Does the integral exist? $\displaystyle\int_{0}^{1}\{\frac{1}{x}\}dx,\quad$ where {x} is the fractional part.
I have broken it into $$\displaystyle\int_{0}^{1}\frac{1}{x}-\lfloor \frac{1}{x} \rfloor dx = \lim_{x\to 0}-\ln{x} -\lim_{n\to\infty} H_n+1$$
This looks very close to a negative Euler-Mascheroni const + 1, although $\gamma$ is defined as $$\lim_{n\to\infty}\quad H_n - \ln{n}$$
Are $$\lim_{x\to\infty}\quad \ln{x} \text{ and } \lim_{x\to 0}\quad -\ln{x}$$ interchangeable in these cases? If so, it would seem then the integral would evaluate to $-\gamma+1$, as is shown from an area approximation from .001 to 1 on my calculator.
Any help is appreciated, thanks!
As explained by @user1952009, use the substitution $u = \dfrac{1}{x}$, $du = \dfrac{-dx}{x^2}$. So $$\int_{0}^{1}\{ \dfrac{1}{x}\} = \int_{1}^{\infty}\dfrac{\{u\}}{u^2}du = $$ (Now use $\{u\} = u - \lfloor u\rfloor$) $$\lim_{n\to\infty} \text{ }\int_{1}^{n}\dfrac{u - \lfloor u\rfloor}{u^2}du = \lim_{n\to\infty}\ln{n}-\sum_{k=1}^{n-1}\int_{k}^{k+1}\dfrac{k}{x^2}dx =$$ $$\lim_{n\to\infty}\ln{n}-\sum_{k=1}^{n-1}\dfrac{1}{k+1} = \lim_{n\to\infty}\ln{n}-H_n+1 = \Large\boxed{-\gamma + 1}$$