An integral equality

Let $\psi^{(1)}$ denote the trigamma function and let $f$ be integrable on $(0, 1)$ . It holds that

$\int_{0}^{1} f \left ( \left \{ \frac{1}{x} \right \} \right )\, \mathrm{d}x = \int_{0}^{1} f(x) \psi^{(1)} (1+x) \; \mathrm{d}x$

Solution

We have successively:

$\begin{align*} \int_{0}^{1} f \left(\left\{\frac{1}{x}\right\}\right) \mathrm{d}x &= \sum_{k=1}^{\infty}\int_{1/(k+1)}^{1/k} f \left(\left\{ \frac{1}{x} \right\} \right) \mathrm{d}x \\ &= \sum_{k=1}^{\infty} \int_{k}^{k+1} f \left( \left\{ u \right\} \right) \, \frac{\mathrm{d} u}{u^{2}} \\ &= \sum_{k=1}^{\infty} \int_{k}^{k+1} f\left( u-k \right) \, \frac{\mathrm{d} u}{u^{2}} \\ &= \sum_{k=1}^{\infty} \int_{0}^{1} f\left( v \right) \: \frac{\mathrm{d} v}{(v+k)^{2}} \\ &= \int_{0}^{1} f\left( v \right) \sum_{k=1}^{\infty} \frac{1}{(v+k)^{2}} \: \mathrm{d} v \\ &= \int_{0}^{1} f(v) \psi^{(1)}(v+1) \: \mathrm{d}v \end{align*}$

where the interchange between the infinite sum and the integration is allowed by the uniform bound

$\left|\sum_{k=1}^{N} \frac{1}{(v+k)^{2}} \right| < \frac{\pi^{2}}{6} \quad , \quad N \geq 1, \, 0 \leq v \leq 1$

The result follows.

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2 Comments

Let $\psi^{(1)}$ denote the trigamma function , $\psi^{(0)}$ the digamma function and let $f$ be integrable on $(0, 1)$ . It holds that

$\int_{0}^{1} f \left( \gamma + \psi^{(0)} \left( \left\{ \frac{1}{x}\right\} \right) + 1)\right)\, \mathrm{d}x = \int_{0}^{1} f(x) \, \mathrm{d}x$

Proof: We have successively , using the above

$\begin{align*} \mathcal{J} &= \int_{0}^{1} f \left( \gamma+\psi^{(0)} (x+1) \right) \psi^{(1)} (x+1) \: \mathrm{d}x \\ & = \int_{0}^{1} f \left( \gamma+\psi^{(0)} (x+1) \right) (\gamma+\psi^{(0)} (x+1))' \, \mathrm{d}x \\ & = \int_{0}^{1} f(u) \, \mathrm{d}u \end{align*}$

Application:

$\begin{align*} \int_{0}^{1} \left \{ \frac{1}{x} \right \}\, \mathrm{d}x &= \int_{0}^{1} x \; \psi^{(1)}(1+x) \, \mathrm{d}x \\ &= \left [ x \; \psi^{(0)}(1+x) \right ]_0^1 - \int_{0}^{1} \psi^{(0)}(1+x)\, \mathrm{d}x \\ &=\psi^{(0)} (2) - \left [ \ln \Gamma(1+x) \right ]_0^1 \\ &= 1 - \gamma \end{align*}$

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