The infinite product representation for $\vartheta_{1}(z, q)$ given by $$\vartheta_{1}(z, q) = 2q^{1/4}\sin z\prod_{n = 1}^{\infty}(1 - q^{2n})(1 - 2q^{2n}\cos 2z + q^{4n})\tag{1}$$ I hope you are familiar with the above result. Taking logs we get
\begin{align}\log\vartheta_{1}(z, q) &= \log(2q^{1/4}) + \log\sin z + \log\prod_{n = 1}^{\infty}(1 - q^{2n})\notag\\
&\,\,\,\,\,\,\,\, + \sum_{n = 1}^{\infty}\log(1 - 2q^{2n}\cos 2z + q^{4n})\tag{2}
\end{align}
and differentiating with respect to $z$ we get
\begin{align}
\frac{\vartheta'_{1}(z, q)}{\vartheta_{1}(z, q)} &= \cot z + 4\sum_{n = 1}^{\infty}\frac{q^{2n}\sin 2z}{1 - 2q^{2n}\cos 2z + q^{4n}}\notag\\
&= \cot z + \frac{4}{2i}\sum_{n = 1}^{\infty}\frac{q^{2n}(e^{2iz} - e^{-2iz})}{(1 - q^{2n}e^{2iz})(1 - q^{2n}e^{-2iz})}\notag\\
&= \cot z + \frac{4}{2i}\sum_{n = 1}^{\infty}q^{2n}\left(\frac{e^{2iz}}{1 - q^{2n}e^{2iz}} - \frac{e^{-2iz}}{1 - q^{2n}e^{-2iz}}\right)\notag\\
&= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\frac{q^{2n}e^{2iz}}{1 - q^{2n}e^{2iz}} - \sum_{n = 1}^{\infty}\frac{q^{2n}e^{-2iz}}{1 - q^{2n}e^{-2iz}}\right)\notag\\
&= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\{q^{2n}e^{2iz}\}^{m} - \sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\{q^{2n}e^{-2iz}\}^{m}\right)\notag\\
&= \cot z + \frac{4}{2i}\left(\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}q^{2m}e^{2imz}q^{2m(n - 1)}\right. \notag\\
&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left. - \sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}q^{2m}e^{-2imz}q^{2m(n - 1)}\right)\notag\\
&= \cot z + \frac{4}{2i}\left(\sum_{m = 1}^{\infty}q^{2m}e^{2imz}\sum_{n = 1}^{\infty}q^{2m(n - 1)}\right.\notag\\
&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left. - \sum_{m = 1}^{\infty}q^{2m}e^{-2imz}\sum_{n = 1}^{\infty}q^{2m(n - 1)}\right)\notag\\
&= \cot z + \frac{4}{2i}\left(\sum_{m = 1}^{\infty}\frac{q^{2m}e^{2imz}}{1 - q^{2m}} - \sum_{m = 1}^{\infty}\frac{q^{2m}e^{-2imz}}{1 - q^{2m}}\right)\notag\\
&= \cot z + 4\sum_{n = 1}^{\infty}\frac{q^{2n}}{1 - q^{2n}}\sin 2nz\tag{3}\end{align}
The above proof is taken from one of my blog posts (see equation $(25)$ of that post).
$f'(x)=\frac{\text{d} f(x)}{\text{d}x}=\int_{0}^{\infty} \frac{\partial }{\partial x}\left(\frac{e^{-u}-e^{-u^{\alpha}x}}{u}\right) du$
Taking, $z=u^{\alpha}x$ we get
$=\frac{x^{-1}}{\alpha}(\int_{0}^{\infty} e^{-z} dz) =\frac{x^{-1}}{\alpha}$.....(1)
Taking, $f(x)=y$
So, from (1) we get the differential equation
$\frac{\text{d} y}{\text{d}x}=\frac{x^{-1}}{\alpha}$ with the boundary condition $y(1)=\int_{0}^{\infty} \frac{e^{-u}-e^{-u^{\alpha}}}{u} du =-\frac{(\alpha-1)}{\alpha}\gamma$
Proof: Take $y_{alpha}(1)=\psi(\alpha)=\int_{0}^{\infty} \frac{e^{-u}-e^{-u^{\alpha}}}{u} du$
So, $\frac{\text{d}\psi(\alpha)}{\text{d}x}=\int_{0}^{\infty} \frac{\partial}{\partial \alpha}\left(\frac{e^{-u}-e^{-u^{\alpha}}}{u}\right) du$
$=\int_{0}^{\infty} u^{\alpha-1}\text{ln}(u)e^{-u^{\alpha}} du$
Taking $u^{\alpha}=z$
$=\frac{1}{{\alpha}^2}\int_{0}^{\infty} \text{ln}(z)e^{-z} dz=-\frac{\gamma}{{\alpha}^2}$
As $\Gamma'(1)=-\gamma$ (https://en.m.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant)
So, we get the differential equation
$\frac{\text{d}\psi}{\text{d}\alpha}=\frac{-\gamma}{{\alpha}^2}$ with the boundary condition $\psi(1)=0$ we get
$y(1)=\psi(\alpha)-\psi(1)=\psi(\alpha)=-\frac{\alpha-1}{\alpha}\gamma$
And for, all $\alpha\geq 1$, $y(1)=\frac{-(\alpha-1)\gamma}{\alpha}$.
Where, $\gamma$ is Euler- Mascheroni constant.
So, $f(x)=\frac{\text{ln}x}{\alpha} -\frac{(\alpha-1)\gamma}{\alpha}$.
So, for $\alpha=1$, $f(x)=\text{ln}x$.
Best Answer
Here's how I was able to get a seemingly correct answer. I, being a physicist by training, have played very fast and loose with somewhat unfamiliar mathematics here. I am sure, on some level, this is illegitimate. However, we have...
$$A(\chi)=-\sum_{i=0}^{\infty}(-1)^{i}(2i+1)\chi\int_{0}^{\infty}t^{-\frac{1}{2}}e^{-\frac{(2i+1)^2\chi^2}{t}}dt$$
applying the substitution $u=\frac{(2i+1)^2\chi^2}{t}$ yields;
\begin{align} \tag{1} A(\chi) &=-\sum_{i=0}^{\infty}(-1)^{i}(2i+1)^2\chi^2\int_{0}^{\infty}u^{-\frac{3}{2}}e^{-u}du\\ \tag{2} &=-\sum_{i=0}^{\infty}(-1)^{i}(2i+1)^2\chi^2\Gamma\big(-\frac{1}{2}\big)\\ \tag{3} &=2\chi^2\sqrt{\pi}\sum_{i=0}^{\infty}(-1)^{i}(2i+1)^2 \end{align}
Now, we can write the remaining sum thusly...
\begin{align} \tag{4} &\space\space\space\space\space \sum_{i=0}^{\infty}(-1)^{i}(2i+1)^2\\ \tag{5} &=\sum_{i=0}^{\infty}(-1)^{i}(4i^2+4i+1) \end{align}
Which gives us;
\begin{align} \tag{6} &= 4\sum_{i=0}^{\infty}(-1)^{i}i^2+4\sum_{i=0}^{\infty}(-1)^{i}i+\sum_{i=0}^{\infty}(-1)^{i}\\ \tag{7} &=-4\eta(-2)-4\eta(-1)+\eta(0)=0-1+\frac{1}{2}=-\frac{1}{2} \end{align}
So our final answer, which I have investigated and confirmed numerically, appears to be...
$$\tag{8} A(\chi)=2\chi^2\sqrt{\pi}(-\frac{1}{2})=-\chi^2\sqrt{\pi}$$
Again, I have not been careful about interchanging the integral and sum at the start nor have I been careful in my application of known Dirichlet eta function values to a classically non-convergent sum. So, whatever outrage this may cause among the actual mathematicians is well deserved.