Limit of a continued fraction

Question

Given the continued fraction: $$f(x,N)=\left[2,3,4,...N,x\right]$$ $$f(x,N)=\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cfrac{1}{...+\cfrac{1}{x}}}}}$$ is it possible to find an expression for the integral: $$g(x,N)=\int f(x,N)dx$$ as function of $N$ and $x$? Thanks.

Answer 1

UPDATE

Let's observe that for : $$0+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cfrac{1}{...+\cfrac{1}{N+\cfrac{1}x}}}}}}$$ the $n$-th convergent is known (from MathWorld, see too OEIS A052119) using Bessel functions :

\begin{align} A_n&=\frac{I_n(-2)K_1(2)-I_1(-2)K_n(2)}{I_2(-2)K_1(2)-I_1(-2)K_2(2)}\\ &=2\left(I_n(-2)K_1(2)+I_1(2)K_n(2)\right)\\ &=2\left((-1)^n I_n(2)K_1(2)+I_1(2)K_n(2)\right)\\ \\ B_n&=\frac{I_n(-2)K_0(2)-I_0(-2)K_n(2)}{I_1(-2)K_0(2)-I_0(-2)K_1(2)}\\ &=2\left(-I_n(-2)K_0(2)+I_0(2)K_n(2)\right)\\ &=2\left((-1)^{n-1} I_n(2)K_0(2)+I_0(2)K_n(2)\right)\\ \end{align} (there was a sign error in MathWorld in the above expression, see too OEIS entries A001053 and A001040)

This means that we have (needing the $N+1$-th convergent for the $N$ at the end) : $$[0;1,2,\dotsc,N]=\frac{A_{N+1}}{B_{N+1}}$$

From this we may get the next convergent using Lord_Farin's helpful link :

$$[0;1,2,\dotsc,N,x]=\frac{xA_{N+1}+A_{N}}{xB_{N+1}+B_{N}}$$

To get your continued fraction you only need to revert this and subtract $1$ obtaining : $$[0;2,\dotsc,N,x]=\frac{xB_{N+1}+B_{N}}{xA_{N+1}+A_{N}}-1$$

Allowing the computation of : \begin{align} g(x,N)&=\int f(x,N)\;dx\\ &=\int \frac{xB_{N+1}+B_{N}}{xA_{N+1}+A_{N}}-1\;dx\\ &=\int \frac{B_{N}A_{N+1}-A_{N}B_{N+1}}{A_{N+1}(xA_{N+1}+A_{N})}+\frac {B_{N+1}}{A_{N+1}}-1\;dx\\ &=\frac{B_{N}A_{N+1}-A_{N}B_{N+1}}{A_{N+1}^2}\log(A_{N+1}\;x+A_{N})+\frac {B_{N+1}-A_{N+1}}{A_{N+1}}x\\ \\ g(x,N)&=\boxed{\displaystyle (-1)^{N+1}\frac{\log(A_{N+1}\;x+A_{N})}{A_{N+1}^2}+\frac {B_{N+1}-A_{N+1}}{A_{N+1}}x}\quad\text{(after simplification)}\\ \end{align}

I'll let you reverify all this,