# Compute ; $\Gamma=\lim_{n\to\infty} \int_{t_n}^{t_{n+1}} \frac{(f(x-t_n))^{g(t_{n+1}-x)}}{(f(t_{n+1}-x))^{g(x-t_n)}+(f(x-t_n))^{g(t_{n+1}-x)}} \, dx$

Let the functions be defined as $$g:R\to R$$ and $$f:R \to (1,\infty)$$. Both f and g are continuous. Let $$F_n$$ denote the $$n^{th}$$ Fibonacci number, then for $$n ∈ N$$ let $$\Gamma$$ and $$t_n$$ be defined as follows;

$$t_n=\sqrt [n]{(2n-1)!! F_n}$$

$$\Gamma=\lim_{n\to\infty} \int_{t_n}^{t_{n+1}} \frac{(f(x-t_n))^{g(t_{n+1}-x)}}{(f(t_{n+1}-x))^{g(x-t_n)}+(f(x-t_n))^{g(t_{n+1}-x)}} \, dx$$

My initial attempt involved the recurrence relation of Fibonacci numbers for $$n>1$$;

$$\boxed{F_n=F_{n-1}+F_{n-2}}$$ and $$\boxed{F_0=0, F_1=1}$$. The closed form being; $$\boxed{F_n=\frac{1}{\sqrt 5}\left[(\phi)^n-(\psi)^n\right]}$$

$$\phi$$ being the golden ratio and $$\psi$$ being the conjugate of golden ratio.

Not being familiar with the double factorial notation, here's what I have understood/found;

$$\boxed{n!!=\sum_{k=0}^{\lceil \frac{n}{2} \rceil-1}(n-2k)=n(n-2)(n-4)\cdots}$$ and $$\boxed{n!=n!!(n-1)!!}$$

$$\boxed{t_n=\sqrt[n] {\frac{(2n-1)!!(\phi^n-\psi^n)}{\sqrt5}}}$$ $$\boxed{t_{n+1}=\sqrt[n+1] {\frac{(2n+1)!!(\phi^{n+1}-\psi^{n+1})}{\sqrt5}}}$$

Substituting,

$$t_n=\sqrt [n]{(2n-1)!! F_n}$$ and $$t_{n+1}=\sqrt [n+1]{(2n+1)!! F_{n+1}}$$.

Let $$\Gamma=\lim_{n\to\infty} \beta_n$$

where $$\beta_n$$ is;

$$\boxed{\int_{\sqrt [n]{(2n-1)!! F_n}}^{\sqrt [n+1]{(2n+1)!! F_{n+1}}} \frac{(f(x-\sqrt [n]{(2n-1)!! F_n}))^{g\left(\sqrt [n+1]{(2n+1)!! F_{n+1}}-x\right)} \, dx}{(f(\sqrt [n+1]{(2n+1)!! F_{n+1}}-x))^{g\left(x-\sqrt [n]{(2n-1)!! F_n}\right)}+(f(x-\sqrt [n]{(2n-1)!! F_n}))^{g\left(\sqrt [n+1]{(2n+1)!! F_{n+1}}-x\right)}}}$$

On just a little simplification;

$$\boxed{\int_{\sqrt [n]{(2n-1)!! F_n}}^{\sqrt [n+1]{(2n+1)!! F_{n+1}}} \frac{1}{\left[\frac{\left(f\left(\sqrt [n+1]{(2n+1)!! F_{n+1}}-x\right)\right)^{g\left(x-\sqrt [n]{(2n-1)!! F_n}\right)}}{\left(f\left(x-\sqrt [n]{(2n-1)!! F_n}\right)\right)^{g\left(\sqrt [n+1]{(2n+1)!! F_{n+1}}-x\right)}}\right]+1}\,dx}$$

I'm now left more confused than before, was this the right approach at all? If so, how does one continue from here?

Note - This question did not come with the solution, so I am not certain if the limit converges (if that is necessary)

• Finally, a problem fit for a Saiyan elite!! It's badassery is over 9,000!!! Commented Jul 20, 2023 at 7:58
• The integral can be easily solved via kings rule, namly substitute $x$ for $t_{n+1} + t_n - x$. After some simplification you should see that 2 copy of the integral amount to $t_{n+1} - t_n$. I think you could proceed to calculate the limit Commented Jul 21, 2023 at 16:44

To start with, write $$L_1=\lim_{n\rightarrow \infty} \int_{t_n}^{t_{n+1}}{\frac{\left( f\left( x-t_n \right) \right) ^{g\left( t_{n+1}-x \right)}}{\left( f\left( t_{n+1}-x \right) \right) ^{g\left( x-t_n \right)}+\left( f\left( x-t_n \right) \right) ^{g\left( t_{n+1}-x \right)}}}\mathrm{d}x$$ $$L_2=\lim_{n\rightarrow \infty} \int_{t_n}^{t_{n+1}}{\frac{\left( f\left( t_{n+1}-x \right) \right) ^{g\left( x-t_n \right)}}{\left( f\left( x-t_n \right) \right) ^{g\left( t_{n+1}-x \right)}+\left( f\left( t_{n+1}-x \right) \right) ^{g\left( x-t_n \right)}}}\mathrm{d}x$$ You can varify that $$L_1=L_2$$ via the substitution $$x\rightsquigarrow t_{n+1}+t_n-x$$. Therefore, \begin{align*} \Rightarrow \varGamma =&\frac{1}{2}\left( \varGamma +\varGamma \right) =\frac{1}{2}\left( L_1+L_2 \right) \\ =&\frac{1}{2}\lim_{n\rightarrow \infty} \int_{t_n}^{t_{n+1}}{\frac{\left( f\left( x-t_n \right) \right) ^{g\left( t_{n+1}-x \right)}}{\left( f\left( t_{n+1}-x \right) \right) ^{g\left( x-t_n \right)}+\left( f\left( x-t_n \right) \right) ^{g\left( t_{n+1}-x \right)}}}\mathrm{d}x \\ +&\frac{1}{2}\lim_{n\rightarrow \infty} \int_{t_n}^{t_{n+1}}{\frac{\left( f\left( t_{n+1}-x \right) \right) ^{g\left( x-t_n \right)}}{\left( f\left( x-t_n \right) \right) ^{g\left( t_{n+1}-x \right)}+\left( f\left( t_{n+1}-x \right) \right) ^{g\left( x-t_n \right)}}}\mathrm{d}x \\ =&\frac{1}{2}\lim_{n\rightarrow \infty} \int_{t_n}^{t_{n+1}}{\mathrm{d}x}=\frac{1}{2}\lim_{n\rightarrow \infty} \left( t_{n+1}-t_n \right) \end{align*} Now comes that hard part. I shall investigate $$t_n$$ (Please take a look at gamma function.) \begin{align*} t_n=&\sqrt[n]{\left( 2n-1 \right) !!\mathbf{F}_n}=\sqrt[n]{\frac{\left( 2n-1 \right) !}{\left( 2n \right) !!}\mathbf{F}_n}=\sqrt[n]{\frac{\left( 2n-1 \right) !}{n!2^n}\mathbf{F}_n} \\ =&\frac{1}{2}\sqrt[n]{\frac{\Gamma \left( 2n+1 \right)}{\Gamma \left( n+1 \right)}\mathbf{F}_n}=\frac{1}{2}\sqrt[n]{\frac{\Gamma \left( 2n+1 \right)}{\Gamma \left( n+1 \right)}\mathbf{F}_n}=2\sqrt[n]{\frac{1}{\sqrt{\pi}}\Gamma \left( n+\frac{1}{2} \right) \mathbf{F}_n} \\ =&2\sqrt[n]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+\frac{1}{2} \right) \left( \left( \frac{1+\sqrt{5}}{2} \right) ^n-\left( \frac{1-\sqrt{5}}{2} \right) ^n \right)} \\ =&\left( 1+\sqrt{5} \right) \sqrt[n]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+\frac{1}{2} \right) \left( 1-\left( \frac{1-\sqrt{5}}{1+\sqrt{5}} \right) ^n \right)} \end{align*} I shall establish few facts, also (Heavy useage of string's formula here) \begin{align*} &\lim_{n\rightarrow \infty} \frac{e}{n}\sqrt[n]{\Gamma \left( n+\frac{1}{2} \right)}\\ =&\lim_{n\rightarrow \infty} \sqrt[n]{\frac{\Gamma \left( n+\frac{1}{2} \right)}{\Gamma \left( n \right)}}\lim_{n\rightarrow \infty} \frac{e}{n}\sqrt[n]{\Gamma \left( n \right)} \\ =&\frac{1}{4}\lim_{n\rightarrow \infty} \sqrt[n]{2\sqrt{\pi}\frac{\Gamma \left( 2n \right)}{\left( \Gamma \left( n \right) \right) ^2}}\lim_{n\rightarrow \infty} \frac{e}{n}\sqrt[n]{\Gamma \left( n \right)} \\ =&\frac{1}{4}\lim_{n\rightarrow \infty} \sqrt[n]{2\sqrt{\pi}\frac{\sqrt{4\pi n}\left( \frac{2n}{e} \right) ^{2n}}{\left( \sqrt{2\pi n}\left( \frac{n}{e} \right) ^n \right) ^2}}\lim_{n\rightarrow \infty} \frac{e}{n}\sqrt[n]{\sqrt{2\pi n}\left( \frac{n}{e} \right) ^n} \\ =&\lim_{n\rightarrow \infty} \sqrt[n]{\frac{2}{\pi \sqrt{n}}}=1\Rightarrow \sqrt[n]{\Gamma \left( n+\frac{1}{2} \right)}\sim \frac{n}{e} \end{align*} \begin{align*} &\lim_{n\rightarrow \infty} \left( \sqrt[n]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+\frac{1}{2} \right) \left( 1-\left( \frac{1-\sqrt{5}}{1+\sqrt{5}} \right) ^n \right)}-\sqrt[n]{\Gamma \left( n+\frac{1}{2} \right) \frac{1}{\sqrt{5\pi}}} \right) \\ =&\lim_{n\rightarrow \infty} \sqrt[n]{\frac{1}{\sqrt{5\pi}}}\sqrt[n]{\Gamma \left( n+\frac{1}{2} \right)}\left( \sqrt[n]{\left( 1-\left( \frac{1-\sqrt{5}}{1+\sqrt{5}} \right) ^n \right)}-1 \right) \\ =&\lim_{n\rightarrow \infty} \frac{n}{e}\left( \sqrt[n]{1-\left( \frac{1-\sqrt{5}}{1+\sqrt{5}} \right)}-1 \right) =\lim_{n\rightarrow \infty} \ln \left( 1-\left( \frac{1-\sqrt{5}}{1+\sqrt{5}} \right) ^n \right) e^{\xi _n-1}=0 \end{align*} Hence \begin{align*} \varGamma =&\frac{1}{2}\lim_{n\rightarrow \infty} \left( t_{n+1}-t_n \right) \\ =&\phi \lim_{n\rightarrow \infty} \left( \sqrt[n+1]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+\frac{3}{2} \right) \left( 1-\left( \frac{1-\sqrt{5}}{1+\sqrt{5}} \right) ^{n+1} \right)}-\sqrt[n]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+\frac{1}{2} \right) \left( 1-\left( \frac{1-\sqrt{5}}{1+\sqrt{5}} \right) ^n \right)} \right) \\ =&\phi \lim_{n\rightarrow \infty} \left( \sqrt[n+1]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+1+\frac{1}{2} \right) \left( 1-\left( \frac{1-\sqrt{5}}{1+\sqrt{5}} \right) ^{n+1} \right)}-\sqrt[n+1]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+1+\frac{1}{2} \right)} \right) \\ +&\phi \lim_{n\rightarrow \infty} \left( \sqrt[n]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+\frac{1}{2} \right)}-\sqrt[n]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+\frac{1}{2} \right) \left( 1-\left( \frac{1-\sqrt{5}}{1+\sqrt{5}} \right) ^n \right)} \right) \\ +&\phi \lim_{n\rightarrow \infty} \left( \sqrt[n+1]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+1+\frac{1}{2} \right)}-\sqrt[n]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+\frac{1}{2} \right)} \right) \\ =&\phi \lim_{n\rightarrow \infty} \left( \sqrt[n+1]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+1+\frac{1}{2} \right)}-\sqrt[n]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+\frac{1}{2} \right)} \right) \\ =&\phi \lim_{n\rightarrow \infty} \left( \sqrt[n+1]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+1+\frac{1}{2} \right)}-\sqrt[n+1]{\Gamma \left( n+1+\frac{1}{2} \right)} \right) \\ +&\phi \lim_{n\rightarrow \infty} \left( \sqrt[n]{\Gamma \left( n+\frac{1}{2} \right)}-\sqrt[n]{\frac{1}{\sqrt{5\pi}}\Gamma \left( n+\frac{1}{2} \right)} \right) \\ +&\phi \lim_{n\rightarrow \infty} \left( \sqrt[n+1]{\Gamma \left( n+1+\frac{1}{2} \right)}-\sqrt[n]{\Gamma \left( n+\frac{1}{2} \right)} \right) \\ =&\phi \lim_{n\rightarrow \infty} \sqrt[n+1]{\Gamma \left( n+1+\frac{1}{2} \right)}\left( \sqrt[n+1]{\frac{1}{\sqrt{5\pi}}}-1 \right) -\phi \lim_{n\rightarrow \infty} \sqrt[n]{\Gamma \left( n+\frac{1}{2} \right)}\left( \sqrt[n]{\frac{1}{\sqrt{5\pi}}}-1 \right) \\ +&\phi \lim_{n\rightarrow \infty} \left( \sqrt[n+1]{\Gamma \left( n+1+\frac{1}{2} \right)}-\sqrt[n]{\Gamma \left( n+\frac{1}{2} \right)} \right) \\ =&\phi \lim_{n\rightarrow \infty} \frac{n+1}{e}\left( \frac{1}{n+1}\ln \left( \frac{1}{\sqrt{5\pi}} \right) \right) e^{\xi}-\phi \lim_{n\rightarrow \infty} \frac{n}{e}\left( \frac{1}{n}\ln \left( \frac{1}{\sqrt{5\pi}} \right) \right) e^{\theta} \\ +&\phi \lim_{n\rightarrow \infty} \left( \sqrt[n+1]{\Gamma \left( n+1+\frac{1}{2} \right)}-\sqrt[n]{\Gamma \left( n+\frac{1}{2} \right)} \right) \\ =&\frac{\phi}{e}-\frac{\phi}{e}+\frac{1+\sqrt{5}}{2}\lim_{n\rightarrow \infty} \left( \sqrt[n+1]{\Gamma \left( n+1+\frac{1}{2} \right)}-\sqrt[n]{\Gamma \left( n+\frac{1}{2} \right)} \right) \\ =&\phi \lim_{n\rightarrow \infty} \left( \sqrt[n+1]{\Gamma \left( n+1+\frac{1}{2} \right)}-\sqrt[n]{\Gamma \left( n+\frac{1}{2} \right)} \right) =\frac{\phi}{e} \end{align*} So at the end, we have $$\varGamma =\frac{\phi}{e}$$ Wolfram seems to agree with this result, see
numerical estimation of $$\Gamma$$
numerical astimation of $$\frac{\phi}{e}$$