The Problem Pit: The Pit of Logarithmic Moments

 

The Pit of Logarithmic Moments: Wrestling With $\int_0^\infty x^n e^{-x} (\ln x)^m dx$

There are integrals that lure you in with simplicity. They look like exercises, the kind of thing you’d solve in two lines on a quiet afternoon. But sometimes, behind the mask of a simple integrand lies a sprawling network of special functions, combinatorics, and asymptotics that refuses to let go once you fall in.

This is the story of how I got trapped in one such pit:

$$I_{n,m} = \int_0^\infty x^n e^{-x} (\ln x)^m \, dx.$$

At first glance, I thought: ah, trivial, it’s just Gamma with a log. Within minutes, I realized I was not climbing out of this problem so easily.

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Step 1: “It’s Just Gamma, Right?”

We all know the Gamma function:

$$\Gamma(s) = \int_0^\infty x^{s-1} e^{-x} dx.$$

Differentiate under the integral sign:

$$\frac{d}{ds} \Gamma(s) = \int_0^\infty x^{s-1} e^{-x} \ln x \, dx.$$

Differentiate $m$ times:

$$\frac{d^m}{ds^m} \Gamma(s) = \int_0^\infty x^{s-1} e^{-x} (\ln x)^m \, dx.$$

So directly,

$$I_{n,m} = \Gamma^{(m)}(n+1).$$

Solved? Not really. All I had done was restate the problem. What is $\Gamma^{(m)}(n+1)$? That question opened the door to the pit.

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Step 2: First Peek — $m=1$ and $m=2$

For $m=1$:

$$I_{n,1} = \Gamma'(n+1).$$

Using the digamma function $\psi(x) = \frac{\Gamma'(x)}{\Gamma(x)}$:

$$I_{n,1} = \Gamma(n+1)\psi(n+1).$$

Since $\Gamma(n+1) = n!$:

$$I_{n,1} = n! \psi(n+1).$$

This already hides harmonic numbers. Recall:

$$\psi(n+1) = H_n - \gamma,$$

where $H_n = 1 + \tfrac12 + \cdots + \tfrac1n$ and $\gamma$ is Euler–Mascheroni.

So:

$$I_{n,1} = n!(H_n - \gamma).$$

That’s neat! But for $m=2$:

$$I_{n,2} = \Gamma''(n+1).$$

Expanding:

$$I_{n,2} = n!\Big(\psi'(n+1) + \psi(n+1)^2\Big).$$

Here $\psi'(x)$ is the trigamma function, and at integers:

$$\psi'(n+1) = \frac{\pi^2}{6} - \sum_{k=1}^n \frac{1}{k^2}.$$

So:

$$I_{n,2} = n!\Big((H_n - \gamma)^2 + \frac{\pi^2}{6} - \sum_{k=1}^n \frac{1}{k^2}\Big).$$

Now we see a pattern: zeta constants, harmonic numbers, factorials.

So what about $m=3$?

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Step 3: The Hydra Appears — $m=3,4$

The third derivative:

$$I_{n,3} = n!\Big(\psi''(n+1) + 3\psi(n+1)\psi'(n+1) + \psi(n+1)^3\Big).$$

Here $\psi''$ is the tetragamma function. At integers, it involves $-2\sum_{k=1}^n \frac{1}{k^3}$. So:

$$I_{n,3} = n!\Big((H_n - \gamma)^3 + 3(H_n - \gamma)\Big(\frac{\pi^2}{6} - \sum_{k=1}^n \frac{1}{k^2}\Big) - 2\sum_{k=1}^n \frac{1}{k^3}\Big).$$

Already messy.

For $m=4$:

$$I_{n,4} = n!\Big(\psi^{(3)}(n+1) + 4\psi(n+1)\psi''(n+1) + 3\psi'(n+1)^2 + 6\psi(n+1)^2\psi'(n+1) + \psi(n+1)^4\Big).$$

At integers:

• $\psi^{(3)}(n+1) = 6\sum_{k=1}^n \frac{1}{k^4} - \frac{\pi^4}{15}.$

So we get:

$$I_{n,4} = n!\Big((H_n - \gamma)^4 + 6(H_n - \gamma)^2\Big(\frac{\pi^2}{6} - \sum_{k=1}^n \tfrac{1}{k^2}\Big) + \cdots \Big).$$

The formulas explode. Each new $m$ brings higher zeta values $\zeta(k)$ and nested harmonic sums.

This is the Hydra: every derivative sprouts more heads.

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Step 4: The Bell Polynomial Revelation

At this point, I realized: these aren’t random; they’re structured.

The derivatives of $\Gamma$ expand like Bell polynomials in the derivatives of $\ln \Gamma$. More precisely:

$$I_{n,m} = n! \, Y_m\big(\psi(n+1), \psi'(n+1), \dots, \psi^{(m-1)}(n+1)\big),$$

where $Y_m$ is the complete exponential Bell polynomial.

For small $m$:

• $Y_1(x_1) = x_1$.

• $Y_2(x_1,x_2) = x_1^2 + x_2$.

• $Y_3(x_1,x_2,x_3) = x_1^3 + 3x_1x_2 + x_3$.

• $Y_4(x_1,x_2,x_3,x_4) = x_1^4 + 6x_1^2x_2 + 4x_1x_3 + 3x_2^2 + x_4$.

And indeed, that matches exactly what I computed!

So the combinatorial skeleton behind these integrals is Bell polynomials in polygamma values.

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Step 5: Stirling Numbers in the Shadows

Then I recalled: Bell polynomials connect to Stirling numbers.

Expanding $\Gamma(n+1+t)$:

$$\Gamma(n+1+t) = n! \sum_{m=0}^\infty S_m(n) \frac{t^m}{m!},$$

where the coefficients $S_m(n)$ are complicated sums involving Stirling numbers of the first kind.

Thus,

$$I_{n,m} = n! \cdot S_m(n),$$

where $S_m(n)$ encodes the log-derivative structure combinatorially.

This was a revelation: the humble integral secretly encodes Stirling numbers.

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Step 6: The Asymptotic Mirage (Again)

I came back to asymptotics.

For large $n$:

• $\psi(n+1) \sim \ln n - \tfrac{1}{2n} + \tfrac{1}{12n^2} - \cdots$.

• $\psi'(n+1) \sim \tfrac{1}{n} + \tfrac{1}{2n^2} - \tfrac{1}{6n^3} + \cdots$.

So:

$$I_{n,m} \sim n!(\ln n)^m + \binom{m}{1}n!(\ln n)^{m-1}\Big(-\gamma + \tfrac{1}{2n}\Big) + \binom{m}{2}n!(\ln n)^{m-2}\Big(\zeta(2) - \tfrac{1}{n}\Big) + \cdots$$

Every layer of correction brings in zeta constants and Bernoulli numbers.

It’s like peeling an infinite onion.

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Step 7: Climbing Out (Barely)

So what did I learn in this pit?

1. Exact Form:

   $$I_{n,m} = n! \, Y_m\big(\psi(n+1), \psi'(n+1), \dots, \psi^{(m-1)}(n+1)\big).$$

2. Combinatorial Backbone: Bell polynomials and Stirling numbers orchestrate the chaos.

3. Asymptotics:

   $$I_{n,m} \sim n!(\ln n)^m \quad (n\to\infty),$$

   with corrections from $\gamma$, $\zeta(2), \zeta(3), \dots$.

4. Moral: What looked like a toy integral is a portal into special functions, combinatorics, and asymptotic analysis.

The pit was deep — and I barely crawled out.

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