오늘의 계산 53 - An Integral

Here we want to calculate an integral related to our earlier calculation.

Problem. Prove that \begin{equation} \label{eqn:wts} \int_{0}^{1} \left( \frac{1}{\log x} + \frac{1}{1-x} \right)^2 \, dx = \log (2\pi) - \frac{3}{2}. \end{equation}

1^st Proof. Note that

$$ \frac{1}{1-x} + \frac{1}{\log x} = \int_{0}^{1} \frac{1 - x^{t}}{1-x} \, dt. $$

Then by Tonelli's theorem, we have

\begin{align*} \int_{0}^{1} \left( \frac{1}{\log x} + \frac{1}{1-x} \right)^2 \, dx &= \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{(1 - x^{s})(1 - x^{t})}{(1-x)^2} \, dsdtdx \\ &= \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{(1 - x^{s})(1 - x^{t})}{(1-x)^2} \, dxdsdt. \end{align*}

Expanding the integrand into series and applying Tonelli's theorem again,

\begin{align*} &= \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \sum_{n=1}^{\infty} n x^{n-1} (1 - x^{s})(1 - x^{t}) \, dxdsdt \\ &= \int_{0}^{1} \int_{0}^{1} \sum_{n=1}^{\infty} n \left( \frac{1}{n} - \frac{1}{n+s} - \frac{1}{n+t} + \frac{1}{n+s+t} \right) \, dsdt \\ &= \sum_{n=1}^{\infty} \Bigg( 1 + n(n+2) \log(n+2) - 2n(n+2)\log(n+1) + n(n+2) \log n \Bigg) \\ &= \lim_{N\to\infty} \sum_{n=1}^{N} \Bigg( 1 + n(n+2) \log(n+2) - 2n(n+2)\log(n+1) + n(n+2) \log n \Bigg). \end{align*}

Simplifying the inner summation, we have

\begin{align*} & \sum_{n=1}^{N} \Bigg( 1 + n(n+2) \log(n+2) - 2n(n+2)\log(n+1) + n(n+2) \log n \Bigg) \\ &= N + \sum_{n=3}^{N+2} (n^2-2n) \log n - \sum_{n=2}^{N+1} (2n^2 - 2) \log n + \sum_{n=1}^{N} (n^2+2n) \log n \\ &= N + (N^2 + 2N) \log(N+2) - (N^2+4N+1)\log(N+1) \\ &\quad + \sum_{n=1}^{N} (n^2-2n) \log n - \sum_{n=1}^{N} (2n^2 - 2) \log n + \sum_{n=1}^{N} (n^2+2n) \log n \\ &= N + (N^2 + 2N) \log \left( \frac{N+2}{N+1} \right) - (2N+1)\log(N+1) + 2 \log (N!) \\ &= 2N - (2N+1)\log N - \frac{3}{2} + 2 \log (N!) + O\left(N^{-1}\right). \end{align*}

Now it follows from the Stirling's formula that this is equivalent to $\log (2\pi)- \frac{3}{2}$, completing the proof.

2^nd Proof.Let $I$ denote the integral. By the substitution $x = e^{-t}$, we have

\begin{align*} I &= \int_{0}^{\infty} \left\{ \frac{1}{(1-e^{-t})^{2}} - \frac{2}{t(1-e^{-t})} + \frac{1}{t^2} \right\} e^{-t} \, dt \\ &= \int_{0}^{\infty} \left\{ \frac{e^{t}}{(e^{t} - 1)^{2}} - \frac{1}{t^2} \right\} \, dt + \int_{0}^{\infty} \left\{ \frac{1 + e^{-t}}{t^2} - \frac{2}{t(e^{t}-1)} \right\} \, dt. \end{align*}

It is easy to observe that the first integral is

\begin{align*} \int_{0}^{\infty} \left\{ \frac{e^{t}}{(e^{t} - 1)^{2}} - \frac{1}{t^2} \right\} \, dt &= \left[ \frac{1}{t} - \frac{1}{e^{t} - 1} \right]_{0}^{\infty} = -\frac{1}{2}. \end{align*}

We thus focus on the second integral. Associated to it, we introduce

$$ F(s) = \int_{0}^{\infty} \left\{ \frac{1 + e^{-t}}{t^2} - \frac{2}{t(e^{t}-1)} \right\} e^{-st} \, dt. $$

By the twice differentiation, we have

\begin{align*}F''(s) &= \int_{0}^{\infty} \left\{ 1 + e^{-t} - \frac{2t}{(e^{t}-1)} \right\} e^{-st} \, dt \\ &= \frac{1}{s} + \frac{1}{s+1} - 2\sum_{n=1}^{\infty} \frac{1}{(n+s)^2} \\ &= \frac{1}{s} + \frac{1}{s+1} - 2\psi_{1}(s+1). \end{align*}

Integrating and using the condition $F'(+\infty) = 0$, we have

$$ F'(s) = \log s + \log(s+1) - 2\psi_{0}(s+1). $$

Here we used the estimate $\psi_{0}(s) \sim \log s$ as $s \to \infty$. Integrating again, we have

$$ F(s) = s \log s + (s+1)\log(s+1) - 2s - 1 - 2\log\Gamma(s+1) + C. $$

To determine the constant $C$, we rearrange the terms as

$$ F(s) = \left\{ (s+1)\log\left(\frac{s+1}{s}\right) - 1 \right\} + 2\left\{ \left(s+\frac{1}{2}\right)\log s - s - \log\Gamma(s+1) \right\} + C. $$

Then by the Stirling's formula, we have

$$ 0 = F(+\infty) = -\log(2\pi) + C $$

and thus $C = \log (2\pi)$. Therefore

$$I = -\frac{1}{2} + F(0) = \log(2\pi) - \frac{3}{2}$$

as desired.

Remark. We remark that the first proof and the second proof corresponds to ordinary summation and Abel summation, respectively. Also, a technique similar to the first proof shows that

\begin{align*} & \int_{0}^{\infty} \left( \frac{1}{\log x} + \frac{1}{1-x} \right)^3 \, dx \\ &= \sum_{n=1}^{\infty} \Bigg[ \frac{n+1}{2} + \left(\frac{n^4}{4}+\frac{7 n^3}{4}+3 n^2+\frac{3 n}{2}\right) \log n \\ &\qquad \qquad -\left(\frac{3 n^4}{4}+\frac{21n^3}{4}+\frac{39 n^2}{4}+\frac{21 n}{4}\right) \log (n+1) \\ &\qquad \qquad +\left(\frac{3 n^4}{4}+\frac{21 n^3}{4}+\frac{21 n^2}{2}+6n\right) \log (n+2) \\ &\qquad \qquad -\left(\frac{n^4}{4}+\frac{7 n^3}{4}+\frac{15 n^2}{4}+\frac{9 n}{4}\right) \log(n+3) \Bigg] \\ &= \lim_{N\to\infty} \Bigg[ -\frac{31}{24} + \frac{3 N^2}{2}-\frac{6 N^2+6N+1}{2} \log N + 6 \sum_{n=1}^{N} n \log n \Bigg] \\ &= -\frac{31}{24} + 6 \log A, \end{align*}

where $A$ stands for the Glaisher-Kinkelin constant.