$\displaystyle\int_{0}^{\frac{\pi}{2}}\ln(1+\alpha^N\tan(x)^N)\mathrm{d}x\quad$ where $N\in\mathbb{N}$

Question

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I wanted to calculate the following integrals $$\displaystyle\int_{0}^{\frac{\pi}{2}}\ln(1+\tan(x)^N)\mathrm{d}x\qquad\text{for }N\in\mathbb{N}$$ and I used the Feymann method applied to the function $$\color{blue}{\displaystyle F(\alpha):=\int_{0}^{\frac{\pi}{2}}\ln(1+\alpha^N\tan(x)^N)\mathrm{d}x}$$ Two cases can be distinguished:

$N$ is even

if $N= 2n$ the integral is easy: $$\color{green}{\int_{0}^{\frac{\pi}{2}}\ln(1+\alpha^{2n}\tan(x)^{2n})\mathrm{d}x=\frac{\pi}{2}\sum_{k=1}^{n}\ln\left(\alpha^2+2\sin\left(\frac{2k-1}{2n}\pi\right)\alpha+1\right)}$$ So $$\int_{0}^{\frac{\pi}{2}}\ln(1+\tan(x)^{2n})\mathrm{d}x=\frac{\pi}{2}\sum_{k=1}^{n}\ln\left(2+2\sin\left(\frac{2k-1}{2n}\pi\right)\right)$$

$N$ is odd

if $N= 2n+1$ the integral much more difficult:

I couldn't find the closed form, but I was able to find the formula for the case of $n=0$ and $n=1$:

$$\int_{0}^{\frac{\pi}{2}}\ln\left(1+a\tan\left(x\right)\right)\mathrm{d}x=\frac{\pi}{4}\ln\left(a^{2}+1\right)+\Im\left[\text{Li}_2\left(ia\right)\right]-\ln(a)\arctan(a)$$ $$\int_{0}^{\frac{\pi}{2}}\ln\left(1+a^3\tan\left(x\right)^3\right)\mathrm{d}x=\frac{\pi}{2}\left(\frac{1}{2}\ln\left(a^{6}+1\right)-\frac{2}{3}\ln\left(\frac{a^{2}-\sqrt{3}a+1}{a^{2}+\sqrt{3}a+1}\right)\right)-\frac{\Im\left[\text{Li}_{2}\left(ia^{3}\right)\right]}{3}+\ln\left(a\right)\arctan\left(a^{3}\right)$$ So $$\int_{0}^{\frac{\pi}{2}}\ln\left(1+\tan\left(x\right)\right)\mathrm{d}x=\frac{\pi}{4}\ln(2)+C$$ $$\int_{0}^{\frac{\pi}{2}}\ln\left(1+\tan\left(x\right)^3\right)\mathrm{d}x=\frac{\pi}{6}\left(\frac{3}{2}\ln\left(2\right)+4\ln\left(2+\sqrt{3}\right)\right)-\frac{C}{3}$$ Where

• $\Im[\text{Li}_2(z)]$ is the imaginary part of the polylog function
• $C$ is the Catalan constant

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Since the general integral is very complicated, I wanted to see if it was possible to see the pattern of the formula by eye. At first I thought it was: $${\frac{\pi}{2}\left[\frac{1}{2}\ln\left(x^{2N}+1\right)+\frac{2}{N}\sum_{k=1}^{n}\ln\left(\frac{x^{2}+2\sin\left(\frac{\left(2k-1\right)\pi}{N}\right)x+1}{x^{2}-2\sin\left(\frac{\left(2k-1\right)\pi}{N}\right)x+1}\right)\right]+\left(-1\right)^{n}\left(\frac{\Im[\text{Li}_2\left(ix^{N}\right)]}{N}-\ln\left(x\right)\arctan\left(x^{N}\right)\right)}$$

Unfortunately this is not correct. Could anyone help me? Even just calculating the subsequent cases with $N=5,7$ etc...

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Usefull formulas

$$\begin{align} f(\alpha)=&F'(\alpha)\\ =&\frac{2n+1}{2a}\pi-\frac{2a\ln\left(a\right)+\pi}{2a\left(a^{2}+1\right)}-2\sum_{k=1}^{n}\int_{0}^{\infty}\frac{1-\cos\left(\frac{2k-1}{2n+1}\pi\right)x}{x^{2}-2\cos\left(\frac{2k-1}{2n+1}\pi\right)x+1}\cdot\frac{1}{a^{2}+x^{2}}\mathrm{d}x \end{align}$$

Where, for $\theta\in(0,2\pi]$ and $a>0$:

$${\int_{0}^{\infty}\frac{1-\cos\left(\theta\right)x}{x^{2}-2\cos\left(\theta\right)x+1}\cdot\frac{1}{a^{2}+x^{2}}\mathrm{d}x=\frac{\cos\left(2\theta\right)\frac{\pi}{2}a+\frac{\pi}{2a}+\sin\left(\theta\right)\left(a^{2}-1\right)\left(\pi-\theta\right)-\cos\left(\theta\right)\left(a^{2}+1\right)\ln\left(a\right)}{a^{4}+2\cos\left(2\theta\right)a^{2}+1}}$$

Answer 1

I finally found the formula on my own:

Let, for brevity

• $N=2n+1$
• $\theta=\dfrac{2k\pi}{2n+1}$ for brevity. $$\color{blue}{\int_{0}^{\frac{\pi}{2}}\ln\left(1+a^{N}\tan\left(x\right)^{N}\right)dx=}$$ $$\color{blue}{\frac{\pi}{4}\ln\left(a^{2N}+1\right)-\sum_{k=1}^{n}\theta\operatorname{tanh^{-1}}\left(\frac{2\sin\left(\theta\right)a}{a^{2}+1}\right)+\left(-1\right)^{n}\left(\frac{\Im[\text{Li}_{2}\left(ia^{N}\right)]}{N}-\ln\left(a\right)\arctan\left(a^{N}\right)\right)}$$

So in particular:

$$\int_{0}^{\frac{\pi}{2}}\ln\left(1+\tan\left(x\right)^{N}\right)dx=\frac{\pi}{4}\ln\left(2\right)+\sum_{k=1}^{n}\theta\operatorname{gd^{-1}}\left(\theta\right)+\left(-1\right)^{n}\frac{C}{N}$$

Answer 2

The even and odd results of $$I_n=\int_0^{\frac\pi2} \ln (1+\tan^N x)\ dx $$ are respectively \begin{align} I_{2m} =& \ m\pi \ln2+2\pi\sum_{k=1}^{[\frac m2]}\ln \cos\frac{(m-2k+1)\pi}{4m}\\ I_{2m+1} =& \ \left(m+\frac14\right)\pi\ln2+\frac{(-1)^m C}{2m+1} +\pi\sum_{k=0}^{m-1}\left[\ln \cos a_k+{2(-1)^{m+k}}\ln\tan a_k \right]\\ \end{align} with $a_k= \frac{\pi(2k+1)}{4(2m+1)}$. In particular

\begin{align} \int_0^{\frac\pi2}\ln(1+\tan x)\ dx =& \ \frac\pi4\ln2 +C\\ \int_0^{\frac\pi2}\ln(1+\tan^2 x)\ dx =& \ \pi\ln2 \\ \int_0^{\frac\pi2}\ln(1+\tan^3 x)\ dx =& \ \frac\pi4\ln2+ \frac{2\pi}3\ln(2+\sqrt3) -\frac13 C \\ \int_0^{\frac\pi2}\ln(1+\tan^4 x)\ dx =& \ \pi\ln(2+\sqrt2)\\ \int_0^{\frac\pi2}\ln(1+\tan^5 x)\ dx =& -\frac{3\pi}4\ln2+\pi\ln\left(1+\sqrt5+\sqrt{2(5+\sqrt5)}\right)\\ &\ +\frac\pi{10}\ln\tan\frac\pi{20}-\frac{3\pi}{10}\ln\tan\frac{3\pi}{20} +\frac1{5}C\\ \int_0^{\frac\pi2}\ln(1+\tan^6 x)\ dx =& \ \pi\ln6\\ \end{align}