above calculation is classical, e.g. see (1.5.4) in Titchmarsh: The theory of the Riemann zeta-function. 


