오늘의 계산 44 - An integral related to the Dirichlet beta function

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Problem. Find the value of the integral \begin{equation*} \int_{0}^{1} \frac{\log \log (1/x)}{1+x^2} \; dx. \end{equation*}

Solution. By the substitution \( x = e^{-t} \), we find that \begin{align*} \int_{0}^{1} \frac{(\log (1/x))^s}{1+x^2} \; dx &= \int_{0}^{\infty} \frac{t^s e^{-t}}{1 + e^{-2t}} \; dt \\ &= \int_{0}^{\infty} \sum_{n=0}^{\infty} (-1)^n t^s e^{-(2n+1)t} \; dt \\ &= \sum_{n=0}^{\infty} (-1)^n \, \frac{\Gamma(s+1)}{(2n+1)^{s+1}} \\ &= \Gamma(s+1)L(s+1, \chi_4), \end{align*} where \( L(s, \chi_4) \) is the Dirichlet L-function of the unique non-principal character \( \chi_4 \) to the modulus 4. Often it is denoted as \( \beta(s) \) and called the Dirichlet beta function. Thus differentiating with respect to \( s \) and plugging \( s = 0 \), we obtain \begin{equation}\int_{0}^{1} \frac{\log \log (1/x)}{1+x^2} \; dx = \psi_0(1) \beta(1) + \beta'(1), \label{eqn:int_1} \end{equation} and the problem reduces to find the value of \( \beta(1) \) and \( \beta'(1) \). Note that \( \beta(1) = \frac{\pi}{4} \) is just the Gregory series. For \( \beta'(1) \), we first notice that the following functional equation holds. \begin{equation} \beta(s)=\left(\frac{\pi}{2}\right)^{s-1} \Gamma(1-s) \cos \left( \frac{\pi s}{2} \right)\,\beta(1-s). \label{eqn:func_eqn} \end{equation} This follows from the general theory of Dirichlet L-functions, and one can consult with any analytic number theory textbook to find its proof. Therefore it is sufficient to calculate \( \beta'(0) \). For \( 0 < s \), we have \begin{align*} -\beta'(s) &= \sum_{n=1}^{\infty} \left[ \frac{\log(4n+1)}{(4n+1)^s} - \frac{\log(4n-1)}{(4n-1)^s} \right] \\ &= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ \log \left( \frac{4n+1}{4n-1} \right) - \frac{1}{2n} \right] + 2^{-2s-1}\zeta(s+1) \\ & \qquad + \sum_{n=1}^{\infty} \left( \frac{1}{(4n+1)^s} - \frac{1}{(4n)^s} \right) \log (4n+1) \\ & \qquad + \sum_{n=1}^{\infty} \left( \frac{1}{(4n)^s} - \frac{1}{(4n-1)^s} \right) \log (4n-1) \\ & =: A(s) + 2^{-2s-1}\zeta(s+1) + B(s) + C(s). \end{align*} We first estimate \( B(s) \). As \( n \to \infty \), we have \begin{equation*} \log \left( \frac{4n}{4n+1} \right) = -\frac{1}{4n} + O\left( \frac{1}{n^2} \right), \quad \log \left( \frac{4n}{4n-1} \right) = \frac{1}{4n} + O\left( \frac{1}{n^2} \right). \end{equation*} Thus when \( s \to 0 \), \begin{align*} B(s) &= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ \exp\left( s \log \left( \frac{4n}{4n+1} \right) \right) - 1 \right] \left[ \log (4n) - \log \left(\frac{4n}{4n+1} \right) \right] \\ &= \sum_{n=1}^{\infty} \frac{1}{(4n)^s} \left[ - \frac{s}{4n} + O \left(\frac{s^2}{n^2} \right) \right] \left[ \log (4n) + O \left(\frac{1}{n} \right) \right] \\ &= -s 2^{-2s-2} \sum_{n=1}^{\infty} \frac{1}{n^{s+1}} \log (4n) + O(s) \\ &= s 2^{-2s-2} \left[ \zeta'(s+1) - \zeta(s+1) \log 4 \right] + O(s). \end{align*} Similar consideration also shows that \begin{equation*} C(s) = s 2^{-2s-2} \left[ \zeta'(s+1) - \zeta(s+1) \log 4 \right] + O(s). \end{equation*} Thus we have \begin{equation*} 2^{-2s-1}\zeta(s+1) + B(s) + C(s) = 2^{-2s-1} \left[ \zeta(s+1) + s \zeta'(s+1) - s \zeta(s+1) \log 4 \right] + O(s). \end{equation*} But since \begin{equation*} \zeta(s+1) = \frac{1}{s} + \gamma + O(s), \end{equation*} we have \begin{equation} \lim_{s\downarrow 0} \left( 2^{-2s-1}\zeta(s+1) + B(s) + C(s) \right) = \frac{\gamma}{2} - \log 2. \label{eqn:inter_1} \end{equation} For \( A(s) \), the summands are positive with possible finite exceptional terms. Thus the Monotone Convergence Theorem guarantees that \begin{equation*} \lim_{s\downarrow 0} A(s) = \sum_{n=1}^{\infty} \left[ \log \left( \frac{4n+1}{4n-1} \right) - \frac{1}{2n} \right]. \end{equation*} Let \( L \) denote this limiting series. Then by Stirling's formula, \begin{align} e^{L} & \stackrel{N\to\infty}{\sim} \prod_{n=1}^{N} \left( \frac{4n+1}{4n-1} \right) e^{-1/2n} \sim \frac{e^{-\gamma/2}}{\sqrt{N}} \prod_{n=1}^{N} \left( \frac{n+(1/4)}{n-(1/4)} \right) \nonumber \\ & \sim \frac{e^{-\gamma/2}}{\sqrt{N}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} \frac{\Gamma\left(N+\frac{5}{4}\right)}{\Gamma\left(N+\frac{3}{4}\right)} \sim \frac{e^{-\gamma/2}}{\sqrt{N}} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} \frac{\left( \frac{N + (5/4)}{e} \right)^{N+\frac{5}{4}}}{\left( \frac{N + (3/4)}{e} \right)^{N+\frac{3}{4}}} \nonumber \\ & \sim e^{-\gamma/2} \frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)} = 4 e^{-\gamma/2} \frac{\pi \sqrt{2}}{\Gamma\left(\frac{1}{4}\right)^2}, \label{eqn:inter_2} \end{align} where the last line follows from the Euler's reflection formula. Combining \eqref{eqn:inter_1} and \eqref{eqn:inter_2}, we obtain \begin{equation} -\beta'(0) = \log (2 \pi \sqrt{2}) - 2 \log \Gamma\left(\frac{1}{4}\right) . \label{eqn:beta_prime_0} \end{equation} Now taking logarithmic differntiation to the functional equation \eqref{eqn:func_eqn}, we have \begin{equation*} \frac{\beta'(s)}{\beta(s)} = \log\left(\frac{\pi}{2}\right) - \psi_0 (1-s) - \frac{\pi}{2} \tan \left( \frac{\pi s}{2} \right) - \frac{\beta'(1-s)}{\beta(1-s)}. \end{equation*} Taking \( s = 0 \), we have \begin{equation*} \frac{\beta'(0)}{\beta(0)} = \log\left(\frac{\pi}{2}\right) + \gamma - \frac{\beta'(1)}{\beta(1)} \quad \Longrightarrow \quad \beta'(1) = \beta(1) \left[ \log\left(\frac{\pi}{2}\right) + \gamma - \frac{\beta'(0)}{\beta(0)} \right]. \end{equation*} But again by \eqref{eqn:func_eqn}, we have \( \beta(0) = \frac{1}{2} \). Therefore, by \eqref{eqn:beta_prime_0} \begin{equation*} \beta'(1) = \frac{\pi}{4} \left[ \gamma + 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right], \end{equation*} and hence \eqref{eqn:int_1} becomes \begin{equation*} \int_{0}^{1} \frac{\log \log (1/x)}{1+x^2} \; dx = \frac{\pi}{4} \left[ 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right]. \end{equation*}

References

1. \( \int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2} \)