Question

How do you determine if the following integral is convergent or divergent using the comparison theorem? $\int\limits_{0}^{\infty }{\dfrac{\arctan \left( x \right)}{2+{{e}^{x}}}dx}$
(a) Comparing with proper terms
(b) Guessing the solution
(c) Integrate with the limit to find a value
(d) Get a random solution

Answer 1

Hint: We are given an integral and we are to check its convergent by the comparison test. We will try to take care of the denominator and numerator differently. So, for the numerator we can compare it with $\dfrac{\pi }{2}$and for the denominator it can be compared with the term $\dfrac{1}{{{e}^{x}}}$. Thus by comparing and finding the integration, we will get our desired answer.

Complete step by step answer:
To start with, we have to get rid of the arctan. This can be done by checking that whenever $x\to \infty $ , we get, $\arctan \left( x \right)\to \dfrac{\pi }{2}$ .
This means that $\arctan \left( x \right)$on $\left[ \left. 0,\infty \right) \right.$ is less than or equal to $\dfrac{\pi }{2}$,
Then, we can write,
$\int\limits_{0}^{\infty }{\dfrac{\arctan \left( x \right)}{2+{{e}^{x}}}dx}\le \dfrac{\pi }{2}\int\limits_{0}^{\infty }{\dfrac{1}{2+{{e}^{x}}}dx}$
Again, If we see the term, it can be a bit tough to get the integrated value of the problem.
Now, we will proceed with the $2+{{e}^{x}}$ part.
To start with, clearly,
$2+{{e}^{x}}>{{e}^{x}}$
Which implies
$\dfrac{1}{2+{{e}^{x}}}<\dfrac{1}{{{e}^{x}}}$
So, now it can be said that,
$\dfrac{\pi }{2}\int\limits_{0}^{\infty }{\dfrac{1}{2+{{e}^{x}}}dx}\le \dfrac{\pi }{2}\int\limits_{0}^{\infty }{\dfrac{1}{{{e}^{x}}}dx}$
Hence, we have,
 $\int\limits_{0}^{\infty }{\dfrac{\arctan \left( x \right)}{2+{{e}^{x}}}dx}\le \dfrac{\pi }{2}\int\limits_{0}^{\infty }{\dfrac{1}{{{e}^{x}}}dx}$
And the part on the right hand side is also easily integrable.
So, we will now try to integrate the term $\dfrac{\pi }{2}\int\limits_{0}^{\infty }{\dfrac{1}{{{e}^{x}}}dx}$.
$\dfrac{\pi }{2}\int\limits_{0}^{\infty }{\dfrac{1}{{{e}^{x}}}dx}=\underset{t\to \infty }{\mathop{\lim }}\,\dfrac{\pi }{2}\int\limits_{0}^{t}{{{e}^{-x}}dx}$
Calculating the integration,
$\Rightarrow \underset{t\to \infty }{\mathop{\lim }}\,\dfrac{\pi }{2}\left[ -{{e}^{-x}} \right]_{0}^{t}$
Putting the values,
$\Rightarrow \underset{t\to \infty }{\mathop{\lim }}\,\dfrac{\pi }{2}\left[ -{{e}^{-t}}-\left( -{{e}^{-0}} \right) \right]$
Simplifying,
$\Rightarrow \underset{t\to \infty }{\mathop{\lim }}\,\dfrac{\pi }{2}\left[ -{{e}^{-t}}+1 \right]$
And now, as we know, $\underset{t\to \infty }{\mathop{\lim }}\,\left( {{e}^{-t}} \right)=0$ ,
Then by putting the values we get,
 $\Rightarrow \dfrac{\pi }{2}\left[ 0+1 \right]$
More simplifying,
$\Rightarrow \dfrac{\pi }{2}$
So, at the end, with the help of the comparison test, we can conclude that,
Our given term, $\int\limits_{0}^{\infty }{\dfrac{\arctan \left( x \right)}{2+{{e}^{x}}}dx}\le \dfrac{\pi }{2}$.
As it is converging to the given point $\dfrac{\pi }{2}$, we can say that the integral is convergent

Hence the solution is, (a) Comparing with proper terms .

Note:
In the given problem we have used the comparison test to get to know if the given integral is convergent or not. If there is a natural candidate g that is larger than f, it will be useful only if its integral converges; we would try to find such g if we suspected that the integral of f converges. The Comparison test essentially says that the same is also true for non-negative functions and improper integrals.