Question

If the integral can be defined as \[{{\text{I}}_{{\text{m,n}}}} = \int_0^1 {{{\text{x}}^{\text{m}}}{{\left( {{\text{log x}}} \right)}^{\text{n}}}{\text{dx, }}} \]then on simplification \[{{\text{I}}_{{\text{m,n}}}}\] is equal to/ can also be expressed as which of the four given options?
$
  {\text{A}}{\text{. }}\dfrac{{\text{n}}}{{{\text{n + 1}}}}{{\text{I}}_{{\text{m,n - 1}}}} \\
  {\text{B}}{\text{. - }}\dfrac{{\text{n}}}{{{\text{m + 1}}}}{{\text{I}}_{{\text{m,n - 1}}}} \\
  {\text{C}}{\text{. - }}\dfrac{1}{{{\text{m + 1}}}}{{\text{I}}_{{\text{m,n - 1}}}} \\
  {\text{D}}{\text{. }}\dfrac{{\text{m}}}{{{\text{n + 1}}}}{{\text{I}}_{{\text{m,n - 1}}}} \\
$

Answer 1

Hint – Check out properties of definite integrals. Apply integration by parts rule to simplify the expression then you will get the solution.

Complete step-by-step answer:
Given Data, \[{{\text{I}}_{{\text{m,n}}}} = \int_0^1 {{{\text{x}}^{\text{m}}}{{\left( {{\text{log x}}} \right)}^{\text{n}}}{\text{dx, }}} \]
If the given function consists of two parts of the function that are multiplied by each other, then for easily computing it we use the integration by parts rule.
Let’s say there is a function a.b, then the integral of a.b i.e. $\int{a.b} = a\int {b} - \int ({{a’} \int {b}}) $.
Comparing \[{{\text{I}}_{{\text{m,n}}}} = \int_0^1 {{{\text{x}}^{\text{m}}}{{\left( {{\text{log x}}} \right)}^{\text{n}}}{\text{dx, }}} \]and $\int{a.b} = a\int {b} - \int ({{a’} \int {b}}) $, let us say
a = \[{\left( {{\text{log x}}} \right)^{\text{n}}}\]and b = \[{{\text{x}}^{\text{m}}}\]
⟹$a’= \dfrac{d}{dx}({log x})^{n} = n({log x})^{n - 1}\dfrac{1}{x}( \dfrac{d}{dx}(\because {log x} = \dfrac{1}{x} )$
Apply integration by parts rule to the given equation, we get
\[{{\text{I}}_{{\text{m,n}}}} = {\left( {{\text{log x}}} \right)^{\text{n}}}\int_0^1 {{{\text{x}}^{\text{m}}}} - \int_0^1 {\dfrac{{\text{d}}}{{{\text{dx}}}}{{\left( {{\text{logx}}} \right)}^{\text{n}}}\int_0^1 {{{\text{x}}^{\text{m}}}} {\text{dx}}} \]
⟹\[{{\text{I}}_{{\text{m,n}}}} = \left[ {{{\left( {{\text{log x}}} \right)}^{\text{n}}}\dfrac{{{{\text{x}}^{{\text{m + 1}}}}}}{{{\text{m + 1}}}}} \right]_0^1{\text{ }} - \int_0^1 {{\text{n}}{{\left( {{\text{logx}}} \right)}^{{\text{n - 1}}}}\dfrac{1}{{\text{x}}}\dfrac{{{{\text{x}}^{{\text{m + 1}}}}}}{{{\text{m + 1}}}}{\text{dx}}} {\text{ }}\]
\[{\text{ - - - }}\left( {{\text{Using the formulae }}\int_0^1 {{{\text{a}}^{\text{b}}}} = \dfrac{{{{\text{a}}^{{\text{b + 1}}}}}}{{{\text{b + 1}}}}{\text{ and }}\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{logk}}} \right) = \dfrac{{\text{d}}}{{{\text{dx}}}}\left( {{\text{logk}}} \right)\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\text{k}} \right){\text{ respectively}}} \right)\]
After integrating the terms under limits, we have to substitute the upper limit in the result and subtract the lower limit substituted result from it. Then on further simplification we get,
⟹\[{{\text{I}}_{{\text{m,n}}}} = 0 - \dfrac{{\text{n}}}{{{\text{m + 1}}}}\int_0^1 {{{\text{x}}^{\text{m}}}{{\left( {{\text{logx}}} \right)}^{{\text{n - 1}}}}{\text{dx }}} {\text{ }}\]
(The first term becomes zero, as x takes the value 0, the entire term becomes zero when multiplied by x.)
Comparing this term to the given, we get
⟹${{\text{I}}_{{\text{m,n}}}} = $$ - \dfrac{{\text{n}}}{{{\text{m + 1}}}}{{\text{I}}_{{\text{m,n - 1}}}}$
Hence, Option B is the correct answer.

Note – In order to solve these types of problems the key is to make the given function simple by rearranging or applying known identities to determine the answer. For understanding which identity we should use, just go through the options then try to implement identities.