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The equation $$ \int_{0}^{2\pi}\log\left(% \left[1 + \sin\left(x\right)\right]^{1 + \cos\left(x\right)} \over 1 + \cos\left(x\right) \right)\,{\rm d}x = 0 $$. Has been bothering me for a few days now. Note that the case from $0$ to $\pi/2$ has already been dealth with earlier. Some of the problems with the integral is the singularities at $x=\pi$ and $x=3\pi/2$, hower as the right and left limits agree the integral does not diverge.

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Now I want to obtain the answer through substitutions and algebraic manipulation. I want to avoid all series representations (Catalans Constant) and alike.

I think one has to split the integrals to avoid the singularities and then show they are alike without explicitly evalutating them. Note

$$ \int_0^{2\pi} f(x) \mathrm{d}x = \int_0^\pi f(x) \mathrm{d}x + \int_\pi^{3\pi/2} f(x) \mathrm{d}x + \int_{3\pi/2}^{2\pi} f(x) \mathrm{d}x $$ Explicit calculations one have $$ \int_0^{\pi/2} f(x)\mathrm{d}x = -1 + 2\log 2 \quad \, \quad \int_{\pi}^{3\pi/2} f(x)\mathrm{d}x = 1 \\ \int_{\pi/2}^{\pi} f(x)\mathrm{d}x = 1 + 4 K-2 \log 2 \quad \, \quad \int_{3\pi/2}^{2\pi} f(x)\mathrm{d}x = -1 - 4 K $$ Where $K$ is Catalan's constant. Now the sum is of course zero, but is it possible to show this without evaluating 4 integrals? The two first integrals I have been able to show $$ \begin{align*} \int_{\pi}^{3\pi/2} f(x)\,\mathrm{d}x & = \frac{1}{2}\int_{\pi}^{3\pi/2} f(x) + f(\pi-x)\,\mathrm{d}x \\ & = \frac{1}{2}\int_{\pi}^{3\pi/2} \log(1+\cos x)\sin x+ \log(1+\sin x)\cos x\,\mathrm{d}x \\ & = \frac{1}{2}\int_0^{-1} \log(1+u) + \log(1+u)\,\mathrm{d}x = 1 \end{align*} $$ Is it perhaps possible to find $$ \int_{3\pi/2}^{2\pi} f(x)\mathrm{d}x + \int_{\pi/2}^{\pi} f(x)\mathrm{d}x = -2 \log 2 $$ ?

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The last can be found elementarily without problems.

$$\begin{align} I &= \int_{\pi/2}^\pi f(x)\,dx\\ &= \int_{\pi/2}^\pi (1+\cos x)\log (1+\sin x) - \log (1+\cos x)\,dx\tag{$x = y+\pi/2$}\\ &= \int_0^{\pi/2} (1-\sin y)\log (1+\cos y) - \log (1-\sin y)\,dy\\ II &= \int_{3\pi/2}^{2\pi} f(x)\,dx\\ &= \int_{3\pi/2}^{2\pi} (1+\cos x)\log (1+\sin x) - \log (1+\cos x)\,dx \tag{$y = 2\pi-x$}\\ &= \int_0^{\pi/2} (1+\cos y)\log (1-\sin y) - \log (1+\cos y)\,dy\\ I+II &= \int_0^{\pi/2} \cos y \log (1-\sin y) - \sin y\log (1+\cos y)\,dy\\ &= \int_0^{\pi/2} \cos y \log (1-\sin y)\,dy\\ &\qquad + \int_0^{\pi/2} (-\sin y)\log (1+\cos y)\,dy\\ &= \int_0^1 \log (1-u)\,du + \int_1^0 \log (1+v)\,dv\\ &= \int_0^1 \log t\,dt - \int_1^2 \log t\,dt\\ &= \left[t\log t-t\right]_0^1 - \left[t\log t-t\right]_1^2\\ &= -1 - (2\log 2 - 2) + (-1)\\ &= -2\log 2. \end{align}$$

If we further use

$$\begin{align} III &= \int_\pi^{3\pi/2} (1+\cos x)\log (1+\sin x) - \log (1+\cos x)\,dx\\ &= \int_0^{\pi/2} (1-\cos y)\log (1-\sin y) - \log (1-\cos y)\,dy\\ I+II+III &= \int_0^{\pi/2} \log (1-\sin y) - \log (1-\cos y) - \sin y\log (1+\cos y)\,dy, \end{align}$$

the first two summands cancel (use $z = \pi/2-y$ on one of them), leaving

$$I+II+III = -\int_0^{\pi/2} \sin y \log (1+\cos y)\,dy = -\int_0^{\pi/2}\cos x \log (1+\sin x)\,dx.$$

Adding that to

$$\int_0^{\pi/2} (1+\cos x)\log (1+\sin x) - \log (1+\cos x)\,dx$$

we see that the entire integral is

$$\int_0^{\pi/2} \log (1+\sin x) - \log (1+\cos x)\,dx$$

which again cancels by symmetry, and we found

$$\int_0^{2\pi} \log \frac{(1+\sin x)^{1+\cos x}}{1+\cos x}\,dx = 0$$

by symmetry without evaluating a single integral explicitly.

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  • $\begingroup$ Beautiful demonstration ! $\endgroup$ – Claude Leibovici Jan 21 '14 at 14:44
  • $\begingroup$ Beautiful =) I'll wait a day or two too see if there are any other answers. But you will get the accept anyways, exactly what I was looking for. $\endgroup$ – N3buchadnezzar Jan 21 '14 at 15:16
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$$ \begin{align} &\int_0^{2\pi}\log\left(\frac{(1+\sin(x))^{1+\cos(x)}}{1+\cos(x)}\right)\,\mathrm{d}x\tag{1}\\ &=\int_0^{2\pi}\Big[(1+\cos(x))\log(1+\sin(x))-\log(1+\cos(x))\Big]\,\mathrm{d}x\tag{2} \end{align} $$ Since $\cos(\pi-x)=-\cos(x)$ and $\sin(\pi-x)=\sin(x)$, we get $$ \begin{align} &\int_0^{2\pi}(1+\cos(x))\log(1+\sin(x))\,\mathrm{d}x\tag{3}\\ &=\int_0^{2\pi}(1-\cos(x))\log(1+\sin(x))\,\mathrm{d}x\tag{4}\\ &=\int_0^{2\pi}\log(1+\sin(x))\,\mathrm{d}x\tag{5}\\ &=\int_0^{2\pi}\log(1+\cos(x))\,\mathrm{d}x\tag{6} \end{align} $$ Explanation:
$(4)$: substitute $x\mapsto\pi-x$
$(5)$: average $(3)$ and $(4)$
$(6)$: substitute $x\mapsto\pi/2-x$

Plugging $(6)$ into $(2)$, we get $$ \int_0^{2\pi}\log\left(\frac{(1+\sin(x))^{1+\cos(x)}}{1+\cos(x)}\right)\,\mathrm{d}x=0\tag{7} $$

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    $\begingroup$ I'll repeat myself: French town you know where ;) $\endgroup$ – Daniel Fischer Jan 21 '14 at 17:12

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