Answer
The improper integral converges to $\dfrac{1}{4}$.
Work Step by Step
We are given that $I=\int_{1}^{+\infty} \dfrac{1}{x^5} \ dx$
In order to solve the above integral, we will use the following formula such as:
$\int x^n \ dx=\dfrac{x^{n+1}}{n+1}+C$
Now, we have $\int_{1}^{+\infty} \dfrac{1}{x^5}=\lim\limits_{a \to +\infty}\int_{1}^{a} x^{-5} \ dx$
or, $=\dfrac{-1}{4} \lim\limits_{a \to +\infty}[\dfrac{1}{x^4}]_1^{a}$
or, $=\dfrac{-1}{4} \lim\limits_{a \to +\infty} [\dfrac{1}{a^4}-1]$
or, $=\dfrac{-1}{4}[\dfrac{1}{(\infty)^4}-1]$
or, $=\dfrac{1}{4}$
Therefore, the given improper integral converges to $\dfrac{1}{4}$.
