I am currently reading Titchmarsh's book about the Riemann Zeta function and came across two little problems at p. 266:
$1)$ $$\sum_{n=1}^{\infty} e^{-2n^2\pi x^2 sin\, \delta} = O \left(x^{-1}\delta^{-\frac{1}{2}} \right) $$
This is what I've got so far with the variables beging: $\, x\in (1, G)$ where $G=e^{1/H}$ and $H\geq 1$
\begin{align} \sum_{n=1}^{\infty} e^{-2n^2\pi x^2 sin \delta} \leq& \int_0^{\infty} e^{-2y^2\pi x^2 sin\, \delta} \, dy \\ \leq & \int_0^{\infty} e^{-2y\pi x^2 sin \delta} \, dy\\ = & \frac{1}{\pi \, sin\, \delta} \end{align}
But I do not know how to complete the inequality.
$2)$ $$ O\left( \delta^{-\frac{1}{2}}H^2(G-1) \right) + O\left(\frac{\delta^{-\frac{1}{2}}}{log \, G} \right) = O \left( H \delta^{-\frac{1}{2}} \right) $$
Putting $G=e^{\frac{1}{H}}$ on the left hand side I get for some constants $C,K >0$
$$ C\delta^{-\frac{1}{2}}H^2(G-1) + K\frac{\delta^{-\frac{1}{2}}}{log \, G} = H \delta^{-\frac{1}{2}}(CH (e^{\frac{1}{H}}-1)+K) $$
I do not see why $CH (e^{\frac{1}{H}}-1)+K$ can become constant.
I think those are two little problems but I really do not see what trick gives me the desired result. I appreciate any hints. Thank you in advance.
First question: $$ \sum_{n=1}^{\infty} e^{-2n^2\pi x^2 sin \delta} \leq \int_0^{\infty} e^{-2y^2\pi x^2 sin\, \delta} \, dy = \frac{1}{x\sqrt{\sin{\delta}}} \int_0^{\infty} e^{-2\pi u^2} \, du $$ by putting $u = yx\sqrt{\sin{\delta}}$, and then $\sin{\delta} = O(\delta)$ as $\delta \to 0$.
Second: (as $H \to \infty$?) $$ H(e^{1/H}-1) = H\left(1+\frac{1}{H}+O\left(\frac{1}{H^{2}}\right)-1\right) = 1+ O(H^{-1}) $$ by the Taylor series of $e^{x}$ around $x=0$.