definite integral with fractional part

[jacks]

Evaluation of $\displaystyle \int_{0}^{1}\left\{(-1)^{\lfloor \frac{1}{x}\rfloor}\frac{1}{x}\right\}dx\;,$ Where $x=\lfloor x \rfloor +\{x\}.$

[Tolaso J Kos]

Evaluation of $\displaystyle \int_{0}^{1}\left\{(-1)^{\lfloor \frac{1}{x}\rfloor}\frac{1}{x}\right\}dx\;,$ Where $x=\lfloor x \rfloor +\{x\}.$

Hello Jacks.

\begin{align*}
\int_{0}^{1}\left \{ (-1)^{\left \lfloor 1/x \right \rfloor} \frac{1}{x} \right \}\, {\rm d}x &\overset{u=1/x}{=\! =\! =\!} \int_{1}^{\infty} \left \{ (-1)^{\left \lfloor x \right \rfloor} x \right \} \frac{{\rm d}x}{x^2} \\
&=\sum_{n=1}^{\infty} \left[ \int_{2n-1}^{2n} \left \{ (-1)^{\left \lfloor x \right \rfloor} x \right \} \frac{{\rm d}x}{x^2} + \int_{2n}^{2n+1} \left \{ (-1)^{\left \lfloor x \right \rfloor} x \right \} \frac{{\rm d}x}{x^2} \right]\\
&= \sum_{n=1}^{\infty} \left [ \int_{2n-1}^{2n} \frac{\left \{ -x \right \}}{x^2}\, {\rm d}x + \int_{2n}^{2n+1} \frac{\left \{ x \right \}}{x^2} \, {\rm d}x \right ]\\
&= \sum_{n=1}^{\infty} \left [ \int_{2n-1}^{2n} \frac{1-\left \{ x \right \}}{x^2} \, {\rm d}x+ \int_{2n}^{2n+1} \frac{\left \{ x \right \}}{x^2} \, {\rm d}x \right ]\\
&= \sum_{n=1}^{\infty} \int_{2n-1}^{2n} \frac{{\rm d}x}{x^2} - \sum_{n=1}^{\infty} \left [ \int_{2n-1}^{2n} \left ( x-2n+1 \right )\frac{{\rm d}x}{x^2} - \int_{2n}^{2n+1} (x-2n) \frac{{\rm d}x}{x^2} \right ]\\
&= \sum_{n=1}^{\infty} \left [ \frac{1}{2n-1} - \frac{1}{2n} \right ] - \sum_{n=1}^{\infty} \left [ \int_{2n-1}^{2n}\frac{{\rm d}x}{x} - (2n-1) \int_{2n-1}^{2n}\frac{{\rm d}x}{x^2} - \int_{2n}^{2n+1} \frac{{\rm d}x}{x} +2n \int_{2n}^{2n+1} \frac{{\rm d}x}{x^2} \right ]\\
&= \sum_{n=1}^{\infty} \left [ \frac{1}{2n-1} - \frac{1}{2n} \right ] - \sum_{n=1}^{\infty} \left [ \log \frac{4n^2}{(2n-1)(2n+1)} - \frac{1}{2n} + \frac{1}{2n+1} \right ] \\
&=\sum_{n=1}^{\infty}\left [ \frac{1}{2n-1} - \frac{1}{2n} \right ] + \sum_{n=1}^{\infty} \left [ \frac{1}{2n} - \frac{1}{2n+1} \right ] - \log \prod_{n=1}^{\infty} \left ( 1+ \frac{1}{4n^2-1} \right ) \\
&=\cancel{\log 2- \log 2} +1- \log \frac{\pi}{2} \\
&=1- \log \frac{\pi}{2}
\end{align*}

because

\begin{align*}
\sum_{n=1}^{\infty} \left [ \frac{1}{2n-1} - \frac{1}{2n} \right ] &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}\\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} 1^n}{n}\\
&= \log (1+1) \\
&=\log 2
\end{align*}

and similarly the second sum. Now that product is actually very famous but if we want a proof we are using the Stirling approximation of

\begin{equation} n! \sim \sqrt{2\pi n} \left ( \frac{n}{e} \right )^n \end{equation}

thus:

\begin{align*}
\log \prod_{n=1}^{\infty} \left ( 1+ \frac{1}{4n^2-1} \right ) &=\lim_{n \rightarrow +\infty} \log \frac{4^n (n!)^2 }{1\cdot 3^2 \cdot 5^2 \cdots (2n-1)^2\cdot (2n+1)} \\
&= \lim_{n \rightarrow +\infty} \log \frac{4^{2n} (n!)^4}{\left ( (2n)! \right)^2 \left ( 2n+1 \right )}\\
&\overset{(1)}{=} \lim_{n \rightarrow +\infty} \log \frac{4^{2n} (2\pi n)^2 \left ( \frac{n}{e} \right )^{4n}}{(4\pi n) \left ( \frac{2n}{e} \right )^{4n} (2n+1)}\\
&= \lim_{n \rightarrow +\infty} \log \frac{\left ( 2\pi n \right )^2}{(4\pi n) (2n+1)} \\
&=\log \frac{\pi}{2}
\end{align*}

[jacks]

Thanks Admin Got it.