integral $\int_{0}^{1}\left( \left\lfloor{\frac{2}{x}} \right\rfloor-2 \left\lfloor{\frac{1}{x}} \right\rfloor \right)dx$ Evaluate $$\int_{0}^{1}\left( \left\lfloor{\frac{2}{x}} \right\rfloor-2 \left\lfloor{\frac{1}{x}} \right\rfloor \right)dx$$
My Attempt
$$I_{1}=\int_{0}^{1}\left\lfloor{\frac{2}{x}} \right\rfloor dx$$
Put $x=2t$
$$I_{1}=\int_{0}^{\frac{1}{2}}2\left\lfloor{\frac{1}{t}} \right\rfloor dt=\int_{0}^{\frac{1}{2}}2\left\lfloor{\frac{1}{x}}\right\rfloor dx$$
let $$I_{2}=\int_{0}^{1} \left\lfloor{\frac{1}{x}} \right\rfloor dx=\int_{0}^{\frac{1}{2}}\left\lfloor{\frac{1}{x}}\right\rfloor dx+\int_{\frac{1}{2}}^{1}\left\lfloor{\frac{1}{x}}\right\rfloor dx$$
Given integral 
$$
\begin{align}
\int_{0}^{1}\left(\left\lfloor{\frac{2}{x}}\right\rfloor-2\left\lfloor{\frac{1}{x}}\right\rfloor \right)dx&=I_{1}-2I_{2}\\
&=2\int_{0}^{\frac{1}{2}}\left\lfloor{\frac{1}{x}}\right\rfloor dx-2\int_{0}^{\frac{1}{2}}\left\lfloor{\frac{1}{x}}\right\rfloor dx-2\int_{\frac{1}{2}}^{1}\left\lfloor{\frac{1}{x}}\right\rfloor dx\\
&=-2\int_{\frac{1}{2}}^{1}\left\lfloor{\frac{1}{x}}\right\rfloor dx\\
&=(-2)(1)=-2.
\end{align}
$$
But answer given is $\ln(\frac{4}{e})$
What mistake am I making?

The function $\lfloor \frac{2}{x} \rfloor$ is unbounded near $x = 0$ so if you try to split your integral, at best you can hope to interpret $I_1 = \int_0^1 \lfloor \frac{2}{x} \rfloor \, dx$ as an improper integral. Unfortunately, this integral actually diverges (it behaves like $\int_0^1 \frac{1}{x} \, dx$) and so is $I_2$ (this also follows from the calculations below) so you can't do all the manipulations as you are subtracting $+\infty$ from $+\infty$...
Let's see how the integrand (which I will denote by $f(x)$) behaves. Divide the interval $\left( 0, 1 \right]$ into intervals
$$\cdots, \left( \frac{2}{k+1}, \frac{2}{k} \right],\cdots,\left( \frac{2}{4}, \frac{2}{3} \right], \left( \frac{2}{3}, 1 \right]. $$


*

*If $x \in \left( \frac{2}{k+1}, \frac{2}{k} \right]$ then $\frac{2}{x} \in [k,k+1)$ so $\lfloor \frac{2}{x} \rfloor = k$. 

*If $x \in \left( \frac{2}{k+1}, \frac{2}{k} \right]$ then $\frac{1}{x} \in [\frac{k}{2},\frac{k+1}{2})$. Now, if $k$ is even then $2\lfloor \frac{1}{x} \rfloor = 2 \frac{k}{2} = k$ while if $k$ is odd then $2 \lfloor \frac{1}{x} \rfloor = 2\frac{k-1}{2} = k - 1$.

*Hence, we have for all $x \in (0,1]$
$$ f(x) = \lfloor \frac{2}{x} \rfloor - 2 \lfloor \frac{1}{x} \rfloor = \begin{cases} 1 & x \in \left( \frac{1}{l+1}, \frac{2}{2l + 1} \right], l \geq 1, \\
0 & \textrm{otherwise}. \end{cases} $$


Hence,
$$ \int_0^1 f(x) \, dx = \lim_{n \to \infty} \int_{\frac{2}{2n+1}}^1 f(x) \, dx = \lim_{n \to \infty} \left( \sum_{l=1}^n \left( \frac{2}{2l+1} - \frac{1}{l+1} \right) \right) \\
= 2 \cdot \sum_{l=1}^{\infty} \left( \frac{1}{2l+1} - \frac{1}{2l+2} \right) = 2 \sum_{n=3}^{\infty} \frac{(-1)^{n+1}}{n} = 2 \left( \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} - \left( 1 - \frac{1}{2} \right) \right) = 2 \left( \ln(2) - \frac{1}{2} \right) = 2 \ln(2) - 1 = \ln(4) - \ln(e) = \ln \left( \frac{4}{e} \right) $$
where I used the Taylor series of $\ln(1 + x)$ which converges at $x = 2$ to $\ln(2)$ to evaluate the infinite sum. 
