I'm trying to solve this integral equation: $$\int_0^\infty \frac{t^{zi}+t^{-zi}}{e^t}dt=0$$ For the least value of $z$. This is, if $S=\{z_0,...,z_n\}$ is a set of solutions, and $|\lambda|=\mathrm{inf}\{|z_0|,...,|z_n|\}$, then $\lambda\in S$ is the desired solution.
After extensive transformations, I managed to reduce the integral on the LHS to an infinite polynomial $p$: $$p(z)=1+\sum_{n\in\mathbb{Z_+}}\frac{\mathcal{D}(n)}{2\cdot n!}\left(\Pi_n(z)+\Pi_n(-z)\right)$$
Where $$\Pi_n(z)=\prod_{s=0}^n(zi-s)$$ $n$ runs through all positive intergers, and $\mathcal{D}(n)$ is the number of derangements, hence: $$\mathcal{D}(n)=n!\sum_{k=0}^n\frac{(-1)^k}{k!}$$ So the problem reduces to finding the absolute infimum of a set of roots of an infinite complicated polynomial $p$... But how can this be done? Any help, alternative solution, or insight on the problem will be appreciated.
Edit
I should've probably clarified this in the first edit: I've found two solution approximations by brute force that satisfy the lower bound constraints: $$z_0=\lambda$$ $$z_1=-\lambda$$ Where $\lambda\in\mathbb{R}$ shows $3.4397<|\lambda|<3.4398$. This solves the equation to an extent, however, I need the error on $\lambda$ to be as small as possible, so I need a good approximant for it (this is the reason why the polynomial equivalent of the equation was introduced). I've never solved for the root of an infinite-degree polynomial before, so i'm not quite sure if this is even possible, but it is the only way to get an approximant that i could think about.
$$I=\int_0^\infty \frac{t^{zi}+t^{-zi}}{e^t}\,dt=\Gamma(1+iz)+\Gamma(1-iz)$$
So, considering that we need to find the first zero of function $$f(z)=\Gamma(1+iz)+\Gamma(1-iz)$$ consider as a very first estimate $$z_0=\pi$$ and use the $\color{red}{\text{first}}$ iterate of a Newton-like method of order $n$; this gives an explicit closed form expression for each approximate.
Converted to decimals, the results are
$$\left( \begin{array}{cc} n & z_1^{(n)} & \text{method}\\ 2 & 3.3637901 & \text{Newton}\\ 3 & 3.4227503 & \text{Halley}\\ 4 & 3.4365242 & \text{Householder}\\ 5 & 3.4392040 & \text{no name}\\ 6 & 3.4396546 & \text{no name}\\ 7 & 3.4397239 & \text{no name}\\ 8 & 3.4397338 & \text{no name}\\ \cdots & \cdots \\ \infty &3.4397353 & \text{solution} \\ \end{array} \right)$$