Question

How do you determine if the improper integral converges or diverges $\int\limits_{0}^{\infty }{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}$ ?
(a) Splitting the integration
(b) Finding the limit value
(c) Integration with limit infinity
(d) None of these

Answer 1

Hint: In this given problem we are to find the convergence of the given integral $\int\limits_{0}^{\infty }{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}$. So, as we have one limit as infinity, we are to split the integration into different limits. Then finding the integrations along with finding their limit will give us our desired result.

Complete step by step answer:
We are given the integral, $\int\limits_{0}^{\infty }{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}$.
We are to check if the integral converges or not.
So, let us suppose, $f\left( x \right)=\dfrac{1}{{{\left( x-2 \right)}^{2}}}$ ,
Now we can easily see, the function is not continuous in the interval of integration.
Then, to get the results of the integration we have to split the integral.
As the function has a point of discontinuity on 2, we will split the integral from that point.
$\int\limits_{0}^{\infty }{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}=\int\limits_{0}^{2}{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}+\int\limits_{2}^{A}{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}+\int\limits_{A}^{\infty }{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}$ with A > 2.
Now, we have to proceed with one part at one time and check if the limits of the given integrations exist or not. If they turn out to be infinity then they diverge or otherwise converge.
To start with,
$\int\limits_{0}^{2}{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}=\underset{t\to 2-}{\mathop{\lim }}\,\int\limits_{0}^{t}{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}$
Now, the integration of $\int{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx=-\dfrac{1}{x-2}dx}$as, $\int{\dfrac{1}{{{x}^{2}}}dx=-\dfrac{1}{x}dx}$, we get,
$\Rightarrow \int\limits_{0}^{2}{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}=\underset{t\to 2-}{\mathop{\lim }}\,\left[ -\dfrac{1}{x-2} \right]_{0}^{t}$
Hence, at the end,
$\Rightarrow \int\limits_{0}^{2}{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}=\underset{t\to 2-}{\mathop{\lim }}\,\left( \dfrac{1}{2-t}-\dfrac{1}{2} \right)=+\infty $
Again, to start with,
$\int\limits_{2}^{A}{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}=\underset{t\to 2+}{\mathop{\lim }}\,\int\limits_{t}^{A}{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}$
Now, the integration of $\int{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx=-\dfrac{1}{x-2}dx}$as, $\int{\dfrac{1}{{{x}^{2}}}dx=-\dfrac{1}{x}dx}$, we get,
$\Rightarrow \int\limits_{2}^{A}{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}=\underset{t\to 2+}{\mathop{\lim }}\,\left[ -\dfrac{1}{x-2} \right]_{t}^{A}$
Hence, at the end,
$\Rightarrow \int\limits_{2}^{A}{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}=\underset{t\to 2+}{\mathop{\lim }}\,\left( \dfrac{1}{2-A}+\dfrac{1}{t-2} \right)=+\infty $
And finally, to start with,
$\int\limits_{A}^{\infty }{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}=\underset{t\to \infty }{\mathop{\lim }}\,\int\limits_{A}^{t}{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}$
Now, the integration of $\int{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx=-\dfrac{1}{x-2}dx}$as, $\int{\dfrac{1}{{{x}^{2}}}dx=-\dfrac{1}{x}dx}$, we get,
$\Rightarrow \int\limits_{A}^{\infty }{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}=\underset{t\to \infty }{\mathop{\lim }}\,\left[ -\dfrac{1}{x-2} \right]_{A}^{t}$
Hence, at the end,
$\Rightarrow \int\limits_{A}^{\infty }{\dfrac{1}{{{\left( x-2 \right)}^{2}}}dx}=\underset{t\to \infty }{\mathop{\lim }}\,\left( \dfrac{1}{2-t}+\dfrac{1}{A-2} \right)=\dfrac{1}{A-2}$
Thus the integral is clearly not convergent.

Hence the solution is, (a) Splitting the integration.

Note:
When integrating a function over two intervals where the upper bound of the first is the same as the first, the integrands can be combined. Integrands can also be split into two intervals that hold the same conditions. If the upper and lower bound are the same, the area is 0. If an interval is backwards, the area is the opposite sign.