Does this integral have a closed form: $\int_0^1 \frac{x^{\beta-1}}{1-x}\log\frac{1-y x^\delta}{1-y}\mathrm dx$?

Consider the following integral:
$$G(\beta,\delta,y) = \int_0^1 \frac{x^{\beta-1}}{1-x}\log\frac{1-y x^\delta}{1-y}\mathrm dx,$$
with $\delta>0$, $\Re\beta>0$, $y\neq1$.

Does it have a closed form in terms of standard special functions?

If not, what is its asymptotic behaviour near the point $y=1$? Is there a closed-form expansion of the type
$$ G(\beta,\delta,y) = \frac{G_0(\beta,\delta)}{1-y} + G_1(\beta,\delta) + o(1), \qquad (y\to1-)$$
or something like it?

Is there a closed form if $\frac{1}{\delta}$ is a positive integer? Does anything useful happen if $y$ is a root of unity?

Context. I encountered this integral while trying to answer this question and also this question, but could not reduce the integral to a simpler form, except for the very specific values of $\beta=\frac14$ and $\delta=\frac14$. When $\delta=\frac14$, a substitution $x=y^4$ leads to an integral with logarithmic terms and the fraction $\frac{1}{1-y^4}$ that can be expanded in partial fractions, and the individual terms can be integrated rather more easily, so $G(\frac14,\frac14,y)$ can be expressed in terms of logs and polylogs.

Numerically, I find the singular behaviour to be $\propto(\log(1-y))^2$.

Answer

This is not a full answer. However, the leading singular behavior is
indeed $\propto \log(1-y)^2$.

For fixed $\beta, \delta$ with $\Re(\beta) > 0, \delta > 0$ and $|y| < 1$. If one expand the log in the integrand, one get:
$$G(\beta,\delta,y) = \int_0^1 \frac{x^{\beta-1}dx}{1-x}\left\{\log\frac{1-yx^\delta}{1-y}\right\}
= \int_0^1 \frac{x^{\beta-1}dx}{1-x}\left\{\sum_{n=1}^{\infty}\frac{y^n}{n}(1-x^{n\delta})\right\}\tag{*}$$
Notice for $x \in (0,1)$, whenever $n\delta \ge 1$, we have:
$$\left|\frac{1 – x^{n\delta}}{1-x}\right| \le n\delta$$

This implies
$$\sum_{n=\lceil\delta^{-1}\rceil}^{\infty}\left|\frac{y^n}{n}\left\{\frac{x^{\beta-1}}{1-x}(1-x^{n\delta})\right\}\right| \le x^{\Re(\beta)-1}\frac{\delta |y|}{1-|y|}
$$
and hence the partial sums in the expansion is dominated by a Lebseque integrable function.
By Lebesgue’s dominated convergence theorem, we can switch the order of summation and integration.

$$\begin{align}
G(\beta,\delta,y) = &\sum_{n=1}^{\infty}\frac{y^n}{n} \int_0^1
\left( \frac{1 – x^{\beta+n\delta-1}}{1-x} – \frac{1 – x^{\beta-1}}{1-x} \right)
dx\\
= & \sum_{n=1}^{\infty}\frac{y^n}{n}( \psi(\beta+n\delta)-\psi(\beta))\\
= & \sum_{n=1}^{\infty}\frac{y^n}{n}\left(
(\psi(\beta+n\delta) – \psi(n\delta)) + (\psi(n\delta) – \psi(n)) + \psi(n) – \psi(\beta)
\right)
\end{align}
$$
where $\psi(x)$ is the digamma function.

Let $\lambda = \max(|\beta|,\delta^{-1},1)$.
When $\beta$ is not too large and not too far away from the real axis,
$$\begin{align}
\psi(\beta+n\delta) – \psi(n\delta)
= & \beta \psi'(n\delta + \xi \beta)\quad\text{ for some }\xi \in (0,1)\\
= & \frac{\beta}{n\delta} + O(\frac{|\beta|}{n^2\delta^2})\\
\psi(n\delta) – \psi(n) = & (\log(n\delta) – \frac{1}{2n\delta}) – (\log(n) – \frac{1}{2n} ) + O(\frac{\lambda^2}{n^2})\\
= & \log\delta + \frac{\delta-1}{2n\delta} + O(\frac{\lambda^2}{n^2})\\
\psi(n) = & H_{n-1} – \gamma
\end{align}$$
where $H_k$ is the $k^{th}$ harmonic number. We get:

$$\begin{align}
G(\beta,\delta,y) =& \sum_{n=1}^{\infty} \frac{y^{n}}{n}\left\{
H_{n-1} + (\log\delta – \gamma – \psi(\beta)) + \frac{2\beta+\delta-1}{2n\delta} + O(\frac{\lambda^3}{n^2})
\right\}\\
= & \frac12 \log(1-y)^2 + (\gamma + \psi(\beta) – \log\delta)\log(1-y) + O(\lambda^3)
\end{align}$$
The $O(\lambda^3)$ term is a term which remains finite as $y \to 1^{-}$.
If $\lambda$ isn’t too large. i.e. $\beta$ not too big and $\delta$ not too small.
The limit of the $O(\lambda^3)$ term is approximately given by:

$$\lim_{y->1^{-}} O(\lambda^3) \text{-term} \sim \frac{2\beta+\delta-1}{2\delta}\zeta(2) = \frac{(2\beta+\delta-1)\pi^2}{12\delta}$$

Attribution
Source : Link , Question Author : Kirill , Answer Author : achille hui

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