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\begin{align*} I = \int_0^{\pi/2} \left( \int_0^{\pi/2} \frac{\log(\cos(x/2)) - \log(\cos(y/2))}{\cos(x) - \cos(y)} dx\right) \ dy \end{align*}

What I do so far

Let $u = \cos(x/2)$ and $v = \cos(y/2)$. Then $du = -\sin(x/2) dx/2$ and $dv = -\sin(y/2) dy/2$.

$$I = \int_0^{\pi/2} \left( \int_0^{\pi/2} \frac{\log(u) - \log(v)}{u - v} \cdot \frac{-2 du}{2 \sin(x/2)} \right) \cdot \frac{-2 dv}{2 \sin(y/2)}$$

Logu=u

$$I = \int_0^{\pi/2} \left( \int_0^{\pi/2} \frac{\log(u) - \log(v)}{u - v} \cdot \frac{-2 du}{2 \sin(x/2)} \right) \cdot \frac{-2 dv}{2 \sin(y/2)} - \int_0^{\pi/2} \frac{\log(u) - \log(v)}{u - v} \cdot \frac{1}{\sin(x/2)} \cdot \frac{-2 dv}{2 \sin(y/2)}$$

$$\int \frac{\log(u) - \log(v)}{u - v} du = \log \left| \frac{u}{v} \right| + C$$

$$I = \int_0^{\pi/2} \left( \log \left| \frac{u}{v} \right| \cdot \frac{2}{\sin(y/2)} - \frac{\log(u) - \log(v)}{u - v} \cdot \frac{1}{\sin(x/2)} \cdot \frac{2 dv}{2 \sin(y/2)} \right)$$

This looks very scary

Jose Avilez
  • 13,432
Martin.s
  • 1

2 Answers2

3

\begin{align*} &\cos(x/2) = \sqrt{\frac{1 + \cos(x)}{2}} \\ &\int_0^{\pi/2} \int_0^{\pi/2} \frac{\log(\sqrt{\frac{1 + \cos(x)}{2}}) - \log(\sqrt{\frac{1 + \cos(y)}{2}})}{\cos(x) - \cos(y)} \, dx \, dy \\ &\qquad\qquad = \int_0^{\pi/2} \int_0^{\pi/2} \frac{\log(\sqrt{1 + u/2}) - \log(\sqrt{1 + v/2})}{u - v} \, du \, dv \\ &\qquad\qquad = \frac{1}{2} \int_0^1 \int_0^1 \frac{\log(1 + u) - \log(1 + v)}{(u - v) \sqrt{1 - u^2} \sqrt{1 - v^2}} \, du \, dv \\ &\qquad\qquad = \frac{1}{2} \int_0^1 \int_0^1 \frac{1}{((1 + tu)(1 + tv) \sqrt{1 - u^2} \sqrt{1 - v^2})} \, dt \, du \, dv \end{align*}

Change the order of integration

\begin{align*} &\qquad\qquad = \frac{1}{2} \int_0^1 \int_0^1 \frac{1}{((1 + tv) \sqrt{1 - v^2})} ( \\ &\qquad \qquad \int_0^1 \frac{1}{((1 + tu) \sqrt{1 - u^2})} du \biggr) dv \, dt \end{align*}

t=cos(x)

\begin{align*} &\qquad\qquad = \frac{1}{2} \int_0^1 \frac{\arccos^2(t)}{1 - t^2} \, dt \\ &\qquad\qquad \int_0^{\pi/2} \frac{x^2}{\sin(x)} \, dx \\ &\qquad\qquad = -\int_0^{\pi/2} x \log(\tan(x/2)) \, dx \\ &\qquad\qquad = -4 \int_0^{\pi/4} y \log(\tan(y)) \, dy \\ &\qquad\qquad = 8 \int_0^{\pi/4} y \sum_{k = 1}^\infty \frac{\cos(2(2k - 1)y)}{2k - 1} dy \\ &\qquad\qquad = 8 \sum_{k = 1}^\infty \int_0^{\pi/4} y \frac{\cos(2(2k - 1)y)}{2k - 1} dy \\ &\qquad\qquad = \sum_{k = 1}^\infty \left( \pi \frac{(-1)^{k - 1}}{(2k - 1)^2} - \frac{2}{(2k - 1)^3} \right) \\ &\qquad\qquad = \pi \sum_{k = 1}^\infty \frac{(-1)^{k - 1}}{(2k - 1)^2} - 2 \sum_{k = 1}^\infty \frac{1}{(2k - 1)^3} \\ &\qquad\qquad = \pi G - \frac{7}{4} \zeta(3) \end{align*}

Martin.s
  • 1
3

Substitute $(x,y)=\left(2\tan^{-1}s,2\tan^{-1}t\right)$, then reduce to a single integral by splitting up the integrand and exploiting symmetry:

$$\begin{align*} & \int_0^{\tfrac\pi2} \int_0^{\tfrac\pi2} \frac{\log \left(\cos \frac x2\right) - \log\left(\cos \frac y2\right)}{\cos x - \cos y} \, dx \, dy \\ &= \int_0^1 \int_0^1 \frac{\log \frac1{\sqrt{1+s^2}} - \log\frac1{\sqrt{1+t^2}}}{\frac1{1+s^2} - \frac1{1+t^2}} \, \frac4{(1+s^2)(1+t^2)} \, ds\,dt \\ &= \int_0^1 \int_0^1 \frac{\log\left(1+s^2\right)}{s^2 - t^2} \, dt \, ds - \int_0^1 \int_0^1 \frac{\log\left(1+t^2\right)}{s^2 - t^2} \, ds \, dt \\ &= \int_0^1 \frac{\log\left(1+s^2\right)}{2s} \log \frac{1+s}{1-s} \, ds - \int_0^1 \frac{\log\left(1+t^2\right)}{2t} \log \frac{1-t}{1+t} \, dt \\ &= \int_0^1 \log\left(1+s^2\right) \log\frac{1+s}{1-s} \, \frac{ds}s = \boxed{\pi G - \frac74 \zeta(3)} \end{align*}$$

whose (negative) value is shown in the OP here, and explicitly evaluated in this answer.

user170231
  • 25,320