Help to understand portion of a proof proving a function is integrable I want to prove that f is integrable:
$$
\mathrm{f}\left(x\right) \equiv
\left\{\begin{array}{ll}\displaystyle{%
{1 \over \left\vert\, x\,\right\vert\,\log^{2}\left(\,1/\left\vert\, x\,\right\vert\,\right) }\,,\quad} &
\displaystyle{\left\vert\, x\,\right\vert\ \leq\ {1 \over 2}}
\\[2mm]
\displaystyle{0\,,} & \mbox{otherwise}
\end{array}\right.
$$
I am trying to understand the proof, but the first few statements (shown in this image link) got me stuck. First part of the proof
The proof will go on to use u-substitution (the full proof is the last problem from http://larryfenn.com/assets/writing/ra.pdf)
But I don't understand this first part shown above -- why is the integral of
$\,\mathrm{f}$ over $\mathbb{R}$ "is really" the same as using $-1/2$ and $1/2$ as limits ?. And what is the relevance of $\,\mathrm{f}$ being an even function - what does that mean in regards to this problem ?. Thanks.
 A: Let $f$ be the function given by
$$f(x)=\begin{cases}\frac{1}{|x|\log^2\left(\frac1{|x|}\right)}&,|x|\le 1/2\\\\0&,|x|>1/2\end{cases}$$
Note that we can decompose the integral as the sum
$$\int_{-\infty}^\infty f(x)\,dx=\int_{|x|\le 1/2}f(x)\,dx+\int_{|x|\ge 1/2}f(x)\,dx \tag 1$$
Since $f(x)=0$ when $|x|>1/2$, then the second integral on the right-hand side of $(1)$ is $0$.  Hence, we have
$$\int_{-\infty}^\infty f(x)\,dx=\int_{|x|\le 1/2}f(x)\,dx \tag 2$$
Furthermore, since $f(x)$ is an even function, then we can modify $(2)$ and write
$$\int_{-\infty}^\infty f(x)\,dx=2\int_{0}^{1/2} f(x)\,dx \tag 2$$
Finally, 
$$\int_{-\infty}^\infty f(x)\,dx=2\int_0^{1/2}\frac{1}{x\log^2\left(\frac1x\right)}\,dx=\frac{2}{\log(2)}$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{-\infty}^{\infty}\mrm{f}\pars{x}\,\dd x & =
2\int_{0}^{1/2}{\dd x \over x\ln^{2}\pars{1/x}} =
2\int_{0}^{1/2}{\dd x \over x\ln^{2}\pars{x}} =
2\int_{0}^{1/2}{1 \over x}\,\,\,
\overbrace{\int_{0}^{\infty}x^{t}\,t\,\,\dd t}^{\ds{1 \over \ln^{2}\pars{x}}}\
\,\,\,\dd x
\\[5mm] & =
2\int_{0}^{\infty}t\int_{0}^{1/2}x^{t - 1}\,\,\dd x\,\dd t =
2\int_{0}^{\infty}\pars{1 \over 2}^{t}\,\,\dd t =
2\bracks{-\,{1 \over \ln\pars{1/2}}} = \bbx{\ds{2 \over \ln\pars{2}}}
\end{align}
