Clarification of the proof of the main theorem of the paper of Hulse et al

Question

I am trying to understand some open steps in the following article The Sign of Fourier coefficients of Half-integral Weight Cusp Form by Hulse, Kiral, Kuan, and Lim, I 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.

Answer 1

The first two questions follow directly from applying the residue theorem to the Mellin transforms.

In the third one you just have to consider the contribution from the residue at $s=1$.

The paper is research-level, but these kind of questions fit better at math.stackexchange.