Mfrhtyj

As part of solving:

begin{equation}
I_m = int_0^1 lnleft(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 + 1right)lnleft(c_j + 1right),
end{equation}

where $c_j = cosleft(frac{pi}{2m}left(1 + 2jright) right)$

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}{sinleft(frac{pi}{n} right)}- Bleft(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[Gammaleft(1 - frac{1}{n} right)Gammaleft(frac{1}{n} right)- Bleft(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 = expleft({frac{pi + 2pi j}{2m} i} right) mbox{ for } j = 0dots 2m - 1,
end{align}

which we can express as the set

begin{align}
S &= Bigg{ expleft({frac{pi + 2pi cdot 0}{2m} i} right) , :expleft({frac{pi + 2pi cdot 1}{2m} i} right),dots,:expleft({frac{pi + 2pi cdot (2m - 2)}{2m} i} right)\
&qquad:expleft({frac{pi + 2pi cdot (2m - 1)}{2m} i} right)Bigg},
end{align}

which can be expressed as the set of $2$-tuples

begin{align}
S &= left{ left( expleft({frac{pi + 2pi j}{2m} i} right) , :expleft({frac{pi + 2pi(2m - 1 - j )}{2m} i} right)right): bigg|: j = 0 dots m - 1right}\
& = left{ (z_j, cleft(z_jright):|: 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_jright)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 + 2Releft(r_jright)x + left|r_j right|^2right)
end{align}

For our case here $left|r_j right|^2 = 1$ and $Releft(r_jright) = cosleft(frac{pi + 2pi j}{2m} right)= cosleft(frac{pi}{2m}left(1 + 2jright)right) = c_j$

begin{align}
int_0^1 logleft( x^{2m} + 1right):dx &= int_0^1 logleft(prod_{r in S} left(x^2 + 2c_jx+ left|r_j right|^2right)right)\
&= sum_{j = 0}^{m - 1} int_0^1 logleft(x^2 + 2c_jx + 1 right)\
&= sum_{j = 0}^{m - 1} left[2sqrt{1 - c_j^2}arctanleft(frac{x + c_j}{sqrt{1 - c_j^2}}right) + left(x + c_jright)logleft(x^2 + 2c_jx + 1right) - 2x right]_0^1 \
&= sum_{j = 0}^{m - 1} left[ 2sqrt{1 - c_j^2}arctanleft(sqrt{frac{1 - c_j}{1 + c_j}} right) + log(2)c_j + left(log(2) - 2right) + left(c_j + 1right)logleft(c_j + 1right) right] \
&= 2sum_{j = 0}^{m - 1}sqrt{1 - c_j^2}arctanleft(sqrt{frac{1 - c_j}{1 + c_j}} right) + log(2)sum_{j = 0}^{m - 1} c_j + mleft(log(2) - 2right)\
&qquad+ sum_{j = 0}^{m - 1}left(c_j + 1right)logleft(c_j + 1right)
end{align}

Thus,

begin{align}
int_0^1 logleft( x^{2m} + 1right):dx &=sum_{j = 0}^{m - 1}c_jsinleft(frac{pi}{2m}left(1 + 2jright)right) + log(2)sum_{j = 0}^{m - 1} c_j + mleft(log(2) - 2right)\
&qquad+ sum_{j = 0}^{m - 1}left(c_j + 1right)logleft(c_j + 1right)
end{align}

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
$$J_m=x log left(1+x^{2 m}right)-2mint frac{ x^{2 m}+1-1}{x^{2 m}+1},dx=x log left(1+x^{2 m}right)-2mx+2mint frac{dx}{x^{2 m}+1}$$ and
$$int frac{dx}{x^{2 m}+1}=x , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-x^{2 m}right)$$ where appears the Gaussian or ordinary hypergeometric function.

So
$$K_m=int_0^a log(1+x^{2m}),dx=a log left(1+a^{2 m}right)-2ma+2ma , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-a^{2 m}right)$$ and, if $a=1$,
$$I_m=int_0^1 log(1+x^{2m}),dx= log left(2right)-2m+2m , _2F_1left(1,frac{1}{2 m};1+frac{1}{2 m};-1right)$$ 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
$$f(1)=frac{pi }{2}-2$$
$$f(2)=frac{1}{4} left(pi tan left(frac{pi }{8}right)+pi cot left(frac{pi
}{8}right)-4 sqrt{2} log left(sin left(frac{pi }{8}right)right)+4
sqrt{2} log left(cos left(frac{pi }{8}right)right)right)-4$$
$$f(3)=frac{2 left(pi -sqrt{3} log left(sqrt{3}-1right)+sqrt{3} log
left(1+sqrt{3}right)right)}{left(sqrt{3}-1right)
left(1+sqrt{3}right)}-6$$
$$f(4)=frac{1}{4} left(pi tan left(frac{pi }{16}right)+pi cot left(frac{pi
}{16}right)-8 sin left(frac{pi }{8}right) log left(sin left(frac{3
pi }{16}right)right)+8 cos left(frac{pi }{8}right) log left(cos
left(frac{pi }{16}right)right)-8 cos left(frac{pi }{8}right) log
left(sin left(frac{pi }{16}right)right)+8 sin left(frac{pi
}{8}right) log left(cos left(frac{3 pi }{16}right)right)right)-8$$

I did it!

I actually have no idea whether or not this works, but this is how I did it.

$ninBbb N$

Define the sequence ${r_k^{(n)}}_{k=1}^{k=n}$ such that
$$x^n+1=prod_{k=1}^nbig(x-r^{(n)}_{k}big)$$
For some reason (I'm not really sure how), for $min Bbb N$
$$r_k^{(2m)}=expbigg[frac{ipi}{2m}(2k-1)bigg],qquad kin[1,2m]capBbb N$$
$$r_k^{(2m+1)}=expbigg[frac{2pi i(m+1)}{2m+1}k+1bigg],qquad kin[1,2m+1]capBbb N$$
Then we define
$$S_n={r_k^{(n)}:kin[1,n]capBbb N}$$
So we have that
$$frac1{x^n+1}=prod_{rin S_n}frac1{x-r}=prod_{k=1}^nfrac1{x-r_k^{(n)}}$$
Then we assume that we can write
$$prod_{rin S_n}frac1{x-r}=sum_{rin S_n}frac{b(r)}{x-r}$$
Multiplying both sides by $prod_{ain S_n}(x-a)$,
$$1=sum_{rin S_n}b(r)prod_{ain S_n\ aneq r}(x-a)$$
So for any $omegain S_n$,
$$1=b(omega)prod_{ain S_n\ aneq omega}(omega-a)$$
$$b(omega)=prod_{ain S_n\ aneq omega}frac1{omega-a}$$
$$b(r_k^{(n)})=prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
So we know that
$$I_n=int_0^1frac{mathrm{d}x}{1+x^n}=sum_{k=1}^{n}b(r_k^{(n)})int_0^1frac{mathrm{d}x}{x-r_k^{(n)}}$$
$$I_n=sum_{k=1}^{n}b(r_k^{(n)})logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|$$
$$I_n=sum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
So we have
$$int_0^1log(1+x^n)mathrm{d}x=log2-n+nsum_{k=1}^{n}logbigg|frac{r_k^{(n)}-1}{r_k^{(n)}}bigg|prod_{p=1\ pneq k}^nfrac1{r_k^{(n)}-r_p^{(n)}}$$
along with a plethora of other identities...

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 $logprod_{i}a_i=sum_ilog a_i$ property to see that
$$log(1+x^n)=logprod_{k=1}^{n}(x-r_k^{(n)})=sum_{k=1}^{n}log(x-r_k^{(n)})$$
So
$$I_n=int_0^1log(1+x^n)mathrm dx=sum_{k=1}^{n}int_0^1log(x-r_k^{(n)})mathrm dx$$
This last integral boils down to
$$begin{align}
int_0^1log(x-a)mathrm dx=&alogfrac{a}{1+a}+log(1-a)-1\
=&logfrac{a^a(1-a)}{e(1+a)^a}
end{align}$$
So
$$I_n=sum_{rin S_n}logfrac{r^r(1-r)}{e(1+r)^r}$$
And you know how I love product representations, so we again use $logprod_{i}a_i=sum_ilog a_i$ to see that
$$
I_n=logprod_{rin S_n}frac{r^r(1-r)}{e(1+r)^r}\
prod_{rin S_n}frac{r^r(1-r)}{(1+r)^r}=exp(n+I_n)
$$
Which I just think is really neat.