I face a problem when I'm dealing with one of the transforms and the problem is as follows: Suppose that we have
$$ f(a+z) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}z^{k} \tag{1} $$
$$ f\left(a+e^{\theta x}\right) = \sum_{k=0}^{\infty} \frac{f^{(k)}(a)}{k!}e^{\theta k x} \tag{2} $$
and I want to express the following series in terms of equation $\text{(1)}$ or $\text{(2)}$
$$\sum_{k=1}^{\infty} \frac{f^{(k)}(a)}{k!} k^{-p} $$ where $ 0 < p < 1 $.
Is there any way that we can do this thing? Or, is it impossible to represent the third series in terms of equation $\text{(1)}$ or $\text{(2)}$
Via the substitutions $u=\log(1/x)$ and $v=ku$, $$\int_0^1 (\log\tfrac 1x)^{p-1}x^{k-1}dx=\int_0^\infty u^{p-1}e^{-ku}du=k^{-p}\int_0^\infty v^{p-1}e^{-v}=\Gamma(p)k^{-p}.$$ This means that \begin{align*} \sum_{k=1}^\infty \frac{f^{(k)}(a)}{k!}k^{-p} &=\frac1{\Gamma(p)}\sum_{k=1}^\infty \frac{f^{(k)}(a)}{k!}\int_0^1(\log\tfrac1x)^{p-1}x^{k-1}dx\\ &=\frac1{\Gamma(p)}\int_0^1(\log \tfrac 1x)^{p-1}\sum_{k=1}^\infty \frac{f^{(k)}(a)}{k!}x^{k-1}dx\\ &=\frac1{\Gamma(p)}\int_0^1(\log \tfrac 1x)^{p-1}\frac{f(a+x)-f(a)}xdx \end{align*} as long as the interchange of summation and integration is justified, which should be the case by Fubini's theorem if $$\sum_{k=1}^\infty \frac{f^{(k)}(a)}{k!}k^{-p}$$ converges absolutely.