$$I=\int_0^{\frac{\pi}{2}} \lbrace\tan x\rbrace\mathrm{d}x$$ So I have this interesting integral and I tried to evaluate it: Let $u=\tan x$ we'll have the following: \begin{align} \int_0^\infty \frac{\lbrace u\rbrace}{1+u^2 }\mathrm{d}u&amp;=\int_0^\infty \frac{u-\lfloor u\rfloor}{1+u^2}\mathrm{d}u\\ &amp;=\lim_{a\to +\infty} \Bigg(\int_0^a \frac{u}{u^2+1}\mathrm{d}u-\int_0^a\frac{\lfloor u\rfloor}{u^2+1}\mathrm{d}u\Bigg) \end{align} And the first part is obvious: $$\int_0^a \frac{u}{u^2+1}\mathrm{d}u=\frac{\ln (a^2+1)}2$$ Therefore: $$I=\lim_{a\to +\infty}\Bigg(\frac{\ln (a^2+1)}2-\underbrace{\int_0^a\frac{\lfloor u\rfloor}{u^2+1}\mathrm{d}u}_{F}\Bigg)$$ And $F$ is the part where I got stuck because $\lfloor x\rfloor$ can't be simplified or approximated. Any thoughts or hints?
How to calculate $$\int_{0}^{\pi/2}{\{\tan(x)\}dx}$$ where $\{x\}$ means the fractional part of $x$. I mentioned the zero points of $\{\tan(x)\}$. $\{\tan(x)\}$ equals to 0 iff $x=\arctan(k)$, $k$ is an integer. I tried to separate it into$$\sum_{k=0}^{\infty}{\int_{\arctan(k)}^{\arctan(k+1)}{(\tan(x)}-k)dx}$$ Then I found it hard to continue. I don't think I am on the right way.
The portion of the tangent to any conic intercepted between the point of contact and the directrix subtends a right angle at the focus of the conic? Time Expressions with "in" I have seen some examples about using "in" with time expressions. They give some examples as shown below:     I will leave for vacation in a month. (Exactly one month from now I   will go on vacation.)      We’ll be there in an hour.  (Exactly one hour will go by and then we   will arrive.)   But these comments confused me. Then I looked the usage of "in" up in a dictionary and it says about usage of "in" with time expressions :"before or at the end of a particular period". I think the comments and the definition in the dictionary contradict each other?    I have examples of mine:     1- Dinner will be ready in ten minutes. (Maybe dinner will be   ready at the 5th minute? or 8th. Does it strictly mean dinner will be   ready at the end of the ten minutes?)      2- My letter arrived in two days. 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If $a$ and $b$ divide $c$, then $lcm(a,b)$ also divides $c$  Proof I've been shown:  Since $c$ is a common multiple of $a,b$, it means that $lcm(a,b) \leq c$.  So: $c=q\cdot lcm(a,b)+r$, $q \in \mathbb Z$ and $0\leq r \leq lcm(a,b)$  Since $c$ and $lcm(a,b)$ are common multiples of $a,b$, then so is $r$. But $m$ is the least common divisor, then it means $r=0$.   [Proof taken from: http://www.math.utah.edu/~wood/4400/hw1sol.pdf]  I cannot understand the last step... Why does it imply that $r=0$?   Thank you! The equivalence between Cauchy integral and Riemann integral for bounded functions Definitions  Suppose $P\colon a=x_0&lt;x_1&lt;\dotsb&lt;x_n=b$ is a partition of $[a,b]$. Let $\Delta x_k=x_k-x_{k-1}$ and $\lVert P\rVert$ denotes $\max_{0&lt;k\le n}\Delta x_k$.  The Cauchy integral of a function $f$ on closed interval $[a,b]$ equals to $I$ if and only if for each $\epsilon&gt;0$, there's some $\delta&gt;0$, for each partition $P$ of $[a,b]$ such that $\lVert P\rVert&lt;\delta$, we have $\left\lvert\sum_{k=1}^nf(x_k)\Delta x_k-I\right\rvert&lt;\epsilon$.  Problem  If $f$ is bounded on $[a,b]$ whose Cauchy integral equals to $I$, then $f$ is Riemann-integrable and $\int_a^bf=I$.  Background  It's an exercise from our calculus(analysis) problemset book, and there's a hint: consider the partitions whose $x_k-x_{k-1}$ is a constant for different $k$'s, and try to estimate the Riemann sum for each of these partitions through the Cauchy integral.  I have no idea about such estimation. After drawing some pictures, I discouraged. I googled on the Internet and found an . I realized that it's a quite different approach and with some advanced techniques (such as the analysis of a positive measure set -- discontinuities). 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