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.


2 Answers 2


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

$$\int_{-\infty}^\infty f(x)\,dx=2\int_{0}^{1/2} f(x)\,dx \tag 2$$


$$\int_{-\infty}^\infty f(x)\,dx=2\int_0^{1/2}\frac{1}{x\log^2\left(\frac1x\right)}\,dx=\frac{2}{\log(2)}$$

  • $\begingroup$ Thank you! This explanation is helpful. I am still not sure why in the solution I provided, they used an integral from -1/2 to 0. If you could reference that part it will be extra helpful. But, your answer makes more sense to me anyway, and I can continue with that understanding. $\endgroup$
    – PBJ
    Nov 18, 2016 at 1:08
  • $\begingroup$ And thanks so much for the QUICK response too! So helpful. $\endgroup$
    – PBJ
    Nov 18, 2016 at 1:13
  • $\begingroup$ Is there a (minor) typo in equation (1)? the lower limit should be a less-than sign, and not less-than-or-equal-to? $\endgroup$
    – PBJ
    Nov 18, 2016 at 1:28
  • $\begingroup$ You're welcome. My pleasure. -Mark $\endgroup$
    – Mark Viola
    Nov 18, 2016 at 2:39
  • $\begingroup$ As for equation $(1)$, the value of an integral does not depend on the value at a point. So, it doesn't matter whether $1/2$ is included or not. $\endgroup$
    – Mark Viola
    Nov 18, 2016 at 2:40

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}


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