Question

How do you evaluate $\int{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}$ from 0 to infinity ?

Answer 1

We can solve the given integral by first using the derivative of substitution method, where you can substitute x with tan(y) . Then you can use integration by parts to solve the integral and get the final answer.

Complete step by step solution:
According to the problem, we are asked to find the integral 0 to infinity of the equation $\int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}$ that is

$\int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}$--- ( 1 )

Therefore, by using the substitution, where we substitute x with tan(y), we get
$x=\tan y$ $\Rightarrow \arctan x=y$ ---- (2)

$\therefore dx={{\sec }^{2}}y$---- (3)
Limits also change when we substitute $x=\tan y$. For lower limit, if , $x=0$,we get $y=0$. For upper limit if $x=\infty$, we get $y=\dfrac{\pi }{2}$

Using the equation 2 and equation 3, if we substitute them in equation 1, we get:
$\begin{aligned} & \Rightarrow \int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{y.\tan y.{{\sec }^{2}}y}{{{\left( 1+{{\tan }^{2}}y \right)}^{2}}}}dy \\\ & \\\ \end{aligned}$

Here, we can now substitute $1+{{\tan }^{2}}x={{\sec }^{2}}$ in the above equation.

$\begin{aligned} & \Rightarrow \int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{y.\tan y.{{\sec }^{2}}y}{{{\sec }^{4}}y}}dy \\\ & \\\ \end{aligned}$

Now, we can substitute $\tan x=\dfrac{\sin x}{\cos x}$ in the above equation.

$\Rightarrow \int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{y.\dfrac{\sin y}{\cos y}}{{{\sec }^{2}}y}}dy$

$\Rightarrow \int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}=\int\limits_{0}^{\dfrac{\pi }{2}}{y.\sin y.\cos y}dy$

But since, $\sin 2a=2\sin a\cos a$, we get:

$\Rightarrow \int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}=\dfrac{1}{2}\int\limits_{0}^{\dfrac{\pi }{2}}{y.\sin 2y}dy$

Now we apply the integration by parts to solve equation 4. Therefore, we get,
\Rightarrow \int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}=\dfrac{1}{2}\left\\{ \left[ y.\left( \dfrac{-\cos 2y}{2} \right) \right]_{0}^{\dfrac{\pi }{2}}+\int\limits_{0}^{\dfrac{\pi }{2}}{1.\dfrac{\cos 2y}{2}}dy \right\\}

Since the integral of sinx is -cosx and the derivative of x is 1.
\Rightarrow \int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}=\dfrac{1}{4}\left\\{ \left[ \dfrac{-\pi .1}{2} \right]+\left[ \dfrac{\sin 2t}{2} \right]_{0}^{\dfrac{\pi }{2}} \right\\}

$\Rightarrow \int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}=\dfrac{\pi }{8}+\dfrac{1}{4}\left[ 0 \right]$

$\Rightarrow \int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}=\dfrac{\pi }{8}$ ---- Final answer

So, we have found the derivative of the given equation $\int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}$ as $\int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}=\dfrac{\pi }{8}$.

Therefore, the solution of the given equation $\int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}$ is $\int\limits_{0}^{\infty }{\left( x\cdot \arctan x \right)\dfrac{dx}{{{\left( 1+{{x}^{2}} \right)}^{2}}}}=\dfrac{\pi }{8}$.

Note: In this problem we should be careful while using substitutions like x = tant. You should be careful to change the dx and also the upper and lower limits. Also, remember the formulas for integration by parts and smaller integrals.