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 $\propto(\log(1-y))^2$.

  • $\begingroup$ Do you need full expansion near $y=1$ or the first coefficients would be sufficient? $\endgroup$ – Start wearing purple Jun 5 '13 at 18:43
  • $\begingroup$ Just the part that doesn't tend to $0$ as $y\to1$. It also matters whether the expansion is uniform in $\beta$ and $\delta$. $\endgroup$ – Kirill Jun 5 '13 at 20:15

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:

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

  • 1
    $\begingroup$ Thank you very much for this answer. When you sum over $n$ to get the $O(\lambda^3)$ term, can the terms hidden in $O(\lambda^3/n^2)$ not have non-zero contributions to the sum as $y\to1^-$? I.e., in the last equation should there not be also extra terms $\propto \zeta(3)$, $\zeta(4)$ coming from the expansion of digamma in powers of $n$? $\endgroup$ – Kirill Jun 7 '13 at 17:00
  • 1
    $\begingroup$ The terms hidden in $O(\lambda^3/n^2)$ can have non-zero contribution to the sum as $y \to 1^{-}$. However, it is sort of pointless to extract only the lowest order one. You do need all orders of them to calculate the correct contribution from those terms with small $n$... $\endgroup$ – achille hui Jun 7 '13 at 17:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.