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Fractional integral

Let \{ \cdot \} denote the fractional part. Prove that

    \[\int_{0}^{\pi/2}\sin 2x \{\ln^{2n-1} \tan x \} \, \mathrm{d}x\]

for the different values of the integer number n.

Solution

Let \mathcal{J} denote the integral,

    \begin{align*} \mathcal{J} &= \int_{0}^{\pi/2} \sin 2x \left \{ \ln^{2n-1} \tan x \right \}\, \mathrm{d}x \\ &\!\!\!\!\!\overset{u=\pi/2-x}{=\! =\! =\! =\! =\! =\!} \int_{0}^{\pi/2} \sin 2 \left ( \frac{\pi}{2} - u\right ) \left \{ \ln^{2n-1} \tan \left ( \frac{\pi}{2} - u \right ) \right \}\, \mathrm{d}u \\ &= \int_{0}^{\pi/2} \sin 2u \left \{ \ln^{2n-1} \frac{1}{\tan u} \right \} \, \mathrm{d}u\\ &=\int_{0}^{\pi/2} \sin 2u \left \{ - \ln^{2n-1} \tan u \right \}\, \mathrm{d}u \\ &= \frac{1}{2} \int_{0}^{\pi/2} \sin 2u \left ( \left \{ \ln^{2n-1} \tan u \right \} + \left \{ -\ln^{2n-1} \tan u \right \} \right ) \, \mathrm{d}u \\ &= \frac{1}{2} \int_{0}^{\pi/2} \sin 2u \, \mathrm{d}u \\ &= \frac{1}{2} \end{align*}

since \left \{ x \right \} + \left \{ -x \right \} = 1 if x \notin \mathbb{Z} whereas \left \{ x \right \} + \left \{ -x \right \} = 0 if x \in \mathbb{Z}. Therefore,

    \[\left \{ \ln^{2n-1} \tan u \right \} + \left \{ -\ln^{2n-1} \tan u \right \} =1\]

except of a countable set \displaystyle A = \left\{ x_k \in \left ( 0, \frac{\pi}{2} \right ) \bigg| \ln^{2n-1} \tan x_k \in \mathbb{Z} \right\} whose measure is 0.

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