Solveeit Logo
Login

Question

Question: How do you evaluate the integral \(\int {\dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}dx} \) from \(e\)...

How do you evaluate the integral ∫1x(ln⁡x)2dx\int {\dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}dx} from ee to ∞\infty ?

Explanation

Solution

In this problem we have given some integral with some limit. And we asked to evaluate the given integral. This problem can be evaluated by using a substitution method and also we are going to work basic terms of integration to evaluate the given integral. More over the given integral is an improper integral.

Complete step-by-step solution:
The integral is ∫1x(ln⁡x)2dx\int {\dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}dx} from ee to ∞\infty .
Let I=∫e∞1x(ln⁡x)2dx−−−−−(1)I = \int\limits_e^\infty {\dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}dx} - - - - - (1)
Take u=ln⁡xu = \ln x and du=dxxdu = \dfrac{{dx}}{x} this can also be written as xdu=dxxdu = dx
Equation (1) can be written as,
I=∫e∞1x(u)2xduI = \int\limits_e^\infty {\dfrac{1}{{x{{\left( u \right)}^2}}}xdu} ,
Now xx in the numerator and denominator get cancelled and if we substitute x=ex = e in u=ln⁡xu = \ln x then we get u=1u = 1
Also if we substitute x=∞x = \infty in u=ln⁡xu = \ln x then we get u=∞u = \infty
⇒I=∫u=1u=∞duu2\Rightarrow I = \int\limits_{u = 1}^{u = \infty } {\dfrac{{du}}{{{u^2}}}}
We know that, 1u2=u−2\dfrac{1}{{{u^2}}} = {u^{ - 2}}
⇒I=∫u=1u=∞u−2du\Rightarrow I = \int\limits_{u = 1}^{u = \infty } {{u^{ - 2}}du}
On integrating we get
⇒I=[−u−1]1∞\Rightarrow I = {\left[ { - {u^{ - 1}}} \right]_1}^\infty
We know that u−1=1u{u^{ - 1}} = \dfrac{1}{u}
⇒I=[−1u]1∞\Rightarrow I = {\left[ { - \dfrac{1}{u}} \right]_1}^\infty
Now applying the limits we get
⇒I=−1∞−(−11)\Rightarrow I = - \dfrac{1}{\infty } - \left( { - \dfrac{1}{1}} \right)
Any number divided by infinity is zero. So the first term in the right hand side of the above equation becomes zero.
⇒I=0−(−11)−−−−−(2)\Rightarrow I = 0 - \left( { - \dfrac{1}{1}} \right) - - - - - (2)
⇒I=1\Rightarrow I = 1

Therefore, ∫e∞1x(ln⁡x)2dx=1\int\limits_e^\infty {\dfrac{1}{{x{{\left( {\ln x} \right)}^2}}}dx} = 1

Note: An indefinite integral is a function that takes the anti -derivative of another function. The indefinite integral is an easier way to symbolize taking the anti – derivative. The indefinite integral is related to the definite integral, but the two are not the same.
One of the simplest ways to define infinity is as a limit of numbers growing without bound. Infinity is then replaced by a limit of an expression using normal numbers. Hence the limit for any is zero and we say any number divided by infinity is zero. And also in equation (2) minus sign is multiplied by minus sign so its result is positive.