Generalization of integral using special function

Question

Is it possible to find a closed formula for the following integrals for any $k\in \mathbb{N}^{*}$: $$ I=\int_{0}^{1} \left[\ln\left(\frac{p}{1-p}\right)\right]^{k}p^{2}dp ,\quad J=\int_{0}^{1} \left[\ln\left(\frac{p}{1-p}\right)\right]^{k}pdp ,\quad K=\int_{0}^{1} \left[\ln\left(\frac{p}{1-p}\right)\right]^{k}dp. $$ I know that $I,J,K$ exist because the function is well defined and can be extended to $0$ and $1$ by continuity.

My question: is it possible to write $I,J,K$ as function of some special function like gamma or beta function? if yes can you give my the right expression.

Answer 1

I realized that using beta function is not a good idea. Here is an alternate solution:

Use the substitution $u = \frac{p}{1-p}$. Then $p = \frac{u}{1+u}$ and $dp = \frac{du}{(1+u)^{2}}$. So we have

\begin{align*} \int_{0}^{1} p^{\nu-1} \log^{k} \left( \frac{p}{1-p} \right) \, dp &= \int_{0}^{\infty} \frac{u^{\nu-1}\log^{k} u}{(1+u)^{\nu+1}} \, du \\ &= \int_{0}^{1} \frac{u^{\nu-1}\log^{k} u}{(1+u)^{\nu+1}} \, du + \int_{1}^{\infty} \frac{u^{\nu-1}\log^{k} u}{(1+u)^{\nu+1}} \, du. \end{align*}

For the second integral, we use the substitution $u \mapsto u^{-1}$ to obtain

\begin{align*} \int_{0}^{1} p^{\nu-1} \log^{k} \left( \frac{p}{1-p} \right) \, dp &= \int_{0}^{1} \frac{u^{\nu-1}\log^{k} u}{(1+u)^{\nu+1}} \, du + (-1)^{k} \int_{0}^{1} \frac{\log^{k} u}{(1+u)^{\nu+1}} \, du. \end{align*}

Plugging $u = e^{-t}$ and utilizing the following generalized binomial series,

$$ \frac{1}{(1 + z)^{\beta+1}} = \sum_{j=0}^{\infty} \binom{j+\beta}{j} (-1)^{j} z^{j}, $$

it follows that

\begin{align*} \int_{0}^{1} p^{\nu-1} \log^{k} \left( \frac{p}{1-p} \right) \, dp &= (-1)^{k}\int_{0}^{\infty} \frac{t^{k}e^{-\nu t} }{(1+e^{-t})^{\nu+1}} \, dt + \int_{0}^{\infty} \frac{t^{k} e^{-t}}{(1+e^{-t})^{\nu+1}} \, dt \\ &= \sum_{j=0}^{\infty} \binom{j+\nu}{j} (-1)^{j} \left\{ (-1)^{k}\int_{0}^{\infty} t^{k}e^{-(\nu+j) t} \, dt + \int_{0}^{\infty} t^{k} e^{-(1+j)t} \, dt \right\} \\ &= \sum_{j=0}^{\infty} \binom{j+\nu}{j} (-1)^{j} \left\{ \frac{(-1)^{k} k!}{(j+\nu)^{k+1}} + \frac{k!}{(j+1)^{k+1}} \right\} \\ &= \frac{k!}{\nu!} \sum_{j=0}^{\infty} (-1)^{j} (j+1)_{\nu} \left\{ \frac{(-1)^{k}}{(j+\nu)^{k+1}} + \frac{1}{(j+1)^{k+1}} \right\}, \\ \end{align*}

where

$$ (x)_{n} = \frac{\Gamma(x+n)}{\Gamma(x)} $$

stands for the Pochhammer symbol. If $\left[{n\atop k}\right]$ denotes the (unsigned) Stirling's number of the first kind, then we have the following identity

$$ (x)_{n} = \sum_{k=0}^{n} \left[{n\atop k}\right] x^{k} $$ and hence

\begin{align*} \int_{0}^{1} p^{\nu-1} \log^{k} \left( \frac{p}{1-p} \right) \, dp &= \frac{k!}{\nu!} \left\{ (-1)^{k} \sum_{j=1}^{\infty} (-1)^{j-1} \frac{(j)_{\nu}}{(j+\nu-1)^{k+1}} + \sum_{j=1}^{\infty} (-1)^{j-1} \frac{(j)_{\nu}}{j^{k+1}} \right\} \\ &= \frac{k!}{\nu!} \left\{ (-1)^{k+\nu+1} \sum_{j=1}^{\infty} (-1)^{j-1} \frac{(j-\nu+1)_{\nu}}{j^{k+1}} + \sum_{j=1}^{\infty} (-1)^{j-1} \frac{(j)_{\nu}}{j^{k+1}} \right\} \\ &= \frac{k!}{\nu!} \sum_{l=0}^{\nu} \left[ {\nu \atop l} \right] \{ 1 + (-1)^{k-l+1} \} \eta(k-l+1), \end{align*}

where $\eta(z) = (1 - 2^{1-z}) \zeta(z)$ is the Dirichlet eta function. Since the only terms with $k - l + 1$ even survives, we can even write it as a rational combination of even powers of $\pi$.

You can check this numerically by the following code in Mathematica:

Eta[z_] := If[z == 1, Log[2], (1 - 2^(1 - z)) Zeta[z]];

{k, n} = {3, 4};
NIntegrate[p^(n - 1) Log[p/(1 - p)]^k, {p, 0, 1}] // Chop
N[k!/n! Sum[
    Abs[StirlingS1[n, l]] (1 + (-1)^(k - l + 1)) Eta[k - l + 1], 
    {l, 0, n}]]
Clear[n, k];