# What's the behaviour of $I_n=\int_0^1\left(\frac1{\log x}+\frac1{1-x}\right)^ndx$, as $n \to \infty$?

Set $$I_n :=\int_0^1\left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n \:\mathrm{d}x \qquad n=1,2,3,\cdots.$$

We have $$I_1 =\gamma, \quad I_2 =\log (2 \pi) - \frac 32, \quad I_3 = 6 \log A - \frac{31}{24}, \quad I_4 = 2 \log A + \frac{5 \zeta(3)}{2\pi^{2}}- \frac{49}{72}, \quad ...$$ where $A$ is the Glaisher-Kinkelin constant defined by $$$$\displaystyle A :=\lim_{n\to\infty}\frac{1^22^2\cdots n^n}{e^{-n^2/4}n^{\frac{n^2+n}{2}+\frac{1}{12}}}=1.28242712\cdots.$$$$

I wonder if there is a "simple" equivalent for $I_n$ as $n$ tends to $+\infty$?

Edit. I had designed the following integral $$\displaystyle \int_0^1\left(\frac{1}{\log x} + \frac{1}{1-x}\right)^2 \mathrm{d}x$$ which I submitted to American Mathematical Monthly (March 2012, problem 11629), the problem was then spread and came in this forum with different interesting solutions (I). An interesting general formula for $I_n$ has been found (II), but I don't think the latter formula is tractable for the above question on asymptotics.

• You've done a few others of this form. Can you link them in your question? – Semiclassical Jul 24 '14 at 3:34
• @Mhenni Benghorbal Yes: $$\frac 12 < \frac{1}{\log(1-x)} + \frac{1}{x} < 1, \quad 0<x<1.$$ Thanks. – Olivier Oloa Jul 24 '14 at 8:15
• @Mhenni Benghorbal Please see the edit in the question. – Olivier Oloa Jul 24 '14 at 9:26
• @OlivierOloa: Thanks for these links. – Mhenni Benghorbal Jul 24 '14 at 9:31
• @Venus There is no duplicate, since in the question above, we were looking for an equivalent as $n$ tends to $+\infty$, which is very different from a closed-form, the latter doesn't give a simple equivalent of the integral. So you should remove the inappropriate term "duplicate" here. Thanks. – Olivier Oloa Dec 20 '14 at 18:55

## 1 Answer

The main idea is that, near the largest point of the integrand (which occurs at $x=0$), we have

$$\frac{1}{\log x} + \frac{1}{1-x} \approx \frac{1}{\log x} + 1.$$

So we split the integral up as

$$\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx = \left[\int_0^{1/e} + \int_{1/e}^1 \right] \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx.$$

For the second integral we have the bound

$$0 < \int_{1/e}^{1} \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx < (e-1)^{-n}$$

and for the first integral we write

\begin{align} \int_0^{1/e} \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx &= \int_0^{1/e} \exp\left\{n \log\left(\frac{1}{\log x} + \frac{1}{1-x}\right)\right\}dx \\ &= \int_0^{1/e} \exp\left\{n \log\left(\frac{1}{\log x} + 1\right) + n f(x)\right\}dx, \end{align}

where

$$f(x) = \log\left(1+\frac{x}{\left(\frac{1}{\log x}+1\right)(1-x)}\right) = O(x) \quad \text{as x \to 0.}$$

We therefore expect that the largest contribution comes from a neighborhood of radius $1/n$ of the point $x=0$. The rest of the integral is bounded by

$$0 < \int_{1/n}^{1/e} \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx < \left(-\frac{1}{\log n} + \frac{n}{n-1}\right)^n = O(e^{-n/\log n}).$$

For $0 \leq x \leq 1/n$ we have

$$\exp\{nf(x)\} = 1 + O(nx),$$

so, after combining our estimates, we arrive at

\begin{align} \int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx &= \int_0^{1/n} \exp\left\{n \log\left(\frac{1}{\log x} + 1\right)\right\}dx \\ &\qquad + O\left(n \int_0^{1/n} x \exp\left\{n \log\left(\frac{1}{\log x} + 1\right)\right\}dx\right)\\ &\qquad + O(e^{-n/\log n}). \end{align}

Now we make the change of variables $-y = \log\left(\frac{1}{\log x} + 1\right)$, transforming our integrals into

\begin{align} &\int_0^{1/n} \exp\left\{n \log\left(\frac{1}{\log x} + 1\right)\right\}dx \\ &\qquad = \int_0^{-\log(1-1/\log n)} e^{-ny} \exp\left\{\frac{1}{e^{-y}-1}\right\} \frac{e^{-y}}{(1-e^{-y})^2}\,dy \end{align}

and

\begin{align} &\int_0^{1/n} x\exp\left\{n \log\left(\frac{1}{\log x} + 1\right)\right\}dx \\ &\qquad = \int_0^{-\log(1-1/\log n)} e^{-ny} \exp\left\{\frac{2}{e^{-y}-1}\right\} \frac{e^{-y}}{(1-e^{-y})^2}\,dy. \end{align}

We can expand the factor which is independent of $n$ as

\begin{align} \exp\left\{\frac{1}{e^{-y}-1}\right\} \frac{e^{-y}}{(1-e^{-y})^2} &= \exp\left\{-\frac{1}{y}-\frac{1}{2}\right\} \sum_{j=-2}^{\infty} c_j y^j \\ &= \exp\left\{-\frac{1}{y}-\frac{1}{2}\right\} \left( \frac{1}{y^2} -\frac{1}{12 y} - \frac{23}{288} + \frac{427 y}{51840} + \cdots \right), \end{align}

so on truncating the expansion at $j=J-1$ we obtain

\begin{align} \int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx &= \sum_{j=-2}^{J-1} \frac{c_j}{\sqrt{e}} \int_0^{-\log(1-1/\log n)} e^{-ny} e^{-1/y} y^j \,dy \\ &\qquad + O\left(\int_0^{-\log(1-1/\log n)} e^{-ny} e^{-1/y} y^J \,dy\right) \\ &\qquad + O\left(n \int_0^{-\log(1-1/\log n)} e^{-ny} e^{-2/y} y^{-2} \,dy \right) \\ &\qquad + O(e^{-n/\log n}). \end{align}

All that remains is to reattach the tails of the integrals. By doing so we add an error of no more than

\begin{align} &\int_{-\log(1-1/\log n)}^\infty e^{-ny} e^{-1/y} y^j \,dy \\ &\qquad = \int_{-\log(1-1/\log n)}^\infty e^{-ny/2} \left(e^{-ny/2} e^{-1/y} y^j\right) \,dy \\ &\qquad \leq \sqrt{\int_{-\log(1-1/\log n)}^\infty e^{-ny}\,dy} \cdot \sqrt{\int_{-\log(1-1/\log n)}^\infty e^{-ny} e^{-2/y} y^{2j}\,dy} \\ &\qquad = \sqrt{\frac{2}{n}} \left(1-\frac{1}{\log n}\right)^{n/2} \sqrt{\int_{-\log(1-1/\log n)}^\infty e^{-ny} e^{-2/y} y^{2j}\,dy} \\ &\qquad < \sqrt{\frac{2}{n}} \left(1-\frac{1}{\log n}\right)^{n/2} \sqrt{\int_{0}^\infty e^{-ey} e^{-2/y} y^{2j}\,dy} \\ &\qquad = O\left(n^{-1/2} e^{-n/(2\log n)}\right) \end{align}

for $n > e$ by the Cauchy-Schwarz inequality, with the same bound when $e^{-1/y}$ is replaced with $e^{-2/y}$. Thus we have

\begin{align} \int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx &= \sum_{j=-2}^{J-1} \frac{c_j}{\sqrt{e}} \int_0^\infty e^{-ny} e^{-1/y} y^j \,dy \\ &\qquad + O\left(\int_0^\infty e^{-ny} e^{-1/y} y^J \,dy\right) \\ &\qquad + O\left(n \int_0^\infty e^{-ny} e^{-2/y} y^{-2} \,dy \right) \\ &\qquad + O\left(n^{-1/2} e^{-n/(2 \log n)}\right). \end{align}

To normalize these integrals a little bit we can substitute $y = u/\sqrt{n}$ to get

\begin{align} \int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx &= \sum_{j=-2}^{J-1} \frac{c_j}{\sqrt{e}} n^{-(j+1)/2} \int_0^\infty e^{-\sqrt{n}(u+1/u)} u^j \,du \\ &\qquad + O\left(n^{-(J+1)/2} \int_0^\infty e^{-\sqrt{n}(u+1/u)} u^J \,du\right) \\ &\qquad + O\left(n^{3/2} \int_0^\infty e^{-\sqrt{n}(u+2/u)} u^{-2} \,du \right) \\ &\qquad + O\left(n^{-1/2} e^{-n/(2 \log n)}\right). \end{align}

With the basic Laplace method estimate

$$\int_0^\infty e^{-\sqrt{n}(u+\alpha/u)} u^j \,du \sim \alpha^{1/4} \sqrt{\frac{\pi}{n}} e^{-2\sqrt{\alpha n}}$$

we may conclude that the series we've written is actually an asymptotic series, i.e.

$$\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx \sim \sum_{j=-2}^{\infty} \frac{c_j}{\sqrt{e}} n^{-(j+1)/2} \int_0^\infty e^{-\sqrt{n}(u+1/u)} u^j \,du.$$

Here we recognize that

$$\int_0^\infty e^{-\sqrt{n}(u+1/u)} u^j \,du = 2 K_{-j-1}\left(2\sqrt{n}\right),$$

where $K_\nu$ is the modified Bessel function of the second kind (see eq. 10.32.10, DLMF).

One asymptotic series for our integral is thus $$\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx \sim \frac{2}{\sqrt{e}}\sum_{j=-2}^{\infty} c_j n^{-(j+1)/2} K_{-j-1}\left(2\sqrt{n}\right),$$ where the coefficients $c_j$ are defined by $$\exp\left\{\frac{1}{e^{-y}-1}\right\} \frac{e^{-y}}{(1-e^{-y})^2} = \exp\left\{-\frac{1}{y}-\frac{1}{2}\right\} \sum_{j=-2}^{\infty} c_j y^j.$$

We can transform this asymptotic series into one in terms of elementary functions by using the known asymptotic expansion for $K_\nu(r)$ as $r \to \infty$ (DLMF reference). In our case we have

$$K_{-j-1}\left(2\sqrt{n}\right) \sim \frac{\sqrt{\pi}}{2} n^{-1/4} e^{-2\sqrt{n}} \sum_{k=0}^{\infty} \frac{a_k(j+1)}{2^k} n^{-k/2},$$

where $a_0(j+1) = 1$ and

$$a_k(j+1) = \frac{1}{k!8^k}\prod_{m=1}^{k} \left(4(j+1)^2-(2m-1)^2\right).$$

Substituting this into our asymptotic expansion yields

\begin{align} \frac{2}{\sqrt{e}}\sum_{j=-2}^{\infty} c_j n^{-(j+1)/2} K_{-j-1}\left(2\sqrt{n}\right) &= \sqrt{\frac{\pi}{e}} n^{-1/4} e^{-2\sqrt{n}} \sum_{j=-2}^{\infty} \sum_{k=0}^{\infty} \frac{c_j a_k(j+1)}{2^k} n^{-(j+k+1)/2} \\ &= \sqrt{\frac{\pi}{e}} n^{1/4} e^{-2\sqrt{n}} \sum_{\ell = 0}^{\infty} \left(\sum_{j+k+2=\ell} \frac{c_j a_k(j+1)}{2^k}\right)n^{-\ell/2}, \end{align}

So we conclude that

As $n \to \infty$, \begin{align} &\int_0^1 \left(\frac{1}{\log x} + \frac{1}{1-x}\right)^n\,dx \\ &\qquad \sim \sqrt{\frac{\pi}{e}} n^{1/4} e^{-2\sqrt{n}} \sum_{\ell = 0}^{\infty} \left(\sum_{j+k+2=\ell} \frac{c_j a_k(j+1)}{2^k}\right)n^{-\ell/2} \\ &\qquad = \sqrt{\frac{\pi}{e}} n^{1/4} e^{-2\sqrt{n}} \left(1 + \frac{5}{48} n^{-1/2} - \frac{479}{4608} n^{-1} + \frac{15313}{3317760} n^{-3/2} + \cdots \right), \end{align} where the indices of the inner sum range over $j=-2,-1,0,1,2,\ldots$ and $k=0,1,2,3,\ldots$.

• Very nice job! I'm impressed. – Olivier Oloa Jul 25 '14 at 10:29