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During our search of real rooted entire function approximations to Riemann $\Xi$ function, we need to calculate the upper and lower bounds of $$f_m(x):=\sum_{j=1}^m b_j(x):=\sum_{j=1}^m\frac{\pi^j}{\Gamma(j)(x^2+(j+1/4)^2)},\quad m>>\pi.\tag{1}$$

Since $b_j(x)$ is a positive and log-cancave sequence, it has uni-modal. Thus $$b_1(x)<b_2(x)<\cdots\leq b_{\lfloor j_0\rfloor}(x)\geq b_{\lfloor j_0\rfloor+1}>...b_m(x)\tag{2}$$

where $j_0=j_0(x)$ is solution to the cubic equation:

$$\frac{b_{j+1}(x)}{b_{j}(x)}=\frac{\pi(x^2+(j+1/4)^2)}{j(x^2+(j+1+1/4)^2)}=1\tag{3}$$

Thus $$f_m(x):=\sum_{j=1}^m b_j(x)=\sum_{j=1}^{\lfloor j_0\rfloor} b_j(x)+\sum_{j=\lfloor j_0\rfloor+1}^m b_j(x)\tag{4}$$

Bsed on the monotonic property of $b_j(x)$ in these two regions , we have $$b_1(x)+\int_{1}^{\lfloor j_0\rfloor} b_t(x)\mathrm{d}t\leq\sum_{j=1}^{\lfloor j_0\rfloor} b_j(x)\leq b_{\lfloor j_0\rfloor}(x)+\int_{1}^{\lfloor j_0\rfloor} b_t(x)\mathrm{d}t\tag{5}$$

$$b_m(x)+\int_{\lfloor j_0\rfloor+1}^m b_t(x)\mathrm{d}t\leq\sum_{j=\lfloor j_0\rfloor+1}^m b_j(x)\leq b_{\lfloor j_0\rfloor+1}(x)+\int_{\lfloor j_0\rfloor+1}^m b_t(x)\mathrm{d}t\tag{6}$$

We can then add (5) to (6) and obtain the bounds:

Questions

  1. Can we close the gap in the integration of $b_t(x)$ from $t=1$ to $t=m$?

  2. Since $j_0$ is a complicated function of $x^2$, can we find the explicit $m-$dependence of the bounds?

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    $\begingroup$ I think that the largest term is always one of the first four, it isn't a good idea to use an integral on the left side. $\endgroup$ Feb 13, 2016 at 21:51

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