# A fractional part integral giving $\frac{F_{n-1}}{F_n}-\frac{(-1)^n}{F_n^2}\ln\left(\!\frac{F_{n+2}-F_n\gamma}{F_{n+1}-F_n\gamma}\right)$

I’ve been asked to elaborate on the following evaluation:

\begin{align}\\ \displaystyle {\large\int_0^{1}} \!\cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots + \cfrac 1 { 1 + \psi (\left\{1/x\right\}+1)}}}} \:\mathrm{d}x & = \dfrac{F_{n-1}}{F_{n}} – \dfrac{(-1)^{n}}{F_{n}^2} \ln \!\left(\!\dfrac{F_{n+2}-F_{n}\gamma}{F_{n+1}-F_{n}\gamma} \right)\\\\ \end{align}
where $\left\{x\right\}=x-\lfloor x\rfloor$ denotes the fractional part of $x$, $\gamma$ is the Euler-Mascheroni constant, $F_{n}$ are the Fibonacci numbers, $\psi:=\Gamma’/\Gamma$ is the digamma function and where the continued fraction has a total of $n$ horizontal bars.

#### Solutions Collecting From Web of "A fractional part integral giving $\frac{F_{n-1}}{F_n}-\frac{(-1)^n}{F_n^2}\ln\left(\!\frac{F_{n+2}-F_n\gamma}{F_{n+1}-F_n\gamma}\right)$"

Here is a general approach.

Recall that the digamma function $\displaystyle \psi : = \Gamma’/\Gamma$
admits the following expansion, coming from the Weierstrass infinite product representation of the $\Gamma$ function,
$$\psi(x+1) = -\gamma + \sum_{k=1}^{\infty} \left( \frac{1}{k} – \frac{1}{x+k} \right), \quad x >-1,$$ where $\gamma$ is the Euler-Mascheroni constant. By differentiation, one obtains
$$\psi'(x+1) = \sum_{k=1}^{\infty} \frac{1}{(x+k)^2}, \quad x>-1.$$

Theorem 1. Let $f$ be integrable on $(0,1)$. Then
$$\displaystyle \int_{0}^{1} f \left(\left\{1/x\right\}\right) \mathrm{d}x = \int_{0}^{1} f(x) \: \psi'(x+1) \mathrm{d}x \tag1$$
where $\left\{x\right\}=x-\lfloor x\rfloor$ denotes the fractional part of $x$, $\psi’$ being the derivative of the digamma function
$\displaystyle \psi : = \Gamma’/\Gamma$.

Proof. One may write
\begin{align*}
\displaystyle \int_{0}^{1} f \left(\left\{1/x\right\}\right) \mathrm{d}x
&= \sum_{k=1}^{\infty}\int_{\frac{1}{k+1}}^{\frac{1}{k}} f \left(\left\{1/x\right\}\right) \mathrm{d}x \\
&= \sum_{k=1}^{\infty} \int_{k}^{k+1} f\left(\left\{ u \right\}\right) \: \frac{\mathrm{d} u}{u^{2}} \\
&= \sum_{k=1}^{\infty} \int_{k}^{k+1} f\left( u-k \right) \: \frac{\mathrm{d} u}{u^{2}} \\
&= \sum_{k=1}^{\infty} \int_{0}^{1} f\left( v \right) \: \frac{\mathrm{d} v}{(v+k)^{2}} \\
&= \int_{0}^{1} f\left( v \right) \sum_{k=1}^{\infty} \frac{1}{(v+k)^{2}} \: \mathrm{d} v \\
&= \int_{0}^{1} f(v) \: \psi'(v+1) \: \mathrm{d}v,
\end{align*}
where the interchange between the infinite sum and the integration is allowed by the
uniform bound: $$\quad \left|\, \sum_{k=1}^{N} \frac{1}{(v+k)^{2}} \, \right| \, < \, \frac{\pi^{2}}{6}, \quad N \geq 1, \, 0 \leq v \leq 1.$$

One of the consequences of Theorem 1 is that the transformation
$x \rightarrow \gamma+\psi ( \left\{1/x\right\}+1)$ leaves the Lebesgue measure on (0,1) invariant.

Theorem 2. Let $f$ be integrable on $(0,1)$. Then
\begin{align} \displaystyle \int_{0}^{1} f \left(\gamma+\psi ( \left\{1/x\right\}+1)\right)\:\mathrm{d}x = \int_{0}^{1} f(x) \: \mathrm{d}x \tag2 \end{align} where $\left\{x\right\}=x-\lfloor x\rfloor$ denotes the fractional part of $x$, $\gamma$ being the Euler-Mascheroni constant and $\psi:= \Gamma’/\Gamma.$

Proof.
From $(1)$, one gets
\begin{align*} \displaystyle \int_{0}^{1} f \left(\gamma+\psi (\left\{1/x\right\}+1)\right) \:\mathrm{d}x &= \int_{0}^{1} f \left(\gamma+\psi (x+1)\right) \psi’ (x+1) \: \mathrm{d}x \\ & = \int_{0}^{1} f \left(\gamma+\psi (x+1) \right) (\gamma+\psi (x+1))’ \: \mathrm{d}x \\ & = \int_{0}^{1} f(u) \: \mathrm{d}u, \end{align*}
using the change of variables $u=\gamma+\psi (x+1)$
which gives $u(0)=\gamma+\psi(1)=0$ and $u(1)=\gamma+\psi(2)=1$.

Theorem 2 enables one to evaluate a great variety of integrals involving the digamma function in the integrand.

Proposition 1. Let $n=0,1,2,\cdots .$ Then \begin{align} \displaystyle \int_{0}^{1} \left(\psi( \left\{1/x\right\}+1)\right)^{n}\:\mathrm{d}x & = \sum_{k=0}^{n}\frac{(-1)^{k}}{n-k+1}{{n}\choose k}\gamma^{k} \end{align} \tag3 and
\begin{align} \gamma^{n} = (-1)^{n}\sum_{k=0}^{n}\!{{n}\choose k}B_{k}\!\displaystyle \int_{0}^{1} \! \left(\psi( \left\{1/x\right\}+1)\right)^{n-k}\mathrm{d}x. \end{align} \tag4 where $\gamma$ is the Euler constant, $\psi:= \Gamma’/\Gamma$ and $B_{k}$ are the Bernoulli numbers.

Proof. Using $(1)$, we may prove $(3)$ by writing \begin{align*} \int_{0}^{1} \left(\psi( \left\{1/x\right\}+1)\right)^{n} \mathrm{d}x &= \int_{0}^{1} \left(\gamma+\psi( \left\{1/x\right\}+1)-\gamma\right)^{n}\:\mathrm{d}x \\ &=\sum_{k=0}^{n} (-1)^{n-k}\gamma^{n-k}{{n}\choose k} \int_{0}^{1} \left(\gamma+\psi( \left\{1/x\right\}+1)\right)^{k}\:\mathrm{d}x \\ &= \sum_{k=0}^{n} (-1)^{n-k}\gamma^{n-k}{{n}\choose k} \int_{0}^{1} x^{k}\:\mathrm{d}x \\ & = \sum_{k=0}^{n}\frac{(-1)^{k}}{n-k+1}{{n}\choose k}\gamma^{k}.\end{align*}
Identity $(4)$ is then deduced from $(3)$ by
appealing to an inversion combinatorial sum (J. Riordan, Inverse Relations and Combinatorial Identities. The American Mathematical
Monthly, Vol. 71, No. 5, May 1964, p. 495).

Proposition 2. \begin{align}\\ \displaystyle \int_0^{1} \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots + \cfrac 1 { 1 + \psi (\left\{1/x\right\}+1)}}}} \mathrm{d}x & = \dfrac{F_{n-1}}{F_{n}} – \dfrac{(-1)^{n}}{F_{n}^2} \ln \left( \dfrac{F_{n+2}-F_{n}\gamma}{F_{n+1}-F_{n}\gamma} \right) \tag5 \\\\ \end{align}
where $\left\{x\right\}=x-\lfloor x\rfloor$ denotes the fractional part of $x$, $\gamma$ is the Euler constant, $F_{n}$ are the Fibonacci numbers, $\psi:=\Gamma’/\Gamma$ is the digamma function and where the continued fraction has $n$ horizontal bars.

Proof. Using $(2)$, we just have to evaluate the corresponding elementary rational function.
One may prove by a simple induction that
\begin{align} \displaystyle \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {\ddots + \cfrac 1 { 1 + \, x \,}}}}
= \dfrac{F_{n}x+F_{n+1}}{F_{n+1}x+F_{n}} \end{align} where the continued fraction has $n$ horizontal bars, leading easily to $(5)$.