Compute $\sum\limits_{j = 0}^{m - 1} \left(c_j + 1\right)\ln\left(c_j + 1\right)$ where $c_j = \cos\left(\frac{\pi}{2m}\left(1 + 2j\right) \right)$

Question

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}

Answer 1

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 $$f(1)=\frac{\pi }{2}-2$$ $$f(2)=\frac{1}{4} \left(\pi \tan \left(\frac{\pi }{8}\right)+\pi \cot \left(\frac{\pi }{8}\right)-4 \sqrt{2} \log \left(\sin \left(\frac{\pi }{8}\right)\right)+4 \sqrt{2} \log \left(\cos \left(\frac{\pi }{8}\right)\right)\right)-4$$ $$f(3)=\frac{2 \left(\pi -\sqrt{3} \log \left(\sqrt{3}-1\right)+\sqrt{3} \log \left(1+\sqrt{3}\right)\right)}{\left(\sqrt{3}-1\right) \left(1+\sqrt{3}\right)}-6$$ $$f(4)=\frac{1}{4} \left(\pi \tan \left(\frac{\pi }{16}\right)+\pi \cot \left(\frac{\pi }{16}\right)-8 \sin \left(\frac{\pi }{8}\right) \log \left(\sin \left(\frac{3 \pi }{16}\right)\right)+8 \cos \left(\frac{\pi }{8}\right) \log \left(\cos \left(\frac{\pi }{16}\right)\right)-8 \cos \left(\frac{\pi }{8}\right) \log \left(\sin \left(\frac{\pi }{16}\right)\right)+8 \sin \left(\frac{\pi }{8}\right) \log \left(\cos \left(\frac{3 \pi }{16}\right)\right)\right)-8$$

Answer 2

$\color{brown}{\textbf{Preliminary notes.}}$

Calculation of the required sum looks more complex task than of the issue integral.

Approach with calculations of the sum via the inttegral allows to get close form both of the integral and the sum.

Denote $$I_n=\int\limits_0^1\log(1+x^n)\mathrm dx.\tag1$$ Easy to see that the issue integral is $I_{2m}.$

$\color{brown}{\textbf{Closed form of the integrals.}}$

The first step is the integration by parts: $$I_n = x\log(1+x^n)\biggr|_0^1-n\int_0^1x\dfrac{x^{n-1}}{1+x^n}\mathrm dx,$$ $$I_n = \log2 - n + n\int\limits_0^1\dfrac{\mathrm dx}{1+x^n}.\tag2$$ (see also Claude Leibovici).

Then, the substitution $$x=e^{-t}\tag3$$ gives $$J_n=\int\limits_0^1\dfrac{\mathrm dx}{1+x^n}=\int\limits_0^\infty\dfrac{e^{-t}\,\mathrm dt}{1+e^{-nt}} = \sum\limits_{k=0}^\infty(-1)^k\int\limits_0^\infty e^{-(kn+1)t}\,\mathrm dt=\sum\limits_{k=0}^\infty\dfrac{(-1)^k}{kn+1} ,$$ $$J_n = \int\limits_0^1\dfrac{\mathrm dx}{1+x^n}= \dfrac1{2n}\left(\psi_0\left(\dfrac{n+1}{2n}\right) - \psi_0\left(\dfrac{1}{2n}\right)\right),\tag4$$ where $\psi_0(x)$ is the polygamma function.

Using $(2),(4),$ easy to get the expression for the OP integral in the form of $$\boxed{I_{2m} = \log2 - 2m +\dfrac1{2}\left(\psi_0\left(\dfrac{2m+1}{4m}\right) - \psi_0\left(\dfrac{1}{4m}\right)\right).}\tag5$$

$\color{brown}{\textbf{Testing of the solution.}}$

Solution $(4)$ can be expressed via the elementary functions for a lot of the given values of n.

Immediate calculations allow to check obtained expressions in the simple cases \begin{align} &J_1 = \int\limits_0^1\dfrac{\mathrm dx}{1+x} = \log 2,\\[4pt] &J_2 = \int\limits_0^1\dfrac{\mathrm dx}{1+x^2} = \arctan 1 = \dfrac\pi4,\\[4pt] &J_3 = \int\limits_0^1\dfrac{\mathrm dx}{1+x^3} = \dfrac16\int\limits_0^1\left(\dfrac{2}{1+x} - \dfrac{2x-1}{1-x+x^2} + \dfrac{3}{1-x+x^2}\right)\,\mathrm dx\\ &= \left(\dfrac13\log(1+x) - \dfrac16 \log(1-x+x^2) + \dfrac1{\sqrt3}\arctan\dfrac{2x-1}{\sqrt3}\right)\bigg|_0^1 = \dfrac\pi{3\sqrt3} + \frac13 \log2.\\[4pt] \end{align} More hard cases are \begin{align} &J_4 = \int\limits_0^1\dfrac{\mathrm dx}{1+x^4} = \dfrac{\sqrt2}8\left(\log\dfrac{x^2 + \sqrt2 x + 1}{x^2 - \sqrt2 x + 1} - 2\arctan(1-\sqrt2x) + 2\arctan(\sqrt2x+1)\right)\bigg|_0^1\\[4pt] &= \dfrac{\sqrt2}8\left(\log\dfrac{(x^2 + \sqrt2 x + 1)^2}{1+x^4} + 2\arctan\dfrac{x\sqrt2}{1-x^2}\right)\bigg|_0^1 = \dfrac{\sqrt2}8(\pi+\log(3+2\sqrt2)),\\[4pt] &J_4=\dfrac18\Bigg(\dfrac12\sqrt{\dfrac{2-\sqrt2}{2+\sqrt2}}\pi+\dfrac12\sqrt{\dfrac{2+\sqrt2}{2-\sqrt2}}\pi -2\sqrt2\left(-\log2+\dfrac12\log(2-\sqrt2)\right)\\[4pt] &+2\sqrt2\left(-\log2+\dfrac12\log(2+\sqrt2)\right)\Bigg)\\[4pt] &=\dfrac18\Bigg(\dfrac\pi2\left(\tan\dfrac\pi8+\cot\dfrac\pi8\right) +\sqrt2\log\dfrac{2+\sqrt2}{2-\sqrt2}\Bigg) = \dfrac{\sqrt2}8(\pi+\log(3+2\sqrt2)), \end{align} and $n>4.$

Numeric calculations of the issue integral $I_{2m}$ and its closed form $(5)$ also confirm the correctness of the closed form.

$\color{brown}{\textbf{The alternative approach.}}$

The alternative approach is considered in OP. Let us repeat it with some differences.

Polynomial factorization can be presented in the form of

\begin{align} &1+x^{2m} = \prod\limits_{j=0}^{2m-1}{\large \left(x-e^{\frac{2j+1}{2m}\pi i}\right)} = \prod\limits_{j=0}^{m-1}\left(x^2-2x\cos\frac{2j+1}{2m}\pi+1\right), \end{align} so \begin{align} &I_{2m} = \int\limits_0^1 \ln(1+x^{2m})\,\mathrm dx = \sum\limits_{j=0}^{m-1} T_j,\quad\text{where}\\[4pt] &T_j = \int\limits_0^1 \ln\left(x^2-2x\cos\frac{2j+1}{2m}\pi+1\right)\,\mathrm dx,\\[4pt] &T_j = x\ln\left(x^2-2x\cos\frac{2j+1}{2m}\pi+1\right)\bigg|_0^1 - 2\int\limits_0^1 \dfrac{x\left(x-\cos\frac{2j+1}{2m}\pi\right)}{x^2-2x\cos\frac{2j+1}{2m}\pi+1}\,\mathrm dx\\[4pt] & = \ln\left(2-2\cos\frac{2j+1}{2m}\pi\right) - 2-2\int\limits_0^1 \dfrac{x\cos\frac{2j+1}{2m}\pi-1}{x^2-2x\cos\frac{2j+1}{2m}\pi+1}\,\mathrm dx\\[4pt] & = \ln\left(2-2\cos\frac{2j+1}{2m}\pi\right) - 2 - 2\cos\frac{2j+1}{2m}\pi \int\limits_0^1 \dfrac{x-\cos\frac{2j+1}{2m}\pi}{x^2-2x\cos\frac{2j+1}{2m}\pi+1}\,\mathrm dx\\[4pt] & + 2\int\limits_0^1 \dfrac{\sin^2\frac{2j+1}{2m}\pi}{\left(x-\cos\frac{2j+1}{2m}\pi\right)^2+\sin^2\frac{2j+1}{2m}\pi}\,\mathrm dx\\[4pt] & = \ln\left(2-2\cos\frac{2j+1}{2m}\pi\right) - 2 - \cos\frac{2j+1}{2m}\pi \ln\left(x^2-2x\cos\frac{2j+1}{2m}\pi+1\right)\bigg|_0^1\\[4pt] & + 2\sin\frac{2j+1}{2m}\pi \arctan\dfrac{x-\cos\frac{2j+1}{2m}\pi}{\sin\frac{2j+1}{2m}\pi}\bigg|_0^1\\[4pt] & = \ln\left(2-2\cos\frac{2j+1}{2m}\pi\right) - 2 - \cos\frac{2j+1}{2m}\pi \ln\left(2-2\cos\frac{2j+1}{2m}\pi\right)\\[4pt] & + 2\sin\frac{2j+1}{2m}\pi \left(\arctan\dfrac{1-\cos\frac{2j+1}{2m}\pi}{\sin\frac{2j+1}{2m}\pi} + \arctan\dfrac{\cos\frac{2j+1}{2m}\pi}{\sin\frac{2j+1}{2m}\pi}\right)\\[4pt] & = \ln\left(2-2\cos\frac{2j+1}{2m}\pi\right) - 2 - \cos\frac{2j+1}{2m}\pi \ln\left(2-2\cos\frac{2j+1}{2m}\pi\right)\\[4pt] & + 2\sin\frac{2j+1}{2m}\pi \arctan\dfrac{\sin\frac{2j+1}{2m}\pi} {\sin^2\frac{2j+1}{2m}\pi - \left(1 - \cos\frac{2j+1}{2m}\pi\right)\cos\frac{2j+1}{2m}\pi}\\[4pt] & = \left(1-\cos\frac{2j+1}{2m}\pi\right)\ln\left(2-2\cos\frac{2j+1}{2m}\pi\right) - 2 + 2\sin\frac{2j+1}{2m}\pi\arctan\cot\frac{2j+1}{4m}\pi,\\[4pt] &T_j = \left(1-\cos\frac{2j+1}{2m}\pi\right)\ln\left(2-2\cos\frac{2j+1}{2m}\pi\right) - 2 + \dfrac{2(m-j)-1}{2m}\pi\sin\frac{2j+1}{2m}\pi. \end{align} Taking in account that \begin{align} &\sum\limits_{j=0}^{m-1}\cos\frac{2j+1}{2m}\pi = \Re \sum\limits_{j=0}^{m-1}e^{\frac{2j+1}{2m}\pi i} = \Re {\large \dfrac{1-e^{\pi i}}{1-e^{\frac\pi{m}i}}e^{\frac\pi{2m}i}} =\Re\dfrac {i}{\sin\frac\pi{2m}} = 0,\\[4pt] &\sum\limits_{j=0}^{m-1}\sin\frac{2j+1}{2m}\pi = \Im \sum\limits_{j=0}^{m-1}e^{\frac{2j+1}{2m}\pi i} =\Im\dfrac{i}{\sin\frac\pi{2m}} = \dfrac1{\sin\frac\pi{2m}},\\[4pt] &\sum\limits_{j=0}^{m-1}\frac{2j+1}{2m}\pi\sin\frac{2j+1}{2m}\pi = \dfrac\pi{2\sin\frac\pi{2m}} \end{align} (see also Wolfram Alpha calculations),

one can get $$\boxed{I_{2m} = m(\ln2-2) + \dfrac\pi2\csc\frac\pi{2m} + \sum\limits_{j=0}^{m-1}\left(1\pm\cos\frac{2j+1}{2m}\pi\right) \log\left(1\pm\cos\frac{2j+1}{2m}\pi\right).}\tag6$$

$\color{brown}{\textbf{Closed form for the sum.}}$

From $(5)-(6)$ should $$\color{green}{\boxed{\sum\limits_{j=0}^{m-1}\left(1\pm c_j\right) \log\left(1\pm c_j\right) = \dfrac1{2}\left(\psi_0\left(\dfrac{2m+1}{4m}\right) - \psi_0\left(\dfrac{1}{4m}\right)\right) - (m-1)\ln2 - \dfrac\pi2\csc\frac\pi{2m} ,}}\tag7$$ where $$c_j = \cos\frac{2j+1}{2m}\pi.$$

Numeric calculations confirm the expression $(6)$ for the both variants of the sign "$\pm$".

Also, numeric calculations of the issue sum and its closed form $(7)$ confirm the correctness of the closed form for the both variants of the sign "$\pm$".

Answer 3

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...

Answer 4

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.