Question

How do you evaluate the integral $\int{\dfrac{1}{x}dx}$ from 0 to 1 if it converges?

Answer 1

In this question, we need to evaluate the integral $\int{\dfrac{1}{x}dx}$ from 0 to 1. As we know $\dfrac{1}{x}$ approaches to infinity if x is equal to 0. So we will use improper integral here. We will evaluate $\underset{m\to 0}{\mathop{\lim }}\,\int\limits_{m}^{1}{\dfrac{1}{x}dx}$. We will use the formula of integration that $\int{\dfrac{1}{x}dx}=\ln \left( x \right)$ where ln is natural log. At last we will evaluate the limit and conclude our answer.

Complete step by step answer:
Here we are given the integral as $\int{\dfrac{1}{x}dx}$ from 0 to 1 and we need to evaluate it.
Our integral will look like this $\int\limits_{0}^{1}{\dfrac{1}{x}dx}$.
But we know that if x = 0 then $\dfrac{1}{x}$ will approach infinity. So this integral is improper. Thus we need to take the limit to the lower limit. Let us take $\underset{m\to 0}{\mathop{\lim }}\,\int\limits_{m}^{1}{\dfrac{1}{x}dx}$ which is improper integral and give solution for $\int\limits_{0}^{1}{\dfrac{1}{x}dx}$.
Evaluating $\underset{m\to 0}{\mathop{\lim }}\,\int\limits_{m}^{1}{\dfrac{1}{x}dx}$.
Let us solve the integration first.
We know that $\int{\dfrac{1}{x}dx}=\ln \left( x \right)$ where ln is the natural logarithm. In the definite integral we take the solution f(upper limit) - f(lower limit). So $\underset{m\to 0}{\mathop{\lim }}\,\int\limits_{m}^{1}{\dfrac{1}{x}dx}$ becomes,
$\underset{m\to 0}{\mathop{\lim }}\,\int\limits_{m}^{1}{\dfrac{1}{x}dx}=\underset{m\to 0}{\mathop{\lim }}\,\left[ \ln \left( x \right) \right]_{m}^{1}\Rightarrow \underset{m\to 0}{\mathop{\lim }}\,\left[ \ln \left( 1 \right)-\ln \left( m \right) \right]$.
Now we have that natural log 1 is equal to 0. So we have ln1 = 0 we get,
$\underset{m\to 0}{\mathop{\lim }}\,\int\limits_{m}^{1}{\dfrac{1}{x}dx}=\underset{m\to 0}{\mathop{\lim }}\,\left[ 0-\ln \left( m \right) \right]\Rightarrow -\underset{m\to 0}{\mathop{\lim }}\,\ln \left( m \right)$.
We know ln(0) approaches to infinity therefore we have $\underset{m\to 0}{\mathop{\lim }}\,\int\limits_{m}^{1}{\dfrac{1}{x}dx}=\infty$.
Hence, $\int{\dfrac{1}{x}dx}$ from 0 to 1 is equal to infinity.
Hence the integral $\int{\dfrac{1}{x}dx}$ between 0 to 1 diverges infinity.

Note:
Students should note that if the integral gives a finite answer then we can say that the integral converges but if it gives infinite then the integral diverges. Taking limits for improper integral is must. Take care of signs as well while evaluating the integral between limits. Note that $\int{\dfrac{1}{x}dx}$ gives natural log only.