Question

If $l(m,n)={\int }_0^1{t}^m(1+t{)}^{n}dt$, then the expression for $l(m,n)$ in terms of $l(m+1,n-1)$ is

• A. $\frac{2^n}{m+1}-\frac{n}{m+1}l(m+1,n-1)$
• B. $\frac{n}{m+1}l(m+1,n-1)$
• C. $\frac{2^n}{m+1}+\frac{n}{m+1}l(m+1,n-1)$
• D. $\frac{m}{m+1}l(m+1,n-1)$

Answer 1

The correct option is A $\frac{2^n}{m+1}-\frac{n}{m+1}l(m+1,n-1)$

We have $I_{m,n}={\int }_0^1{t}^m(1+t{)}^{n}dt$
Intergarting by parts considering $(1+t{)}^n$ as first function, we get
$I_{m,n}={\left[\frac{t^{m+1}}{m+1}\right(1+t{)}^n]}_0^1-\frac{n}{m+1}{\int }_0^1{t}^{m+1}(1+t{)}^{n-1}dt{I}_{m,n}=\frac{2^n}{m+1}-\frac{n}{m+1}{l}_{m+1,n-1}$