How to approximate exponential function or under approximate this function I am trying to find the minimum of this function analytically:
$$
G(x)=\frac{c(a.x^3+b)}{x}+2(a.x^3+b)\sum_{m=1}^{\infty}\frac{e^{-\beta^2m^2(T-\frac{c}{x})}-e^{-\beta^2m^2T}}{\beta^2m^2}
$$
where $0<x<1$ and all of the other variables have positive value. But the derivative is so complex and it is not solvable analytically. I tried to use the solution of Basal problem to approximate the main function as follows:
$$
\sum_{m=1}^{\infty}\frac{e^{-\beta^2m^2(T-\frac{c}{x})}-e^{-\beta^2m^2T}}{\beta^2m^2}<\frac{e^{-\beta^2(T-\frac{c}{x})}}{\beta^2}.\frac{\pi^2}{6}
$$
so I can write:
$$
G(x)<\frac{c(a.x^3+b)}{x}+2(a.x^3+b)\frac{e^{-\beta^2(T-\frac{c}{x})}}{\beta^2}.\frac{\pi^2}{6}
$$
Then, I tried to solve the new formula analytically so that I can earn the minimum of upper bound of the $G(x)$. But I couldn't do it again.
Is there any way to get rid of exponential part or do something else to earn the minimum of G(x)?
 A: I don't know if it helps, but I think this can be written as:
$$
G(x) = (ax^3 + b)\left(\frac{c}{x} + 2\int_0^{\frac{c}{x}}\psi\left(\frac{\beta^2(T - t)}{\pi}\right)\,dt\right) \quad \left(x > \frac{c}{T}\right),
$$
or, perhaps a little more simply:
$$
F(y) = \left(\frac{ac^3}{y^3} + b\right)\left(y + 2\int_0^y\psi\left(\frac{\beta^2(T - t)}{\pi}\right)\,dt\right) \quad (0 < y < T),
$$
where $\psi$ is the function defined on p.15 of H. M. Edwards, Riemann's Zeta Function (1974, repr. Dover 2001):
$$
\psi(s) = \sum_{m=1}^\infty e^{-m^2{\pi}s} \quad (s > 0).
$$
Perhaps someone who knows something about theta functions (I don't, I'm afraid) can do something with this?
In terms of this notation, which seems to be standard:
$$
\vartheta(s) = \sum_{m=-\infty}^\infty e^{-m^2{\pi}s} \quad (s > 0),
$$
the formulae are so much neater:
\begin{gather*}
G(x) = (ax^3 + b)\int_0^{\frac{c}{x}}\vartheta\left(\frac{\beta^2(T - t)}{\pi}\right)\,dt \quad \left(x > \frac{c}{T}\right), \\
F(y) = \left(\frac{ac^3}{y^3} + b\right)\int_0^y\vartheta\left(\frac{\beta^2(T - t)}{\pi}\right)\,dt \quad (0 < y < T)
\end{gather*}
that I increasingly suspect this has only got us back to where the problem came from in the first place! If so, I'm sorry!
