Compute $\sum\limits_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right)$ where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$ As part of solving:
\begin{equation}
 I_m = \int_0^1 \ln\left(1 + x^{2m}\right)\:dx.
\end{equation}
where $m \in \mathbb{N}$. I found an unresolved component that I'm unsure how to start:
\begin{equation} 
 G_m = \sum_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right),
\end{equation}
where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$
I'm just looking for a starting point. Any tips would be greatly appreciated. 
By the way, I was able to show (and this was part of the solution too) :
\begin{equation}
 \sum_{j = 0}^{m - 1} c_j = 0
\end{equation}
Edit: For those that may be interested. In collaboration with clathratus, we found that for $n > 1$
\begin{equation}
    \int_{0}^{1} \frac{1}{t^n + 1}\:dt = \frac{1}{n}\left[\frac{\pi}{\sin\left(\frac{\pi}{n} \right)}- B\left(1 - \frac{1}{n}, \frac{1}{n},  \frac{1}{2}\right)\right]
\end{equation}
Or for any positive upper bound $x$:
\begin{align}
 I_n(x) &= \int_{0}^{x} \frac{1}{t^n + 1}\:dt = \frac{1}{n}\left[\Gamma\left(1 - \frac{1}{n} \right)\Gamma\left(\frac{1}{n} \right)- B\left(1 - \frac{1}{n}, \frac{1}{n},  \frac{1}{x^n + 1}\right)\right]
\end{align}
Here though, I was curious to investigate when $n$ was an even integer. This is my work:
Here we will consider $r = 2m$ where $m \in \mathbb{N}$. In doing so, we observe that the roots of the function are $m$ pairs of complex roots $(z, c(z))$ where $c(z)$ is the conjugate of $z$. To verify this:
\begin{align}
 x^{2m} + 1 = 0 \rightarrow x^{2m} = e^{\pi i}
\end{align}
By De Moivre's formula, we observe that:
\begin{align}
 x = \exp\left({\frac{\pi + 2\pi j}{2m} i} \right) \mbox{ for } j = 0\dots 2m - 1,
\end{align}
which we can express as the set
\begin{align}
 S &= \Bigg\{  \exp\left({\frac{\pi + 2\pi \cdot 0}{2m} i} \right) , \:\exp\left({\frac{\pi + 2\pi \cdot 1}{2m} i} \right),\dots,\:\exp\left({\frac{\pi + 2\pi \cdot (2m - 2)}{2m} i} \right)\\
&\qquad\:\exp\left({\frac{\pi + 2\pi \cdot (2m - 1)}{2m} i} \right)\Bigg\},
\end{align}
which can be expressed as the set of $2$-tuples
\begin{align}
 S &= \left\{ \left( \exp\left({\frac{\pi + 2\pi j}{2m} i} \right) , \:\exp\left({\frac{\pi + 2\pi(2m - 1 - j )}{2m} i} \right)\right)\: \bigg|\: j = 0 \dots m - 1\right\}\\
& = \left\{ (z_j, c\left(z_j\right)\:|\: j = 0 \dots m - 1 \right\}
\end{align}
From here, we can factor $x^{2m} + 1$ into the form 
\begin{align}
x^{2m} + 1 &= \prod_{r \in S} \left(x + r_j\right)\left(x + c(r_j)\right) \\
 &= \prod_{i = 0}^{m - 1} \left(x^2 + \left(r_j + c(r_j)\right)x + r_j c(r_j)\right) \\
 &= \prod_{i = 0}^{m - 1}  \left(x^2 + 2\Re\left(r_j\right)x + \left|r_j \right|^2\right)
\end{align}
For our case here $\left|r_j \right|^2 = 1$ and $\Re\left(r_j\right) = \cos\left(\frac{\pi + 2\pi j}{2m} \right)= \cos\left(\frac{\pi}{2m}\left(1 + 2j\right)\right) = c_j$ 
\begin{align}
    \int_0^1 \log\left( x^{2m} + 1\right)\:dx &=  \int_0^1 \log\left(\prod_{r \in S} \left(x^2 + 2c_jx+ \left|r_j \right|^2\right)\right)\\
    &= \sum_{j = 0}^{m - 1} \int_0^1 \log\left(x^2 + 2c_jx + 1 \right)\\
    &= \sum_{j = 0}^{m - 1} \left[2\sqrt{1 - c_j^2}\arctan\left(\frac{x + c_j}{\sqrt{1 - c_j^2}}\right) + \left(x + c_j\right)\log\left(x^2 + 2c_jx + 1\right) - 2x \right]_0^1 \\
    &= \sum_{j = 0}^{m - 1} \left[ 2\sqrt{1 - c_j^2}\arctan\left(\sqrt{\frac{1 - c_j}{1 + c_j}} \right) + \log(2)c_j + \left(\log(2) - 2\right) + \left(c_j + 1\right)\log\left(c_j + 1\right) \right] \\
    &= 2\sum_{j = 0}^{m - 1}\sqrt{1 - c_j^2}\arctan\left(\sqrt{\frac{1 - c_j}{1 + c_j}} \right) + \log(2)\sum_{j = 0}^{m - 1} c_j + m\left(\log(2) - 2\right)\\
    &\qquad+ \sum_{j = 0}^{m - 1}\left(c_j + 1\right)\log\left(c_j + 1\right)
\end{align}
Thus, 
\begin{align}
\int_0^1 \log\left( x^{2m} + 1\right)\:dx &=\sum_{j = 0}^{m - 1}c_j\sin\left(\frac{\pi}{2m}\left(1 + 2j\right)\right)  + \log(2)\sum_{j = 0}^{m - 1} c_j + m\left(\log(2) - 2\right)\\
&\qquad+ \sum_{j = 0}^{m - 1}\left(c_j + 1\right)\log\left(c_j + 1\right)
\end{align}
 A: This does not answer the question as asked in the post.
Consider
$$J_m=\int \log(1+x^{2m})\,dx$$ One integration by parts gives
$$J_m=x \log \left(1+x^{2 m}\right)-2m\int \frac{ x^{2 m}+1-1}{x^{2 m}+1}\,dx=x \log \left(1+x^{2 m}\right)-2mx+2m\int \frac{dx}{x^{2 m}+1}$$ and
$$\int \frac{dx}{x^{2 m}+1}=x \, _2F_1\left(1,\frac{1}{2 m};1+\frac{1}{2 m};-x^{2 m}\right)$$ where appears  the Gaussian or ordinary hypergeometric function. 
So 
$$K_m=\int_0^a \log(1+x^{2m})\,dx=a \log \left(1+a^{2 m}\right)-2ma+2ma \, _2F_1\left(1,\frac{1}{2 m};1+\frac{1}{2 m};-a^{2 m}\right)$$ and, if $a=1$,
$$I_m=\int_0^1 \log(1+x^{2m})\,dx= \log \left(2\right)-2m+2m \, _2F_1\left(1,\frac{1}{2 m};1+\frac{1}{2 m};-1\right)$$ which can write
$$I_m=\log (2)-\Phi \left(-1,1,1+\frac{1}{2 m}\right)$$
where appears  the Lerch transcendent function.
Now, (this is something I never looked at), a few (ugly) expressions for $f_m=\Phi \left(-1,1,1+\frac{1}{2 m}\right)$ before any simplification
$$f(1)=\frac{\pi }{2}-2$$
$$f(2)=\frac{1}{4} \left(\pi  \tan \left(\frac{\pi }{8}\right)+\pi  \cot \left(\frac{\pi
   }{8}\right)-4 \sqrt{2} \log \left(\sin \left(\frac{\pi }{8}\right)\right)+4
   \sqrt{2} \log \left(\cos \left(\frac{\pi }{8}\right)\right)\right)-4$$
$$f(3)=\frac{2 \left(\pi -\sqrt{3} \log \left(\sqrt{3}-1\right)+\sqrt{3} \log
   \left(1+\sqrt{3}\right)\right)}{\left(\sqrt{3}-1\right)
   \left(1+\sqrt{3}\right)}-6$$
$$f(4)=\frac{1}{4} \left(\pi  \tan \left(\frac{\pi }{16}\right)+\pi  \cot \left(\frac{\pi
   }{16}\right)-8 \sin \left(\frac{\pi }{8}\right) \log \left(\sin \left(\frac{3
   \pi }{16}\right)\right)+8 \cos \left(\frac{\pi }{8}\right) \log \left(\cos
   \left(\frac{\pi }{16}\right)\right)-8 \cos \left(\frac{\pi }{8}\right) \log
   \left(\sin \left(\frac{\pi }{16}\right)\right)+8 \sin \left(\frac{\pi
   }{8}\right) \log \left(\cos \left(\frac{3 \pi }{16}\right)\right)\right)-8$$
A: I did it!
I actually have no idea whether or not this works, but this is how I did it. 
$n\in\Bbb N$
Define the sequence $\{r_k^{(n)}\}_{k=1}^{k=n}$ such that
$$x^n+1=\prod_{k=1}^{n}\big(x-r^{(n)}_{k}\big)$$
We then know that $$r_k^{(n)}=\exp\bigg[\frac{i\pi}{n}(2k-1)\bigg]$$
Then we define 
$$S_n=\{r_k^{(n)}:k\in[1,n]\cap\Bbb N\}$$
So we have that 
$$\frac1{x^n+1}=\prod_{r\in S_n}\frac1{x-r}=\prod_{k=1}^n\frac1{x-r_k^{(n)}}$$
Then we assume that we can write
$$\prod_{r\in S_n}\frac1{x-r}=\sum_{r\in S_n}\frac{b(r)}{x-r}$$
Multiplying both sides by $\prod_{a\in S_n}(x-a)$,
$$1=\sum_{r\in S_n}b(r)\prod_{a\in S_n\\ a\neq r}(x-a)$$
So for any $\omega\in S_n$,
$$1=b(\omega)\prod_{a\in S_n\\ a\neq \omega}(\omega-a)$$
$$b(\omega)=\prod_{a\in S_n\\ a\neq \omega}\frac1{\omega-a}$$
$$b(r_k^{(n)})=\prod_{p=1\\ p\neq k}^n\frac1{r_k^{(n)}-r_p^{(n)}}$$
So we know that 
$$I_n=\int_0^1\frac{\mathrm{d}x}{1+x^n}=\sum_{k=1}^{n}b(r_k^{(n)})\int_0^1\frac{\mathrm{d}x}{x-r_k^{(n)}}$$
$$I_n=\sum_{k=1}^{n}b(r_k^{(n)})\log\bigg|\frac{r_k^{(n)}-1}{r_k^{(n)}}\bigg|$$
$$I_n=\sum_{k=1}^{n}\log\bigg|\frac{r_k^{(n)}-1}{r_k^{(n)}}\bigg|\prod_{p=1\\ p\neq k}^n\frac1{r_k^{(n)}-r_p^{(n)}}$$
So we have
$$\int_0^1\log(1+x^n)\mathrm{d}x=\log2-n+n\sum_{k=1}^{n}\log\bigg|\frac{r_k^{(n)}-1}{r_k^{(n)}}\bigg|\prod_{p=1\\ p\neq k}^n\frac1{r_k^{(n)}-r_p^{(n)}}$$
along with a plethora of other identities...
A: Here's another, quicker, method (I also don't know if this one works)
Using the same $r_k^{(n)}$ as last time, we apply the $\log\prod_{i}a_i=\sum_i\log a_i$ property to see that 
$$\log(1+x^n)=\log\prod_{k=1}^{n}(x-r_k^{(n)})=\sum_{k=1}^{n}\log(x-r_k^{(n)})$$
So 
$$I_n=\int_0^1\log(1+x^n)\mathrm dx=\sum_{k=1}^{n}\int_0^1\log(x-r_k^{(n)})\mathrm dx$$
This last integral boils down to 
$$\begin{align}
\int_0^1\log(x-a)\mathrm dx=&a\log\frac{a}{1+a}+\log(1-a)-1\\
=&\log\frac{a^a(1-a)}{e(1+a)^a}
\end{align}$$
So 
$$I_n=\sum_{r\in S_n}\log\frac{r^r(1-r)}{e(1+r)^r}$$
And you know how I love product representations, so we again use $\log\prod_{i}a_i=\sum_i\log a_i$ to see that 
$$
I_n=\log\prod_{r\in S_n}\frac{r^r(1-r)}{e(1+r)^r}\\
\prod_{r\in S_n}\frac{r^r(1-r)}{(1+r)^r}=\exp(n+I_n)
$$
Which I just think is really neat.
