Is there a limit for this complex sequence?

Question

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 $$\left( \begin{array}{cc} n & \frac{I_n}{A_n} \\ 1 & 0.76498434 \\ 2 & 0.76742187 \\ 3 & 0.76742296 \\ 4 & 0.76718009 \\ 5 & 0.76694124 \\ 6 & 0.76673938 \\ 7 & 0.76657308 \\ 8 & 0.76643577 \\ 9 & 0.76632132 \\ 10 & 0.76622482 \\ 20 & 0.76573433 \\ 30 & 0.76554963 \\ 40 & 0.76545318 \\ 50 & 0.76539397 \\ 60 & 0.76535394 \\ 70 & 0.76532508 \\ 80 & 0.76530328 \\ 90 & 0.76528623 \\ 100 & 0.76527254 \\ 200 & 0.76521029 \\ 300 & 0.76518932 \\ 400 & 0.76517879 \\ 500 & 0.76517246 \\ 600 & 0.76516823 \\ 700 & 0.76516521 \\ 800 & 0.76516294 \\ 900 & 0.76516117 \\ 1000 & 0.76515976 \end{array} \right)$$

Answer 1

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:

$$I_{11} \approx \color{blue}{0.089884326883}393284625...$$

Answer 2

As already noted, $\int_0^\infty=(1-e^{-\pi})^{-1}\int_0^\pi$ in both cases. Denoting $K=\dfrac{1}{2\sinh\pi/2}$, we have \begin{align} \sqrt{n}A_n&=K\sqrt{n}\int_{-\pi/2}^{\pi/2}e^{-x}\cos^{2n}x\,dx \\&=K\int_{-\pi\sqrt{n}/2}^{\pi\sqrt{n}/2}e^{-x/\sqrt{n}}\cos^{2n}(x/\sqrt{n})\,dx \\&\underset{n\to\infty}{\longrightarrow}K\int_{-\infty}^{\infty}e^{-x^2}\,dx=K\sqrt{\pi} \end{align} by DCT, and similarly $\sqrt{n}I_n\underset{n\to\infty}{\longrightarrow}K\displaystyle\int_{-\infty}^{\infty}\log(1+e^{-x^2})\,dx$.

Thus, the limit exists and is equal to $$\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}\log(1+e^{-x^2})\,dx=\frac{1}{\sqrt{\pi}}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}\int_{-\infty}^{\infty}e^{-nx^2}\,dx=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n^{3/2}},$$ which is known to be $\color{blue}{(1-1/\sqrt{2})\zeta(3/2)}$.