Question

If $I_{m,n}={\int }_0^1{x}^m{\left(\log x\right)}^{n}dx$, then $I_{m,n}$it is equal to

• A. $\frac{n}{n+1}{I}_{m,n-1}$
• B. $-\frac{n}{m+1}{I}_{m,n-1}$
• C. $-\frac{1}{m+1}{I}_{m,n-1}$
• D. $\frac{m}{n+1}{I}_{m,n-1}$

Answer 1

The correct option is C $-\frac{1}{m+1}{I}_{m,n-1}$

$I_{m,n}={\int }_0^1{x}^m{\left(\log x\right)}^{n}dx$
$={[{\left(\log x\right)}^n\cdot \frac{x^{m+1}}{m+1}]}_0^1-{\int }_0^{1}n{\left(\log x\right)}^{n-1}\cdot \frac{1}{x}\cdot \frac{x^{m+1}}{(m+1)}dx$
$=0-\frac{n}{(m+1)}{\int }_0^1{x}^m{\left(\log x\right)}^{n-1}dx$
$=\frac{-n}{m+1}\cdot I_{m,n-1}$