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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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