A relation with complex unity and logs.

Let $n \in \mathbb{N}$ then prove that:

$$\int_0^{\pi/2} (\log \cos x+ \log 2 + ix)^n \, {\rm d}x + \int_0^{\pi/2} (\log \cos x + \log 2 - ix)^n \, {\rm d}x=0$$

Solution

 The function $f(z)=\log^n z $ is analytical in the open disk $\mathbb{D}=\{z \in \mathbb{C} \mid |z-1|<1\}$ since ($\mathfrak{Re}(1+z)>0$).

Using Gauss Mean Value Theorem we have that:

$$f(1)= \frac{1}{2\pi}\int_{0}^{2\pi} f\left ( 1+e^{ix} \right )\, {\rm d}x$$

However since $f(1)=0$ then we have successively:

$$\begin{align*}
0 &=\int_{0}^{2\pi} \log^n \left ( 1+\cos x +i \sin x \right )\, {\rm d}x \\
 &= \int_{0}^{2\pi} \log^n \left ( 2\cos \frac{x}{2} e^{ix/2} \right )\, {\rm d}x\\
 &\overset{u=x/2}{=\! =\! =\! =\!}\int_{0}^{\pi} \log^n \left ( 2\cos y \cdot e^{iy} \right )\, {\rm d}y \\
 &= \int_{0}^{\pi/2} \log^n \left ( 2\cos x \cdot e^{iy} \right ) \, {\rm d}y + \int_{\pi/2}^{\pi} \log^n \left ( 2\cos y \cdot e^{iy} \right )\, {\rm d}y \\
 &\overset{y =\pi-x }{=\! =\! =\! =\!} \int_{0}^{\pi/2}\log^n \left ( 2\cos x \cdot e^{ix} \right )\, {\rm d}y + \int_{0}^{\pi/2}\log^n \left ( -2 \cos x \cdot e^{i(\pi-x)} \right ) \, {\rm d}x \\
 &= \int_{0}^{\pi/2}\log^n \left ( 2\cos x \cdot e^{ix} \right )\, {\rm d}x + \int_0^{\pi/2}\log^n \left ( 2\cos x \cdot e^{-ix} \right )\, {\rm d}x
\end{align*}$$

Thus:

$$\int_{0}^{\pi/2}\left ( \log \cos x + \log 2 + ix \right )^n \, {\rm d}x + \int_{0}^{\pi/2}\left ( \log \cos x + \log 2 - i x \right )^n \, {\rm d}x = 0 \tag{1} \label{1}$$

Setting $n=1, \; n=2$ back at \eqref{1} we get that $\displaystyle \int_{0}^{\pi/2} \log^2 \cos x \, {\rm d}x = \frac{\pi}{2}\log^2 2 + \frac{\pi^3}{24}$ while for $n=3$ we get that

$$\int_{0}^{\pi/2} \log^3 \cos x \, {\rm d}x = -\frac{3\pi \zeta(3)}{4}- \frac{\pi^3 \log 2}{8} - \frac{\pi \log^3 2}{2}$$

The above relation can also be found in mathematica.gr