Question

Find the value of $\int\limits_{0}^{1}{\log \left( \dfrac{1-x}{x} \right)}$?

Answer 1

We first use the definite integral formula of $\int\limits_{a}^{b}{f\left( z \right)}=\int\limits_{a}^{b}{f\left( a+b-z \right)}$. We take the replacements of $a=0,b=1,z=x$. We use the logarithm formula of $\log \left( \dfrac{a}{b} \right)=-\log \left( \dfrac{b}{a} \right)$. We add the integrals to find the solution of the integral.

Complete step by step answer:
We are going to use the concept of definite integral where we can use the formula of equality
$\int\limits_{a}^{b}{f\left( z \right)}=\int\limits_{a}^{b}{f\left( a+b-z \right)}$.
For our given integral the upper and lower limits are 1 and 0 respectively.
For the formula of $\int\limits_{a}^{b}{f\left( z \right)}=\int\limits_{a}^{b}{f\left( a+b-z \right)}$, we can use $a=0,b=1,z=x$.
We can replace the value of $x$ with $1+0-x=1-x$.
Therefore, we get \int\limits_{0}^{1}{\log \left( \dfrac{1-x}{x} \right)}=\int\limits_{0}^{1}{\log \left\\{ \dfrac{1-\left( 1-x \right)}{\left( 1-x \right)} \right\\}}.
Simplifying the equation, we get \int\limits_{0}^{1}{\log \left( \dfrac{1-x}{x} \right)}=\int\limits_{0}^{1}{\log \left\\{ \dfrac{1-1+x}{1-x} \right\\}}=\int\limits_{0}^{1}{\log \left( \dfrac{x}{1-x} \right)}.
We now use the logarithmic formula of $\log \left( \dfrac{a}{b} \right)=-\log \left( \dfrac{b}{a} \right)$.
We can use the formula to get $\log \left( \dfrac{x}{1-x} \right)=-\log \left( \dfrac{1-x}{x} \right)$.
Let us assume $I=\int\limits_{0}^{1}{\log \left( \dfrac{1-x}{x} \right)}$ which gives $I=\int\limits_{0}^{1}{\log \left( \dfrac{1-x}{x} \right)}=\int\limits_{0}^{1}{\log \left( \dfrac{x}{1-x} \right)}$.
We now these integrals to get I+I=\int\limits_{0}^{1}{\log \left( \dfrac{1-x}{x} \right)}+\int\limits_{0}^{1}{\log \left( \dfrac{x}{1-x} \right)}=\int\limits_{0}^{1}{\left\\{ \log \left( \dfrac{1-x}{x} \right)+\log \left( \dfrac{x}{1-x} \right) \right\\}}.
We already have that $\log \left( \dfrac{x}{1-x} \right)=-\log \left( \dfrac{1-x}{x} \right)$ which makes the integral as
I+I=\int\limits_{0}^{1}{\left\\{ \log \left( \dfrac{1-x}{x} \right)-\log \left( \dfrac{1-x}{x} \right) \right\\}}=0.
So, $2I=0\Rightarrow I=0$.
The integral value of $\int\limits_{0}^{1}{\log \left( \dfrac{1-x}{x} \right)}=0$.

Note: The definite integral is defined to be exactly the limit and summation and that’s the limit changes remain unchanged for the integrations the variable change keeps the area same.