Question

The value of the integral $\int\limits_0^1 {\dfrac{{{x^b} - 1}}{{\log x}}dx} \left( {b > 0} \right)$ is

Answer 1

Hint: We will first express the integral as $I\left( b \right)$, then use the method of differentiation under the integral sign to simplify the given integral. That is, take the differentiation of the expression with respect to $b$. Then, we will get a simplified expression of $I'\left( b \right)$. Next, integrate the function with respect to $x$.

Complete step-by-step answer:
We have to find the value of the integral $\int\limits_0^1 {\dfrac{{{x^b} - 1}}{{\log x}}dx} $, where $\left( {b > 0} \right)$
Let this integral be denoted by $I\left( b \right)$
That is,
$I\left( b \right) = \int\limits_0^1 {\dfrac{{{x^b} - 1}}{{\log x}}dx} $ (1)
Here, we will follow the method of differentiation under the integral sign to simplify the given integral.
Let us take the derivative of $I\left( b \right)$ with respect to $b$
That is, $I'\left( b \right) = \dfrac{d}{{db}}\int\limits_0^1 {\dfrac{{{x^b} - 1}}{{\log x}}dx} $
According to the Leibniz Rule, $\dfrac{d}{{dx}}\int\limits_a^b {f\left( {x,y} \right)dx} = \int\limits_a^b {{f_x}\left( {x,y} \right)dx} $
That is, the derivative of an integral function of two variables is equal to the integral of partial derivatives of that function.
Then, $I'\left( b \right) = \dfrac{d}{{db}}\int\limits_0^1 {\dfrac{{{x^b} - 1}}{{\log x}}dx} = \int\limits_0^1 {\dfrac{\delta }{{\delta b}}\dfrac{{{x^b} - 1}}{{\log x}}dx} $
Which then simplifies to $I'\left( b \right) = \int\limits_0^1 {\dfrac{{\log \left( x \right){x^b}}}{{\log x}}dx} $ because if \[y = {a^x}\] then \[\dfrac{{dy}}{{dx}} = b\left( {\ln x} \right)\]
Therefore, we get the differentiation as,
$I'\left( b \right) = \int\limits_0^1 {{x^b}dx} $
We can simplify the value of $I'\left( b \right)$ by integrating the function ${x^b}dx$ with respect to $x$
$I'\left( b \right) = \left[ {\dfrac{{{x^{b + 1}}}}{{b + 1}}} \right]_{x = 0}^{x = 1}$
On putting the limits we’ll, get
$I'\left( b \right) = \left[ {\dfrac{{{1^{b + 1}} - {0^{b + 1}}}}{{b + 1}}} \right] = \dfrac{1}{{b + 1}}$
We will now integrate the above expression to find the value of $I\left( b \right)$
$I\left( b \right) = \int {\dfrac{1}{{b + 1}}db} = \log \left( {b + 1} \right) + c$, where $c$ is the constant.
$I\left( b \right) = \log \left( {b + 1} \right) + c$ (2)
We have to find the value of the $c$
We have the function $I\left( b \right) = \int\limits_0^1 {\dfrac{{{x^b} - 1}}{{\log x}}dx} $
Put $b = 0$ in the above equation to find the value of $I\left( b \right)$
$
  I\left( 0 \right) = \int\limits_0^1 {\dfrac{{{x^0} - 1}}{{\log x}}dx} \\
   = 0 \\
 $
Hence, $b = 0$, we get $I\left( b \right) = 0$
Therefore, on substituting the value of $b = 0$ and $I\left( b \right) = 0$ in equation (2)
$
  0 = \log \left( {0 + 1} \right) + c \\
  0 = 0 + c \\
  c = 0 \\
$
Thus, $I\left( b \right) = \log \left( {b + 1} \right)$.

Note: We use differentiation under integral sign to find certain integrals. It allows us to interchange the order of integration and differentiation. To use this method, the function $f\left( {x,t} \right)$ should be continuous and a partial derivative should exist.