Understanding $\int_\limits{0}^{1}\frac{1}{|t-s|^{\frac{1}{3}}}ds$

Question

I am trying to understand the following integral $\int_\limits{0}^{1}\frac{1}{|t-s|^{\frac{1}{3}}}ds =\int_\limits{0}^{t}\frac{1}{(t-s)^{\frac{1}{3}}}ds +\int_\limits{t}^{1}\frac{1}{(s-t)^{\frac{1}{3}}}ds=2\int_\limits{0}^{1} \frac{dy}{y^{\frac{1}{3}}}\leqslant 3 $

I am not understanding the step $\int_\limits{0}^{1}\frac{1}{|t-s|^{\frac{1}{3}}}ds =\int_\limits{0}^{t}\frac{1}{(t-s)^{\frac{1}{3}}}ds +\int_\limits{t}^{1}\frac{1}{(s-t)^{\frac{1}{3}}}ds$.

Question:

Why is there a break in the interval $[0,1]$? What happened to the modulus? How can one integrate with modulus? I do not know this technique and I have not been able to figure it out.

Thanks in advance!

Answer 1

Consider the simpler integral

$$\int_{-1}^1 |x|dx.$$

We don't know the antiderivative of $|x|$ immediately, but we do know that either $|x|=x$ or $|x|=-x$, and we know the antiderivative of both of those. So, we can split up the integral as follows:

$$\int_{-1}^1 |x|dx = \int_{-1}^0 |x|dx + \int_0^1 |x|dx.$$

We know that, if $x<0$, $|x|=-x$, and if $x>0$, $|x|=x$, so we have that

$$\int_{-1}^1 |x|dx = -\int_{-1}^0 xdx + \int_0^1 xdx,$$

each of which can be integrated separately. The integral you pose uses the same technique on a slightly more complicated function.

Answer 2

Taking a simpler integral:

$$\int_{-1}^{1} |x|\,dx$$

Realize that there is no antiderivative of the whole function $|x|$. That's because $|x|$ is a piecewise function and is composed of two different functions with different antiderivatives. Specifically $-x$ and $x$:

$$ |x| = \left\{\begin{aligned} &-x&x<0 \\ &x&x\geq 0 \end{aligned} \right.$$

$$\int |x|\,dx = \left\{\begin{aligned} &-{x^2\over 2}&x<0 \\ &{x^2\over 2}&x\geq 0 \end{aligned} \right.$$

Thus, you have to split the integral into two to actually integrate:

$$\int_{-1}^0 -x\,dx + \int_0^1 x\,dx$$

Answer 3

Also note that here $t\in(0,1)$. You may try to compute what if $t=0$ or $t=1$ or $t>1$.

For $t>1$, there is no singularity in the integrand, and the integrand becomes $\dfrac{1}{(t-s)^{1/3}}$.

For $t=0$, the integrand is $\dfrac{1}{t^{1/3}}$.

Similar situation happens for $t=1$.

For $t\in(0,1)$, it is an improper integral which the singularity lies in the domain of the integration $(0,1)$, and we define such an improper integral to be the sum of two $\displaystyle\int_{0}^{t}f(s)ds$ and $\displaystyle\int_{t}^{1}f(s)ds$ and say that the improper integral exists provided that the latter two integrals exist.

Indeed, one cannot generally deal with the integral of $|\cdot|$, one splits the domain to break out the $|\cdot|$.

Answer 4

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{1}{\dd s \over \verts{t - s}^{1/3}} & = \int_{-t}^{1 - t}\verts{s}^{-1/3}\,\dd s \\[1cm] & = \bracks{t < 0}\int_{-t}^{1 - t}s^{-1/3}\,\dd s + \bracks{0 < t < 1}\pars{% \int_{-t}^{0}\pars{-s}^{-1/3}\,\dd s + \int_{0}^{1 - t}s^{-1/3}\,\dd s} \\[2mm] & + \bracks{t > 1}\int_{-t}^{1 - t}\pars{-s}^{-1/3}\,\dd s \\[1cm] & = \bracks{t < 0}\bracks{{3 \over 4}\,\pars{1 - t}^{4/3} - {3 \over 4}\,\pars{-t}^{4/3}} + \bracks{0 < t < 1}\bracks{{3 \over 2}\,t^{2/3} + {3 \over 4}\,\pars{1 - t}^{4/3}} \\[2mm] & + \bracks{t > 1}\bracks{-\,{3 \over 2}\,\pars{t - 1}^{2/3} + {3 \over 2}\,t^{2/3}} \\[1cm] & = \begin{array}{|l|}\hline\mbox{}\\ \ds{\ {3 \over 4}\bracks{t < 0}\bracks{\pars{1 - t}^{4/3} - \pars{-t}^{4/3}} + {3 \over 4}\bracks{0 < t < 1}\bracks{2\,t^{2/3} + \pars{1 - t}^{4/3}}\ } \\[2mm] \ds{\ +\ {3 \over 2}\bracks{t > 1}\bracks{t^{2/3} -\pars{t - 1}^{2/3}}\ } \\ \mbox{}\\ \hline \end{array} \end{align}