Understanding $\int_\limits{0}^{1}\frac{1}{|t-s|^{\frac{1}{3}}}ds$ 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!
 A: 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. 
A: 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:
$$\int_{-1}^0 -x\,dx + \int_0^1 x\,dx$$
A: 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 $|\cdot|$.  
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}

