Question

How do you show whether the improper integral $\int {\dfrac{1}{{1 + {x^2}}}dx}$ converges or diverges from negative infinity to infinity?

Answer 1

The only way to prove it is by evaluating it. We have to integrate it from negative infinity to positive infinity. Then we have to analyze the final result, if it is coming out to be a definite number then we can say that it converges otherwise it diverges.

Complete answer:
Consider a function f(x) which exhibits a Type I or Type II behaviour on the interval $\left[ {a,b} \right]$ (in other words, the integral $\int\limits_a^b {f\left( x \right)dx}$ is improper). We saw before that this integral is defined as a limit. Therefore, we have two cases:
$\left( 1 \right)$ The limit exists (and is a number), in this case we say that the improper integral is convergent.
$\left( 2 \right)$ The limit does not exist or it is infinite, then we say that the improper integral is divergent.
In the above question, it is given that
$\Rightarrow \int {\dfrac{1}{{1 + {x^2}}}dx}$
Now we will integrate this function,
On integrating the above function, we get
$\Rightarrow \int {\dfrac{1}{{1 + {x^2}}}dx} = {\tan ^{ - 1}}x + C$
We know that ${\lim _{x \to \infty }}{\tan ^{ - 1}}x = \dfrac{\pi }{2}$ and ${\lim _{x \to - \infty }}{\tan ^{ - 1}}x = - \dfrac{\pi }{2}$
We need to split the integral. $0$ is usually easy to work with, so let’s use it.
$\int_{ - \infty }^\infty {\dfrac{1}{{1 + {x^2}}}dx = {{\lim }_{a \to - \infty }}\int\limits_a^0 {\dfrac{1}{{1 + {x^2}}}dx + {{\lim }_{b \to \infty }}\int\limits_0^b {\dfrac{1}{{1 + {x^2}}}dx} } }$
Now, use the above value of integral.
$\Rightarrow {\lim _{a \to - \infty }}\left[ {{{\tan }^{ - 1}}x} \right]_a^0 + {\lim _{b \to \infty }}\left[ {{{\tan }^{ - 1}}x} \right]_0^b$
On putting the limits in function, we get
$\Rightarrow {\lim _{a \to - \infty }}\left[ {{{\tan }^{ - 1}}0 - {{\tan }^{ - 1}}a} \right] + {\lim _{b \to \infty }}\left[ {{{\tan }^{ - 1}}b - {{\tan }^{ - 1}}0} \right]$
We know that ${\tan ^{ - 1}}\infty = \dfrac{\pi }{2}$ and ${\tan ^{ - 1}} - \infty = - \dfrac{\pi }{2}$.
$\Rightarrow \left[ {0 - \left( { - \dfrac{\pi }{2}} \right)} \right] + \left[ {\dfrac{\pi }{2} - 0} \right]$
$\Rightarrow \dfrac{\pi }{2} + \dfrac{\pi }{2}$
$\Rightarrow \pi$
Therefore, the integral converges from negative infinity to infinity.

Note: If the limit exists and is a finite number, we say the improper integral converges. If the limit is $\pm \infty$ or does not exist, we say the improper integral diverges. an improper integral is the limit of a definite integral as an endpoint of the interval(s) of integration approaches either a specified real number or positive or negative infinity; or in some instances as both endpoints approach limits. Such an integral is often written symbolically just like a standard definite integral, in some cases with infinity as a limit of integration.