Closed expression for a given continued fraction

Question

Does anyone know a closed-form expression for the continued fraction $$F(x) = \cfrac{x}{x+\cfrac{x}{x+1+\cfrac{2x}{x+2+\cfrac{3x}{x+3+\cfrac{4x}{x+4+\cdots}}}}}?$$ According to page 181 of An Invitation to Modern Number Theory, by Miller and Takloo-Bighash, one has $F(1) = e-2$. Numerical exploration suggests that $F(0^+) = F(0^-) = \frac{1}{2}$ and $F(\frac{1}{2}) = e^{1/2}-1$, and also $F(\infty) = 1$, and maybe also $$1-F(x) \sim \frac{1}{x} \ (x \to \infty).$$ Also, its domain of convergence on $\mathbb{R}$ appears to inlude $(-\frac{1}{2},\infty)$, but I'm not sure about the negative reals.

Answer 1

Conclusion. In this answer, we will provide a sketch of proof of the identity

\begin{align*} F(x) &= 1 - \int_{0}^{1} \frac{(1-\xi)^{2x}}{(1+\xi)^2} e^{x \xi} \, \mathrm{d}\xi \tag{$x > -\tfrac{1}{2}$}\\ &= x \int_{0}^{1} (1-\xi)^{2x-1} e^{x \xi} \, \mathrm{d}\xi. \tag{$x > 0$} \end{align*}

Before delving into the proof, note that this allows us to find a closed-form for the values of $F(\cdot)$ at positive half-integer arguments. For example,

\begin{align*} F(\tfrac{1}{2}) &= \int_{0}^{1} \frac{1}{2}e^{\xi/2} \, \mathrm{d}\xi = \sqrt{e} - 1, \\ F(1) &= \int_{0}^{1} (1-\xi) e^{\xi} \, \mathrm{d}\xi = e - 2, \\ F(2) &= \int_{0}^{1} 2 (1-\xi)^3 e^{2\xi} \, \mathrm{d}\xi = \frac{3e^2}{4}-\frac{19}{4}. \end{align*}

The formula above also allows us to analyze the asymptotic behavior of $F(x)$ as $x\to\infty$. Indeed, substitute $\xi = s/x$ to write

$$ F(x) = 1 - \frac{1}{x} \int_{0}^{x} \frac{(1-s/x)^{2x}}{(1+s/x)^2} e^{s} \, \mathrm{d}s. $$

Then, as $x \to \infty$, the dominated convergence theorem yields

$$ \lim_{x\to\infty} \int_{0}^{x} \frac{(1-s/x)^{2x}}{(1+s/x)^2} e^{s} \, \mathrm{d}s = \int_{0}^{\infty} e^{-s} \, \mathrm{d}s = 1. $$

Therefore it follows that

$$ F(x) = 1 - \frac{1+o(1)}{x} \qquad\text{as}\qquad x \to \infty. $$

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Preliminary. Let $(a_n)_{n\geq0}$ and $(b_n)_{n\geq 0}$ with $b_0 = 1$ be given, and define $(h_n)_{n\geq0}$ and $(k_n)_{n\geq0}$ by

$$ \left\{ \begin{aligned} h_n &= a_n h_{n-1} + b_n h_{n-2}, & h_{-2} &= 0, & h_{-1} &= 1 \\ k_n &= a_n k_{n-1} + b_n k_{n-2}, & k_{-2} &= 1, & k_{-1} &= 0 \end{aligned} \right. \tag{1} $$

Then it is well-known that

$$ \frac{h_n}{k_n} = a_0 + \cfrac{b_1}{a_1 + \cfrac{b_2 }{a_2 + \dfrac{\ddots }{\ddots \cfrac{b_{n-1}}{a_{n-1} + \cfrac{b_n}{a_n} } }}} $$

Analysis. To make the notation neater, we will analyze $x/F(x)$ instead. Then by the observation above, we find that $F(x)$ can be defined by the relation

$$ \frac{x}{F(x)} = \lim_{n\to\infty} \frac{h_n}{k_n}, \tag{2} $$

where $(h_n)$ and $(k_n)$ are defined by $\text{(1)}$ with the choices $ a_n = x + n $ and $ b_n = nx $.

From this point on, we will assume we know that the limit $\text{(2)}$ converges and then focus on the question of identifying its value. To this end, we consider the power series

$$ y(z) = \sum_{n=0}^{\infty} \frac{c_n}{n!} z^n, \qquad \text{where} \quad c_n = (n+x)c_{n-1} + nxc_{n-2}. \tag{3} $$

1. By noting that

\begin{align*} \frac{c_n}{n!} &= \left(1 + \frac{x}{n}\right) \frac{c_{n-1}}{(n-1)!} + \frac{x}{n-1} \frac{c_{n-2}}{(n-2)!}, \end{align*}

it follows that the sequence $M_{n} = \max_{0\leq j \leq n} |c_j/j!|$ satisfies

$$ M_{n} \leq \left(1 + \frac{|x|}{n} + \frac{|x|}{n-1} \right) M_{n-1}. $$

This allows to deduce that $M_{n} = \mathcal{O}(n^{2|x|})$, and in particular, the radius of convergence of $\text{(3)}$ is at least $1$.

2. Plugging the recurrence relation for $(c_n)$ to $\text{(3)}$,

\begin{align*} y(z) &= c_0 + c_1 z + \sum_{n=2}^{\infty} \frac{c_n}{n!} z^n \\ &= c_0 + c_1 z + \sum_{n=2}^{\infty} \frac{(n+x)c_{n-1} + nxc_{n-2}}{n!} z^n \\ &= c_0 + c_1 z + x \Biggl( \sum_{n=1}^{\infty} \frac{c_n}{(n+1)!} z^{n+1} - c_0 z \Biggr) + z ( y(z) - c_0 ) + xz \sum_{n=0}^{\infty} \frac{c_n}{(n+1)!} z^{n+1}. \end{align*}

Solving this for the sum $\sum_{n=0}^{\infty} \frac{c_n}{(n+1)!} z^{n+1}$, we get

$$ \sum_{n=0}^{\infty} \frac{c_n}{(n+1)!} z^{n+1} = \frac{(1-z)y(z) + c_0 ((x+1)z - 1) - c_1 z}{x(z+1)}. $$

Taking the differential operator $x(z+1)^2\frac{\mathrm{d}}{\mathrm{d}z}$ to both sides and rearranging, we end up with the linear ODE of the form

$$ y'(z) + \frac{x(z+1)^2 + 2}{z^2-1} y(z) = \frac{(2+x) c_0 - c_1}{z^2 - 1}. $$

This can be solved using the integrating factor method:

$$ y(z) = \frac{1}{\mu(z)} \left( \int_{0}^{z} \frac{(2+x) c_0 - c_1}{\xi^2 - 1} \mu(\xi) \, \mathrm{d}\xi + c_0\mu(0) \right), \tag{4} $$

where $\mu(z)$ is the integrating factor given by

$$ \mu(z) = \exp\left( \int \frac{x(z+1)^2 + 2}{z^2-1} \, \mathrm{d}z \right) = \frac{e^{x(z+1)}}{z+1} (1-z)^{2x+1}. $$

3. Now we define $y_h(z)$ and $y_k(z)$ by

$$ y_h(z) = \sum_{n=0}^{\infty} \frac{h_n}{n!} z^n \qquad\text{and}\qquad y_k(z) = \sum_{n=0}^{\infty} \frac{k_n}{n!} z^n. $$

Then by noting that

$$ \begin{cases} h_0 = x, \\ h_1 = x(x+2), \end{cases} \qquad \begin{cases} k_0 = 1, \\ k_1 = x+1, \end{cases} $$

it follows from $\text{(4)}$ that

$$ y_h(z) = \frac{x\mu(0)}{\mu(z)} = \frac{x (z+1)e^{-x z}}{(1-z)^{2x+1}} \qquad\text{and}\qquad y_k(z) = \frac{1}{\mu(z)} \left( \int_{0}^{z} \frac{1}{\xi^2 - 1} \mu(\xi) \, \mathrm{d}\xi + \mu(0) \right). $$

In particular, $y_k(z)$ diverges to $-\infty$ as $z \to 1^-$ for $x > -\frac{1}{2}$. So, assuming $x > -\frac{1}{2}$, we may invoke the following 'regularization trick':

$$ \lim_{n\to\infty} \frac{h_n}{k_n} = \lim_{z \to 1^-} \frac{y_h(z)}{y_k(z)}. $$

From this, it follows that

$$ \frac{x}{F(x)} = \lim_{n\to\infty} \frac{h_n}{k_n} = \frac{x\mu(0)}{\int_{0}^{1} \frac{1}{\xi^2 - 1} \mu(\xi) \, \mathrm{d}\xi + \mu(0)} $$

and therefore

\begin{align*} F(x) = 1 + \frac{1}{\mu(0)} \int_{0}^{1} \frac{1}{\xi^2 - 1} \mu(\xi) \, \mathrm{d}\xi = 1 - \int_{0}^{1} \frac{(1-\xi)^{2x}}{(1+\xi)^2} e^{x \xi} \, \mathrm{d}\xi. \end{align*}

When $x > 0$, we can perform integration by parts to get

$$ F(x) = x \int_{0}^{1} (1-\xi)^{2x-1} e^{x \xi} \, \mathrm{d}\xi. $$