Limit of a continued fraction 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.
 A: 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,
