Explanation of a deduction in "On $L^{\infty}$ norms of holomorphic cusp forms" I am reading the article On $L^{\infty}$ norms of holomorphic cusp forms. I am particularly interested to the following :
 
I don't see how we can get from
$$ |y^k f(z)|     \ll \frac{y^k k^\epsilon(4\pi)^{k-1/2}}{\sqrt{\Gamma(2k)}(2\pi y)^{k-1/2}}\sum_{n\ge 1}(2\pi ny)^{k-1/2+\epsilon}e^{-2\pi ny}$$
that  : $|y^kf(z)|\ll k^{{1/4}+\epsilon}$ for $y\gg k.$
Can someone clarify to me it ?
 A: In the article, the author shows how to go from the sum in 
$$ |y^k f(z)|     \ll \frac{y^k k^\epsilon(4\pi)^{k-1/2}}{\sqrt{\Gamma(2k)}(2\pi y)^{k-1/2}}\sum_{n\ge 1}(2\pi ny)^{k-1/2+\epsilon}e^{-2\pi ny} $$
to the integral equation in the image. So I suppose you are having trouble with the integral. We can do that here. I drop all epsilons as they're annoying and don't matter to convey how this is seen.
We consider
$$ \frac{y^{1/2} 2^k}{\sqrt{\Gamma(2k)}} \frac{1}{y} \int_0^\infty t^{k - \frac{1}{2}}e^{-t} dt = \frac{y^{1/2} 2^k}{\sqrt{\Gamma(2k)}} \frac{1}{y}\int_0^\infty t^{k + \frac{1}{2}}e^{-t} \frac{dt}{t} = \frac{y^{1/2} 2^k}{\sqrt{\Gamma(2k)}} \frac{1}{y}\Gamma(k + \tfrac{1}{2}). \tag{1}$$
We now make ample use of Stirling's approximation 
$$\Gamma(z) \sim \sqrt{\frac{2\pi}{z}} \left(\frac{z}{e}\right)^z.$$
Then $(1)$ becomes
$$ \begin{align}
y^{-1/2} 2^k &\sqrt{\frac{2\pi}{k + \frac{1}{2}}} \left(\frac{k + \frac{1}{2}}{e}\right)^{k + \frac{1}{2}} \left( \sqrt{\frac{2k}{2\pi}} \left( \frac{e}{2k} \right)^{2k} \right)^{1/2} \\
&= y^{-1/2} 2^k (2\pi)^{1/4} \frac{(2k)^{1/4}}{(k + \frac{1}{2})^{1/2}}\left(\frac{k + \frac{1}{2}}{e}\right)^{k + \frac{1}{2}} \left( \frac{e}{2k} \right)^k \\ &= y^{-1/2} (2\pi)^{1/4} (2k)^{1/4} e^{-1/2} \left( \frac{k + \frac{1}{2}}{k}\right)^k. \tag{2}
\end{align}$$
Nothing fancy happened at any step --- it's just a lot of rearrangement and simplification. Notice that
$$ \left( \frac{k+\frac{1}{2}}{k}\right)^k = \left( 1 + \frac{1}{2k} \right)^k \leq 2e^{-1/2}$$
(and is well-approximated by $e^{-1/2}$ for large $k$). Then $(2)$ is bounded by
$$ y^{-1/2} (2\pi)^{1/4} (2k)^{1/4} 2 e^{-1} \ll \frac{k^{1/4}}{y^{1/2}}.$$
As $y \gg k$, it's clear that this term is at most $k^{1/4}$ as claimed.
The other term from approximating the sum is the first summand, given by [again dropping all epsilons]
$$2^k (2\pi y)^k e^{-2\pi y} / \sqrt{\Gamma(2k)} = (4\pi y)^k e^{-2\pi y} / \sqrt{\Gamma(2k)}. \tag{3}$$
As you've included in your question, this is maximized at $y = k/2\pi$.
By Stirling's approximation again, we write
$$ \sqrt{\Gamma(2k)} \sim \left(\frac{2\pi}{2k}\right)^{1/4} \left(\frac{2k}{e} \right)^{k}. $$
Using this in $(3)$, and substituting $y = k/2\pi$, we have
$$(2k)^k e^{-k} \left( \frac{k}{\pi} \right)^{1/4} e^k (2k)^{-k} = \left( \frac{k}{\pi} \right)^{1/4}.$$
This completes the derivation.

In summary, you use Stirling's approximation on every gamma function in sight, and follow it up with algebraic manipulations.
