Let $a$ be a real number, and let $f_X(x)=kx^{-a}$, where $k>0$. Since we are dealing with positive quantities, the only thing that we require in order for $f_X(x)$ to be a density function is $\int_{x_l}^\infty kx^{-a}\,dx=1.\qquad\qquad(\ast)$ The above integration, like many others that arise in probability, is over an infinite interval, so may fail to exist.
We will show that the above integral converges if $a>1$ and diverges otherwise.
Let $I(M)=\int_{x_l}^M kx^{-a}\,dx.$ By definition, if $\lim_{M\to\infty}I(M)$ exists, the integral in $(\ast)$ converges and has value equal to that limit.
Suppose first that $a>1$. Integrating, we find that $I(M)=\int_{x_l}^M kx^{-a}\,dx=\left.\frac{-k}{(a-1)x^{a-1}}\right|_{x_l}^M$ Thus $I(M)=\frac{k}{(a-1)x_{x_l}^{a-1}}-\frac{k}{(a-1)M^{a-1}}. \qquad\qquad(\ast\ast)$ Since $a-1>0$, we can see that $\frac{k}{(a-1)M^{a-1}}\to 0$ as $M\to\infty$. It follows that if $a>1$, then $\int_{x_l}^\infty kx^{-a}\,dx=\frac{k}{(a-1){x_l}^{a-1}}.$
For any $a>1$, and any positive $x_l$, we can find a unique constant of proportionality $k$ such that $\frac{k}{(a-1){x_l}^{a-1}}=1.$ Just take $k=(a-1)x_l^{a-1}$. So everything is fine if $a>1$.
We complete the analysis by showing that if $a \le 1$, then our integral does not converge. There are two somewhat different cases, $a=1$ and $a<1$. Suppose first that $a=1$. Then $I(M)=\int_{x_l}^M kx^{-1}\,dx=\left.k\ln x\right|_{x_l}^M=k\ln M-k\ln x_l.$ As $M\to\infty$, $\ln M\to\infty$, so $I(M)$ does not have a finite limit, and therefore $\int_{x_l}^\infty kx^{-1}\,dx$ does not exist.
Finally, we deal with $a<1$. In this case, $I(M)=\frac{kM^{1-a}}{1-a}-\frac{kx_l^{1-a}}{1-a}.$ As $M\to\infty$, $I(M)\to\infty$, so the integral from $x_l$ to $\infty$ diverges (does not have a finite value).