Question

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

Answer 1

Hint: 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.