TITLE: Is there a limit for this complex sequence?
QUESTION [6 upvotes]: Interested by this question, I tried to work the more general problem of
$$I_n=\int_0^\infty e^{-x}\log\big(1+\sin^{2n}(x)\big)\, dx$$ for which were found expressions of the type
$$I_n=A_n  \,
   _{2n+1}F_{2n}\left(1,1,\color{red}{\textbf{#}};\color{green}{\textbf{@}};-1\right)$$ in which $\color{red}{\textbf{ #}}$ and $\color{green}{\textbf{@}}$ show nice and simple patterns.
The front coefficient is
$$A_n=\int_0^\infty e^{-x} \sin^{2n}(x)\,dx=(-1)^n\,\frac{i \,  n\, \Gamma \left(\frac{i}{2}-n\right)\, \Gamma (2
   n)}{4^n\,\Gamma \left(n+1+\frac{i}{2}\right)}$$


What I wonder is if $$L=\lim_{n\to \infty } \, \frac{I_n}{A_n}=\lim_{n\to \infty } \, \,
   _{2n+1}F_{2n}\left(1,1,\color{red}{\textbf{#}};\color{green}{\textbf{@}};-1\right)$$ does exist or not.


Could we somehow use the fact that, for $k \pi \leq x \leq (k+1)\pi$, $\log\big(1+\sin^{2n}(x)\big)$ looks like a gaussian ?
In the table below, I tabulated some of the numerical values I obtained
$$\left(
\begin{array}{cc}
 n & \frac{I_n}{A_n} \\
 1 & 0.76498434 \\
 2 & 0.76742187 \\
 3 & 0.76742296 \\
 4 & 0.76718009 \\
 5 & 0.76694124 \\
 6 & 0.76673938 \\
 7 & 0.76657308 \\
 8 & 0.76643577 \\
 9 & 0.76632132 \\
 10 & 0.76622482 \\
 20 & 0.76573433 \\
 30 & 0.76554963 \\
 40 & 0.76545318 \\
 50 & 0.76539397 \\
 60 & 0.76535394 \\
 70 & 0.76532508 \\
 80 & 0.76530328 \\
 90 & 0.76528623 \\
 100 & 0.76527254 \\
200 & 0.76521029 \\
 300 & 0.76518932 \\
 400 & 0.76517879 \\
 500 & 0.76517246 \\
 600 & 0.76516823 \\
 700 & 0.76516521 \\
 800 & 0.76516294 \\
 900 & 0.76516117 \\
 1000 & 0.76515976
\end{array}
\right)$$

REPLY [3 votes]: Using the periodic part of the integrated function, we can rewrite the integral:
$$I_n=\sum_{k=0}^\infty \int_{\pi k}^{\pi(k+1)} e^{-x} \ln(1+\sin^{2n} x) dx=\sum_{k=0}^\infty e^{- \pi k} \int_0^\pi e^{-x} \ln(1+\sin^{2n} x) dx= \\ =  \frac{1}{1-e^{-\pi}}\int_0^\pi e^{-x} \ln(1+\sin^{2n} x) dx$$
Now let's deal with the integral itself. To extract the $n$ dependence, we will use integration by parts and several substitutions. First, integration by parts gives us:
$$I_n=\frac{2n}{1-e^{-\pi}} \int_0^\pi e^{-x} \frac{\sin^{2n-1} x \cos x}{1+\sin^{2n} x} dx$$
Now we substitute $\cos x=t$ and obtain:
$$I_n=\frac{2n}{1-e^{-\pi}} \int_{-1}^1 e^{-\arccos t} \frac{(1-t^2)^{n-1} t}{1+(1-t^2)^n} dt$$
Let's separate the integral into two parts $\int_{-1}^1=\int_0^1+\int_{-1}^0$ and use the following relations:
$$\arccos t= \frac{\pi}{2}-\arcsin t \\ \arccos(- t)= \frac{\pi}{2}+\arcsin t$$
This gives us:
$$I_n=\frac{4n e^{-\pi/2}}{1-e^{-\pi}} \int_0^1 \sinh (\arcsin t) \frac{(1-t^2)^{n-1} t}{1+(1-t^2)^n} dt=\frac{2n}{\sinh (\pi/2)} \int_0^1 \sinh (\arcsin t) \frac{(1-t^2)^{n-1} t}{1+(1-t^2)^n} dt$$
Now let's perform a couple of obvious substitutions:
$$t^2=u$$
$$I_n=\frac{n}{\sinh (\pi/2)} \int_0^1 \sinh (\arcsin \sqrt{u}) \frac{(1-u)^{n-1}}{1+(1-u)^n} du$$
$$1-u=v$$
$$I_n=\frac{n}{\sinh (\pi/2)} \int_0^1 \sinh (\arcsin \sqrt{1-v}) \frac{v^{n-1}}{1+v^n} dv$$
$$v=e^{-s}$$

$$I_n=\frac{n}{\sinh (\pi/2)} \int_0^\infty \sinh (\arcsin \sqrt{1-e^{-s}}) \frac{ds}{e^{n s}+1} $$

The function $g(s)=\sinh \left(\arcsin \sqrt{1-e^{-s}}\right)$ starts like $\sqrt{s}$ and then approaches a constant as $s \to \infty$.
For our purpose, we are interested in large $n$, so it makes sense to replase $g(s)$ by its series:
$$g(s)=\sqrt{s} \sum_{k=0}^\infty a_k s^k=\sqrt{s} \left(1+\frac{s}{12}-\frac{s^2}{32}+\frac{13 s^3}{8064}+\frac{2657 s^4}{5806080}-\frac{16243 s^5}{255467520}-\frac{581 s^6}{175177728}+O(s^7) \right)$$
So we have:
$$I_n=\frac{n}{\sinh (\pi/2)} \sum_{k=0}^\infty a_k  \int_0^\infty  \frac{s^{k+1/2} ds}{e^{n s}+1} $$
Changing the variable $ns=q$, we get:
$$I_n=\frac{1}{\sinh (\pi/2) \sqrt{n}} \sum_{k=0}^\infty \frac{a_k}{n^k}  \int_0^\infty  \frac{q^{k+1/2} dq}{e^q+1} $$
Or, using the integral definition of zeta function:

$$I_n=\frac{1}{\sinh (\pi/2) \sqrt{n}} \sum_{k=0}^\infty \left(1-\frac{1}{2^{k+1/2}}\right) \Gamma \left(k+\frac{3}{2}\right) \zeta \left(k+\frac{3}{2}\right)  \frac{a_k}{n^k} $$

Using absolutely the same method with a more simple integral $A_n$ we can obtain an asymptotic series for it as well, and taking the ratio of first terms should give us the same limit metamorphy obtained.

As a numerical example, $n=11$ gives us:
$$I_{11}=0.089884326883595958870...$$
And using the proposed series with $16$ terms gives us:
$$I_{11} \approx \color{blue}{0.089884326883}393284625...$$