9
$\begingroup$

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}

$\endgroup$
12
  • $\begingroup$ @Mason - Yes, thanks for the pickup, I will edit now $\endgroup$
    – user150203
    Dec 27, 2018 at 4:46
  • $\begingroup$ And $G_n$ should be $G_m$? Sorry. I don't have a lot to add to this question except catching pedantic typos... Here's the closed forms for $I_0,I_1,I_2$. Does $I_m$ have a nice closed form? And is it easy to see how $I_m$ and $G_m$ are connected? $\endgroup$
    – Mason
    Dec 27, 2018 at 4:58
  • $\begingroup$ @Mason - Yes, sorry, I'm not on my game at the moment. I will correct. Thanks for the pickup. Have been trying to find the relationship through brute force examination (at present) and haven't been able to find anything. $\endgroup$
    – user150203
    Dec 27, 2018 at 5:02
  • 1
    $\begingroup$ Thanks for the shout-out David ;) I'll try to answer the question $\endgroup$
    – clathratus
    Dec 27, 2018 at 9:08
  • 1
    $\begingroup$ @clathratus - Thank you for posting the question and your work on it. I've been fascinated with the question since your initial solution (the same that Claude) came to. $\endgroup$
    – user150203
    Dec 27, 2018 at 9:13

4 Answers 4

6
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Yes, my collaborator and I came to the same result (very nice closed form indeed). If I may ask - what are your thoughts on the special case I've discussed? $\endgroup$
    – user150203
    Dec 27, 2018 at 8:49
  • $\begingroup$ @DavidG. Your post is very interesting. I need to spend more time with it ! $\endgroup$ Dec 27, 2018 at 8:54
  • $\begingroup$ Please do! I look forward to any comments you may have. Thanks again for your post. $\endgroup$
    – user150203
    Dec 27, 2018 at 9:48
  • $\begingroup$ Mathematica calculates the same result $\endgroup$
    – stocha
    Mar 22, 2019 at 19:11
4
+250
$\begingroup$

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

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 "$\pm$".

$\endgroup$
6
  • 1
    $\begingroup$ Very nice presentation, already before the alternate approach (+1). $\endgroup$ Mar 28, 2019 at 21:46
  • $\begingroup$ @MarkusScheuer Thanks, you are welcome! $\endgroup$ Mar 28, 2019 at 22:16
  • $\begingroup$ @MarkusScheuer What about this? $\endgroup$ Aug 2, 2019 at 15:11
  • $\begingroup$ @MarkusScheuer Interesting integration $\endgroup$ Aug 3, 2020 at 9:58
  • $\begingroup$ @YuriNegoetyanov: Very nice approach. Thanks for pointing at it. (+1) $\endgroup$ Aug 16, 2020 at 15:25
2
$\begingroup$

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...

$\endgroup$
3
  • 1
    $\begingroup$ Cheers @clathratus - I will have to have a look over in detail, but on the surface it looks good. Thanks heaps! $\endgroup$
    – user150203
    Dec 31, 2018 at 6:16
  • $\begingroup$ I say this not as a critique of your post, but of my own... do I have defend the set $r_j$ as having size $n$ ?? i.e. that the $c_j$ are distinct... $\endgroup$
    – user150203
    Dec 31, 2018 at 9:47
  • $\begingroup$ Well any polynomial with degree $n$ is going to have $n$ distinct roots, so it would make sense that $|S_n|=n$ $\endgroup$
    – clathratus
    Dec 31, 2018 at 20:28
1
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.