Proof. Take an arbitrary geometric function: \(a\cdot r^{n-1}\), where \(a\) and \(r\) are constants.

$$ \begin{aligned} \ln (a\cdot r^{n-1}) &= \ln a+\left(n-1\right)\ln r \\ & \sim n\ln r \\ \end{aligned} $$

Since \(\ln r\) is a mere constant, then the resulting complexity is simply \(\Theta(n)\). 

Proof. Given a Gaussian hypergeometric series, the general coefficient is

$$ A_n=\frac{(a)_n (b)_n}{(c)_n n!} = \frac{\prod_{k=0}^{n-1} (a+k) \prod_{k=0}^{n-1} (b+k)}{\prod_{k=0}^{n-1} (c+k) \cdot n!} $$

Applying the logarithmic transformation:

$$ \begin{aligned} \ln A_n &= \ln \left( \frac{\prod_{k=0}^{n-1} (a+k) \prod_{k=0}^{n-1} (b+k)}{\prod_{k=0}^{n-1} (c+k) \cdot n!} \right) \\ & = \ln \left( \prod_{k=0}^{n-1} (a+k) \right) + \ln \left( \prod_{k=0}^{n-1} (b+k) \right) -\ln \left( \prod_{k=0}^{n-1} (c+k) \right) - \ln n! \end{aligned} $$

Turning products into Gamma functions:

$$ \begin{aligned} & \ln \left( \prod_{k=0}^{n-1} (a+k) \right) + \ln \left( \prod_{k=0}^{n-1} (b+k) \right) -\ln \left( \prod_{k=0}^{n-1} (c+k) \right) - \ln n! \\ & = \ln\frac{\Gamma(a+n)}{\Gamma(a)} + \ln\frac{\Gamma(b+n)}{\Gamma(b)} - \ln\frac{\Gamma(c+n)}{\Gamma(c)} -\ln n!\\ & \sim \left(a+n\right)\ln n-\left(a+n\right) + \left(b+n\right)\ln n-\left(b+n\right) - \left(\left(c+n\right)\ln n-\left(c+n\right) \right) -n\ln n \\ & \sim \left(\ln n-1\right)\left(a+b-c+n\right) -n\ln n \\ & \implies \ln(A_n) \in \Theta(\ln n) \end{aligned} $$

 

Proof. Given a q-series, the general coefficient is

$$ (a;q)_k=\prod_{k=0}^{n-1} (1 - a q^k). $$

Applying the logarithm transforms the product into a sum:

$$ \ln \left| \prod_{k=0}^{n-1} (1 - a q^k) \right| = \sum_{k=0}^{n-1} \ln |1 - a q^k|. $$

For \(|q| > 1\), the term \(|aq^k|\) dominates \(1\) for large \(k\), so asymptotically:

$$ \sum_{k=0}^{n-1} \ln|1 - aq^k| \sim \sum_{k=0}^{n-1} \ln |a q^k| = \sum_{k=0}^{n-1} (\ln|a| + k \ln|q|) = n \ln|a| + \frac{n(n-1)}{2} \ln|q| = \Theta(n^2). $$

For \(|q| < 1\), the quadratic term is negative:

$$ \sum_{k=0}^{n-1} \ln|a q^k| = n \ln|a| + \frac{n(n-1)}{2} \ln|q| = -\Theta(n^2), $$

but the magnitude still grows like \(n^2\), so asymptotically we have \(\Theta(n^2)\) in all cases. 

Proof. We proceed by way of mathematical induction, showing the base case (\(n=1\)) and then proving for general \(n\) with \(n-1\) as inductive hypothesis.
Base Step: \(n=1\)

Gamma is defined to be the first case of the Vigneras hierarchy; using Stirling’s approximation and \(\Gamma(x) = (x-1)!\),

$$ \Gamma(x) \sim \sqrt{2\pi} (x-1)^{x-\frac{1}{2}} e^{-(x-1)}. $$

Taking logarithms,

$$ \ln \Gamma(x) \sim \ln(\sqrt{2\pi}) + \left(x-\tfrac{1}{2}\right)\ln(x-1) - (x-1). $$

Since \(\ln(x-1) = \ln x + o(1)\), it follows that

$$ \begin{aligned} \ln \Gamma(x) &\sim x\ln x - x. \\ & \implies \ln G_1(x) \in \Theta(x^1\ln x) \end{aligned} $$

Inductive Step: \(n-1\)

Assume as inductive hypothesis that the Log-Profile of \(G_{n-1}(x)\) has complexity \(\Theta(x^n \ln x)\).
A level \(G_n\) can be written as a finite product of the previous level.

$$ G_n\left(x+1\right)=\prod_{k=0}^{x-1} G_{n-1}(k)=G_{n-1}(0)\cdot G_{n-1}(1)\cdot G_{n-1}(2) ...G_{n-1} (x-1) $$

Taking the logarithm allows us to convert the right into summations:

$$ \begin{aligned} & \ln\left(G_{n}\left(x+1\right)\right)=\sum_{k=1}^{x-1}\ln\left(G_{n-1}(k)\right) \end{aligned} $$

Substitute the summand with the inductive hypothesis and rewrite in terms of an integral:

$$ \begin{aligned} \ln\left(G_{n}\left(x+1\right)\right)& =\sum_{k=1}^{x-1}\ln\left(G_{n-1}\left(k\right)\right) \\ & \sim\sum_{k=1}^{x-1}k^{n-1}\ln k \\ & = \int_{1}^{x-1}\left(k^{n-1}\ln k\right)\text{dk}+\text{o}\left(k^{n-1}\ln k\right)\\ \end{aligned} $$

With Integration by Parts, the resulting integral can be written as:

$$ \begin{aligned} &\frac{k^{n}}{n}\ln k-\int_{ }^{ }\left(\frac{1}{k}\right)\left(k^{n}\right)\text{dk}+\text{o}\left(k^{n-1}\ln k\right) \\ & = \boxed{\frac{k^{n}}{n}\ln k-\frac{k^{n}}{n+1} + \text{o}\left(k^{n-1}\ln k\right)} \\ \end{aligned} $$

Since the only remaining term is \(\text{o}\left(k^{n-1}\ln k\right)\), which must grow slower than \(k^n\ln k\), the complexity of \(G_n\) must be \(\Theta (x^n \ln x)\)

 

Proof. Since induction only holds for discrete cases as shown above, we must generalize the concept of multiple integration to fractional applications. To define the continuous analog of the Vignéras hierarchy, we utilize the Riemann-Liouville fractional integral operator \(J^\alpha\) of order \(\alpha\). For a base function \(f(t) = \ln t\), the operator is defined as:

$$ J^\alpha (\ln x) = \frac{1}{\Gamma(\alpha)} \int_{0}^{x} (x - t)^{\alpha - 1} \ln(t) \, dt $$

To evaluate the asymptotic growth of this integral, we employ the same integration by parts strategy. Integrating the power function and differentiating the logarithm yields the exact analytical closed-form solution:

$$ J^\alpha (\ln x) = \frac{x^\alpha}{\Gamma(\alpha + 1)} \left[ \ln(x) - \psi(\alpha + 1) - \gamma \right] $$

where \(\psi(z)\) is the digamma function and \(\gamma\) is the Euler-Mascheroni constant.

For a fixed fractional order \(\alpha\), both \(\psi(\alpha + 1)\) and \(\gamma\) are finite constants. As \(x \to \infty\), the term \(\ln(x)\) strictly dominates the bracketed expression. Thus, the leading term of the expansion reduces to:

$$ J^\alpha (\ln x) \sim \frac{1}{\Gamma(\alpha + 1)} x^\alpha \ln(x) $$

This completes the proof that \(\ln G_\alpha(x)\) scales at a rate of \(\Theta(x^\alpha \ln x)\). Furthermore, for integer values \(\alpha = n\), the coefficient cleanly collapses to the expected discrete bound of \(x^n \ln x\), matching the previous theorem, as well as known literature. ∎