Hierarchical Integrals

Hierarchical integrals are the inverse operators of hierarchical derivatives. Rank \(0\) accumulates additively, rank \(1\) accumulates multiplicatively (relative scale), and rank \(2\) accumulates in the \(\ln\!\ln\) space (structural / double-exponential form).

Notation rule: we always write derivatives with explicit rank and degree: \(\;D_r^n\). On this page we focus on the most used case: \(\;D_n^{1}\).

Concept DOI: 10.5281/zenodo.17917302
Version DOI (v1.0.0): 10.5281/zenodo.17917303

GOSSA AHMED

Independent Researcher — Hierarchical Calculus

Quick Table of Contents

1) Definition of the hierarchical integral \(\mathcal{I}_n\) 2) Fundamental inverse relations 3) Domain conditions (avoid hidden singularities) 4) Rank 1: \(\mathcal{I}_1\) — verified examples 5) Rank 2: \(\mathcal{I}_2\) — verified examples 6) Mini workflow (how to integrate in practice) 7) Citation

1) Definition: the hierarchical integral \(\mathcal{I}_n\)

The rank-\(n\) hierarchical integral \(\mathcal{I}_n\) is defined as an inverse of the rank-\(n\), degree-1 derivative \(D_n^{1}\) on a suitable domain:

\[ \boxed{ D_{n}^{1}\big(\mathcal{I}_n[f]\big)=f } \]

Constant rule (critical): “Constants” live in the natural accumulation space of the rank:

• Rank 0: \(F \mapsto F + C\).
• Rank 1: \(\ln F \mapsto \ln F + C\) (equivalently \(F \mapsto C_{\times}F\)).
• Rank 2: \(\ln\!\ln F \mapsto \ln\!\ln F + C\) (a structural constant inside the double-exponential form).

Interpretation: The hierarchy is not “new constants”, but new natural spaces where additive constants live.

2) Fundamental inverse relations

2.1) First inverse relation

\[ \boxed{ D_{n}^{1}\!\big(\mathcal{I}_n[f]\big)=f } \]

2.2) Second inverse relation (up to a hierarchical constant)

If \(F\) satisfies \(D_n^{1}F=f\), then \(\mathcal{I}_n[f]\) reproduces \(F\) up to a rank-\(n\) constant:

Rank	“Constant” lives in	Equivalent form
0	\(F\)	\(F + C\)
1	\(\ln F\)	\(F\cdot C_\times\) where \(C_\times=e^C>0\)
2	\(\ln(\ln F)\)	\(\ln(\ln F)\mapsto \ln(\ln F)+C\) (structural constant)

Mnemonic: Sum → Product → Structural (double exponential) → higher ranks.

3) Domain conditions (avoid hidden singularities)

Because higher ranks use iterated logarithms, every integration formula must state its domain.

Rank	Required conditions	Reason
0	Classical domain of \(f\)	No iterated logs
1	\(x>0\), and outputs satisfy \(y(x)>0\)	\(\ln x\), \(\ln y\) are used
2	\(x>1\), and outputs satisfy \(y(x)>1\)	\(\ln(\ln x)\), \(\ln(\ln y)\) must exist

Important: For rank 2, avoid \(x\approx 1\) because \(\ln x\to 0\) causes singular behavior in the integrator.

4) Relative integral \(\mathcal{I}_1\) (rank 1) — verified examples

Since \(D_{1}^{1}y=\dfrac{d\ln y}{d\ln x}\), solving \(D_1^{1}y=f(x)\) gives:

\[ \boxed{ \mathcal{I}_1[f](x)=\exp\!\left(\int \frac{f(x)}{x}\,dx\right) } \]

\(f(x)\)	\(\mathcal{I}_1[f](x)\)	Check
\(1\)	\(x\)	\(D_{1}^{1}(x)=1\)
\(a\) (constant)	\(x^a\)	\(D_{1}^{1}(x^a)=a\)
\(\ln x\)	\(\exp\!\big((\ln x)^2/2\big)\)	\(D_{1}^{1}(\cdot)=\ln x\)
\(x^k\)	\(\exp\!\big(x^k/k\big)\ \ (k\neq0)\)	\(D_{1}^{1}(\cdot)=x^k\)
\(\dfrac{1}{\ln x}\)	\(\ln x\)	\(D_{1}^{1}(\ln x)=\dfrac{1}{\ln x}\)

Rank-1 integration naturally produces controlled relative growth: power-laws and exponentials appear automatically because the integrator is built in \(\ln\)-space.

5) Logarithmic integral \(\mathcal{I}_2\) (rank 2) — verified examples

Domain (rank 2): rank 2 uses \(\ln(\ln x)\), so we assume \(x>1\) (hence \(\ln x>0\)). Also outputs satisfy \(y>1\) so \(\ln(\ln y)\) exists.

Since \(D_{2}^{1}y=\dfrac{d\ln(\ln y)}{d\ln(\ln x)}\), solving \(D_2^{1}y=f(x)\) gives:

\[ \boxed{ \mathcal{I}_2[f](x)= \exp\!\left( \exp\!\left(\int \frac{f(x)}{x\,\ln x}\,dx\right) \right) } \]

\(f(x)\)	\(\mathcal{I}_2[f](x)\)	Check
\(1\)	\(x\)	\(D_{2}^{1}(x)=1\)
\(a\) (constant)	\(\exp\!\big((\ln x)^a\big)\) (up to a structural constant)	\(D_{2}^{1}(\cdot)=a\)
\(\ln\ln x\)	\(\exp\!\big(\exp((\ln\ln x)^2/2)\big)\)	\(D_{2}^{1}(\cdot)=\ln\ln x\)
\(\ln x\)	\(\exp(x)\)	\(D_{2}^{1}(e^x)=\ln x\)
\((\ln\ln x)^n\)	\(\exp\!\Big(\exp\big((\ln\ln x)^{n+1}/(n+1)\big)\Big)\)	\(D_{2}^{1}(\cdot)=(\ln\ln x)^n\)

Why “up to a structural constant”? Rank 2 constants live in \(\ln\!\ln y\), so multiplying the inner exponent by a positive constant does not change \(D_2^1\).

6) Mini workflow (how to integrate in practice)

Step 1: Identify the rank: is the equation natural in \(x\), \(\ln x\), or \(\ln\ln x\)?
Step 2: Rewrite the derivative definition (e.g. \(D_1^1y = d\ln y / d\ln x\)).
Step 3: Convert to an ordinary integral in the correct scale.
Step 4: Integrate and add the correct rank-constant.
Step 5: Invert the transform (exp or double-exp) to recover \(y(x)\).

7) Citation

GOSSA AHMED. Hierarchical Calculus — Hierarchical Integrals.
Official Website. Zenodo DOI: 10.5281/zenodo.17917302 (2025).