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

Show that $\displaystyle{\int_0^1 \log \log \left(\frac{1}{x}\right) \frac{dx}{1+x^2} = \frac{\pi}{2}\log \left(\sqrt{2\pi} \Gamma\left(\frac{3}{4}\right) / \Gamma\left(\frac{1}{4}\right)\right)}$

This question was posted as part of this question:

Solve the integral $\displaystyle{S_k = (-1)^k \int_0^1 (\log(\sin \pi x))^k dx}$

I cannot think of a change of variable nor other integrating methods. May be there is a known method that I am missing.

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@PeterT.off: That isn't true at all. –  Eric Naslund Mar 18 '12 at 12:18
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## 1 Answer

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 both sides with respect to $s$ and plugging $s = 0$, we obtain a representation formula

$$\int_{0}^{1} \frac{\log \log (1/x)}{1+x^2} \; dx = \psi_0(1) \beta(1) + \beta'(1),$$

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.

$$\beta(s)=\left(\frac{\pi}{2}\right)^{s-1} \Gamma(1-s) \cos \left( \frac{\pi s}{2} \right)\,\beta(1-s).$$

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

$$\log \left( \frac{4n}{4n+1} \right) = -\frac{1}{4n} + O\left( \frac{1}{n^2} \right), \quad \log \left( \frac{4n}{4n-1} \right) = \frac{1}{4n} + O\left( \frac{1}{n^2} \right).$$

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

$$C(s) = s 2^{-2s-2} \left[ \zeta'(s+1) - \zeta(s+1) \log 4 \right] + O(s).$$

Thus we have

$$2^{-2s-1}\zeta(s+1) + B(s) + C(s) = 2^{-2s-1} \left[ \zeta(s+1) + s \zeta'(s+1) - s \zeta(s+1) \log 4 \right] + O(s).$$

But since

$$\zeta(1+s) = \frac{1}{s} + \gamma + O(s),$$

we have

$$\lim_{s\downarrow 0} \left( 2^{-2s-1}\zeta(s+1) + B(s) + C(s) \right) = \frac{\gamma}{2} - \log 2.$$

For $A(s)$, the summands are positive with possible finite exceptional terms. Thus the Monotone Convergence Theorem guarantees that

$$\lim_{s\downarrow 0} A(s) = \sum_{n=1}^{\infty} \left[ \log \left( \frac{4n+1}{4n-1} \right) - \frac{1}{2n} \right].$$

Let $L$ denote this limit. 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) \\ & \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}}} \\ & \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}, \end{align*}

where we have used the Euler's reflection formula in the last line. Combining all the efforts, we obtain

$$-\beta'(0) = \log (2 \pi \sqrt{2}) - 2 \log \Gamma\left(\frac{1}{4}\right) .$$

Now taking logarithmic differntiation to the functional equation, we have

$$\frac{\beta'(s)}{\beta(s)} = \log\left(\frac{\pi}{2}\right) - \psi_0 (1-s) - \frac{\pi}{2} \tan \left( \frac{\pi s}{2} \right) - \frac{\beta'(1-s)}{\beta(1-s)}.$$

Taking $s = 0$, we have

$$\frac{\beta'(0)}{\beta(0)} = \log\left(\frac{\pi}{2}\right) + \gamma - \frac{\beta'(1)}{\beta(1)} \quad \Longrightarrow \quad \beta'(1) = \beta(1) \left[ \log\left(\frac{\pi}{2}\right) + \gamma - \frac{\beta'(0)}{\beta(0)} \right].$$

But again by the functional equation, we have $\beta(0) = \frac{1}{2}$. Therefore

$$\beta'(1) = \frac{\pi}{4} \left[ \gamma + 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right],$$

and hence

$$\int_{0}^{1} \frac{\log \log (1/x)}{1+x^2} \; dx = \frac{\pi}{4} \left[ 2 \log 2 + 3 \log \pi - 4 \log \Gamma\left(\frac{1}{4}\right) \right],$$

which is identical to the proposed answer.

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Excellent answer. I was trying to figure out how to evaluate $\beta^'(1)$, but couldn't get very far. –  Eric Naslund Mar 18 '12 at 12:18
Ahhh, a small typo. Each of the zeta terms should be of the form $\zeta(s+1)$, specifically it should be $2^{-2s-1}\zeta(s+1)$ instead of $2^{-2s-1}\zeta(s)$. Otherwise in the line where you take $s\rightarrow 0$, the pole from $s\zeta^'(s+1)$ will not cancel out with anything. ($\zeta^'(s+1)$ has a double pole) –  Eric Naslund Mar 18 '12 at 12:29
@EricNaslund, thanks for pointing out typos! I will fix it right now. –  sos440 Mar 18 '12 at 12:32
Once I saw half way through, I wanted to try myself and so did not check for errors. But thanks to Eric and s0s440 both. –  Kirthi Raman Mar 18 '12 at 16:50
@sos440, I just wanted to add, this is one of my favorite proofs I have read so far on Math Stack Exchange. –  Eric Naslund Apr 14 '12 at 9:33
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