After reading through a series of three problems I, II, and III, I decided to create one of my own and solve it in a close manner to problem III. I don't know if convergence is an issue in this case and if my solution is valid, which is why I'd like an opinion from more people. Here's the problem:
Let $f_{n}(x)=\log\Gamma\left(\frac{x}{f_{n-1}(x)}\right)+1$ with $f_{1}(x)=\log\Gamma(x)+1$, find a closed-form solution to $$\lim\limits_{n\to\infty}\int_{0}^{1}f_{n}(x)\mathrm dx$$
Here's my thought process. Observe that $$\lim\limits_{n\to\infty}f_{n}(x)=\log\Gamma\left(\frac{x}{f_{\infty}(x)}\right)+1$$ $$\implies \frac{x}{f_{\infty}(x)}\Big[\log\Gamma\left(\frac{x}{f_{\infty}(x)}\right)+1\Big]=x\tag*{(a.)}$$
Next, we consider the function $t\log\Gamma(t)+t=x$, and define its inverse $\mathcal{G}(x)=t$. From here, it follows that $$\mathcal{G}(x)=\frac{x}{f_{\infty}(x)}\longrightarrow f_{\infty}(x)=\frac{x}{\mathcal{G}(x)}\tag*{(b.)}$$
Thus, $$\lim\limits_{n\to\infty}\int_{0}^{1}f_{n}(x)\mathrm dx=\int_{0}^{1}\frac{x}{\mathcal{G}(x)}\mathrm dx\tag*{(c.)}$$
We make the substitution $x=t\log\Gamma(t)+t$ such that $\mathrm dx=\left(1+t\psi(t)+\log\Gamma(t)\right)\mathrm dt$ and $\mathcal{G}(x)=t$. The limits of integration won't change since $\lim\limits_{t\to 0^{+}}\big[t\log\Gamma(t)+t\big]\rightarrow 0$ and $\lim\limits_{t\to 1}\big[t\log\Gamma(t)+t\big]\rightarrow 1$. Hence, $(c.)$ is equivalent to $$\int_{0}^{1}\Big[\log\Gamma(t)+1\Big]\Big[1+t\psi(t)+\log\Gamma(t)\Big]\mathrm dt\tag*{(d.)}$$ $$=2\underbrace{\int_{0}^{1}\log\Gamma(t)\mathrm dt}_{\mathcal{I}}+\underbrace{\int_{0}^{1}\log^{2}\Gamma(t)\mathrm dt}_{\mathcal{J}}+\underbrace{\int_{0}^{1}t\psi(t)\log\Gamma(t)\mathrm dt}_{\mathcal{K}}+\underbrace{\int_{0}^{1}t\psi(t)\mathrm dt}_{\mathcal{L}} +1\tag*{(e.)}$$
The first integral $\mathcal{I}$ was evaluated by J. L. Raabe, whereby $$\int_{0}^{1}\log\Gamma(t)\mathrm dt=\log\sqrt{2\pi}$$ One may also easily evaluate $\mathcal{L}$ viz. integration by parts from Raabe's integral, whereby $$\mathcal{L}=-\frac{1}{2}\mathcal{I}$$
The middle two integrals are also related to each other from a quick integration by parts. In 2010, a solution was found where $$\int_{0}^{1}\log^{2}\Gamma(t)\mathrm dt=\frac{\gamma^{2}}{12}+\frac{\pi^{2}}{48}+\frac{1}{3}\gamma\log\sqrt{2\pi}+\frac{4}{3}\log^{2}\sqrt{2\pi}-(\gamma+2\log\sqrt{2\pi})\frac{\zeta^{\prime}(2)}{\pi^{2}}+\frac{\zeta^{\prime\prime}(2)}{2\pi^{2}}\tag*{(f.)}$$
Also, we have that $$\mathcal{K}=-\frac{1}{2}\mathcal{J}$$
So we can combine everything to get that
$$\lim\limits_{n\to\infty}\int_{0}^{1}f_{n}(x)\mathrm dx=\frac{3}{2}\log\sqrt{2\pi}+\frac{1}{2}\mathcal{J}+1\approx 3.3115663417$$
Is this correct?
