I am trying to understand some open steps in the following article The Sign of Fourier coefficients of Half-integral Weight Cusp Form by HuluseHulse, MehmerKiral, Kuan, and Chan Lim, I foundI find the following :
Let $f\in S_{\frac{k}{2}}(\Gamma_0(4))$ be an eigenform of all Hecke operators $T_{\frac{k}{2}}(p^2)$ for $p$ prime, where $k$ is an odd integer. Take the Dirichlet series : $$M(s)=\sum_{t\geq 1, \; t\;\text{square-free}}\frac{a(t)}{t}$$ With the inverse Mellin transform, we get : $$\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}M(s)\Gamma(s)x^sds=\sum_ta(t)e^{-t/x}$$ They assert that the integral on the left-hand side above is $O(x^{3/4+\varepsilon})$ for any $\varepsilon>0.$ I don't understand why ?
I have another questions :
Considering the inverse Mellin transform
$$I=\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}L^{(2)}(s)\Gamma(s)x^sds=\sum_na(n)^2e^{-n/x}$$
and shifting the line of integration to $\Re(s)=\frac{1}{2}$
past the pole at $s=1,$ we get
$$I=(\mathrm{Res}_{s=1}L^{(2)}(f,s))x+\frac{1}{2\pi i}\int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty}L^{(2)}(s)\Gamma(s)x^sds\;\;\;(*)$$ They assert that :
$$\frac{1}{2\pi i}\int_{\frac{1}{2}-i\infty}^{\frac{1}{2}+i\infty}L^{(2)} (s)\Gamma(s)x^sds=O(x^{\frac{1}{2}})\;\;\;\;(1)$$
and $(1)$ combined with $(*)$ implies, that
$$x\ll\sum_na(n)^2e^{-n/x}.\;\;\;\;\;\;(2)$$
I don't see why $(1)$ is true ? and why $(1)\Rightarrow (2).$
Can someone clarify to me it ? Thanks in advance.
