How to find the bound of this sum? Let $t>0,a(t)=\arg(\Gamma(1/4+it))$,$\kappa(n)=\frac{1}{2}x\pi n^2$,we need to calculate the bound,$A(x)$, of the following finite sum:
$$ S(x)=\sum_{1\le n\le x}e^{\kappa(n)}\left(e^{ia(t)}(\kappa(n))^{\frac{3}{4}+it}+e^{-ia(t)}(\kappa(n))^{\frac{3}{4}-it}\right)\tag{1}$$
$$|S(x)|=A(x)(1+o(1))\quad\text{as}\quad x\to\infty\tag{2}$$.
We expect that $A(x)$>0$.
This problem shows up when we search for approximations to the Riemann $\Xi(z)$ function.
 A: To expound my comment,
Consider:
$$
S(x,t) = \sum_{1 \le n \le x} e^{\frac{\pi}{2}xn^2}\left(e^{ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{3/4+it}+e^{-ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{3/4-it} \right)
$$
Where $ a(t)=arg(\Gamma(1/4+it)) $. Then:
$$
S(x,t) = \sum_{1 \le n \le x} e^{\frac{\pi}{2}xn^2}\left(\frac{\pi}{2}xn^2 \right)^{3/4}\left(e^{ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{it}+e^{-ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{-it} \right)
\\ |S(x,t)|\le \sum_{1 \le n \le x} \left|e^{\frac{\pi}{2}xn^2}\left(\frac{\pi}{2}xn^2 \right)^{3/4}\right|\left|\left(e^{ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{it}+e^{-ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{-it} \right)\right|
$$
The next and previous step uses the fact that the absolute value is a subadditive function. Looking at the second of the terms inside the sum:
$$
|R|=\left|\left(e^{ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{it}+e^{-ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{-it} \right)\right|
\\ \le \left|e^{ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{it}\right|+\left|e^{-ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{-it}\right|
\\ = \left|e^{ia(t)}\right|\left|\left(\frac{\pi}{2}xn^2 \right)^{it}\right|+\left|e^{-ia(t)}\right|\left|\left(\frac{\pi}{2}xn^2 \right)^{-it}\right|
$$
Now, using the fact that $ |e^{ix}| = 1 $ when $ x \in \mathbb{R} $:
$$
|R| \le (1)(1)+(1)(1) = 2
$$
Therefore:
$$
|S(x,t)| \le 2 \sum_{1 \le n \le x} \left|e^{\frac{\pi}{2}xn^2}\left(\frac{\pi}{2}xn^2 \right)^{3/4}\right|
\\ = 2 \sum_{1 \le n \le x} e^{\frac{\pi}{2}xn^2}\left(\frac{\pi}{2}xn^2 \right)^{3/4}
$$
Note that, as the inside of this sum is now monotonically increasing in n, the maximum value obtained inside the sum is the xth term. Therefore:
$$
|S(x,t)| \le 2 \sum_{1 \le n \le x} e^{\frac{\pi}{2}x \ x^2}\left(\frac{\pi}{2}x \ x^2 \right)^{3/4}
\\ = 2xe^{\frac{\pi}{2}x^3}\left(\frac{\pi}{2}x^3 \right)^{3/4} = 2\left(\frac{\pi}{2}\right)^{3/4} x^{13/4} e^{\frac{\pi}{2}x^3}
$$
This is a strict upper bound, but will computationally seem to be about x times the actual value. This is because, as $ x \to \infty $, the only term that really ought to matter is the xth term, as it is the largest term by a large margin for large enough x. Therefore a more accurate approximation would be:
$$
|S(x,t)| \sim 2\left(\frac{\pi}{2}\right)^{3/4} x^{9/4} e^{\frac{\pi}{2}x^3}
$$
But this may no longer be an exactly strict upper bound.
Edit:
By rewriting the above R, it may be seen:
$$
R = e^{ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{it}+e^{-ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{-it}
\\ = e^{ia(t)}e^{it \ln \left(\frac{\pi}{2}xn^2 \right)}+e^{-ia(t)} e^{-it \ln \left(\frac{\pi}{2}xn^2 \right)}
\\ = e^{i \left(a(t)+t \ln \left(\frac{\pi}{2}xn^2 \right) \right)}+e^{-i \left(a(t)+t \ln \left(\frac{\pi}{2}xn^2 \right) \right)}
\\ = 2\cos \left(a(t)+t \ln \left(\frac{\pi}{2}xn^2 \right) \right)
\\ = 2\cos(a(t)) \cos \left(\ln \left(\frac{\pi}{2}xn^2 \right) \right) - 2\sin(a(t)) \sin \left(\ln \left(\frac{\pi}{2}xn^2 \right) \right)
$$
So the summand is in fact real, giving rise to a real sum, ehich may not be immediately evident.
