Question

It is given that if $l\left( m,n \right)=\int\limits_{0}^{1}{{{t}^{m}}{{\left( 1+t \right)}^{n}}dt}$ , then the expression for $l\left( m,n \right)$ in terms of $l\left( m+1,n-1 \right)$ is:
(a) $\dfrac{{{2}^{n}}}{m+1}-\dfrac{n}{m+1}.l\left( m+1,n-1 \right)$
(b) $\dfrac{n}{m+1}.l\left( m+1,n-1 \right)$
(c) $\dfrac{2n}{m+1}+\dfrac{n}{m+1}l.\left( m+1,n-1 \right)$
(d) $\dfrac{m}{n+1}.l\left( m+1,n-1 \right)$

Answer 1

Hint: First of all write the integration of $l\left( m+1,n-1 \right)$ by substituting m as m + 1 and n as n – 1 in $l\left( m,n \right)$ . Then integrate $l\left( m,n \right)=\int\limits_{0}^{1}{{{t}^{m}}{{\left( 1+t \right)}^{n}}dt}$ using integration by parts in which take the first term as ${{\left( 1+t \right)}^{n}}$ and the second term as ${{t}^{m}}$ . After integration you will find that one of the expression in the integration is in the form of $l\left( m+1,n-1 \right)$ and then rearrangement can give you $l\left( m,n \right)$ in terms of $l\left( m+1,n-1 \right)$ .

Complete step-by-step answer:
It is given that:
$l\left( m,n \right)=\int\limits_{0}^{1}{{{t}^{m}}{{\left( 1+t \right)}^{n}}dt}$ ……… Eq. (1)
Now, we are going to write the value of $l\left( m+1,n-1 \right)$ by substituting m as m + 1 and n as n – 1 in the above integral.
$l\left( m+1,n-1 \right)=\int\limits_{0}^{1}{{{t}^{m+1}}{{\left( 1+t \right)}^{n-1}}dt}$ ………… Eq. (2)
We are going to integrate the equation (1) using integration by parts.
$l\left( m,n \right)=\int\limits_{0}^{1}{{{t}^{m}}{{\left( 1+t \right)}^{n}}dt}$
The integration by parts is done when two functions are given in product form. In the below, we have shown integration by parts for the function “u” and “v” as follows:
$\int{u\left( t \right).v\left( t \right)dt}=u\left( t \right)\int{v\left( t \right)dt}-\int{u'\left( t \right)}\left( \int{v\left( t \right)dt} \right)dt$
In the above integration we have taken the first term as u(t) and second term as v(t).
Now, using the above formula for the integration by parts we are going to integrate the following integral:
$l\left( m,n \right)=\int\limits_{0}^{1}{{{t}^{m}}{{\left( 1+t \right)}^{n}}dt}$
While applying integration by parts on the above integral we have taken the first term as ${{\left( 1+t \right)}^{n}}$ and the second term as ${{t}^{m}}$ .

& l\left( m,n \right)=\int\limits_{0}^{1}{{{\left( 1+t \right)}^{n}}{{t}^{m}}dt} \\\ & \Rightarrow l\left( m,n \right)={{\left( 1+t \right)}^{n}}\int\limits_{0}^{1}{{{\left( t \right)}^{m}}dt}-\int\limits_{0}^{1}{\left( {{\left( 1+t \right)}^{n}} \right)}'\left( \int{{{\left( t \right)}^{m}}dt} \right)dt.........Eq.(3) \\\ \end{aligned}$$ The integration of $ {{t}^{m}} $ with respect to “t” is shown below: $ \int{{{t}^{m}}dt=\dfrac{{{t}^{m+1}}}{m+1}} $ The derivative of $ {{\left( 1+t \right)}^{n}} $ with respect to t we get, $ \dfrac{d{{\left( 1+t \right)}^{n}}}{dt}=n{{\left( 1+t \right)}^{n-1}} $ Substituting the above differentiation and integration in eq. (3) we get, $$l\left( m,n \right)=\left[ {{\left( 1+t \right)}^{n}}\dfrac{{{\left( t \right)}^{m+1}}}{m+1} \right]_{0}^{1}-\int\limits_{0}^{1}{n{{\left( 1+t \right)}^{n-1}}\left[ \dfrac{{{\left( t \right)}^{m+1}}}{m+1} \right]}dt$$ In the above equation putting upper and the lower limit in the first term of the right hand side of the equation and taking n and $ \dfrac{1}{m+1} $ as common in the integral we get, $$l\left( m,n \right)=\left[ \dfrac{{{2}^{n}}}{m+1}-0 \right]-\dfrac{n}{m+1}\int\limits_{0}^{1}{{{t}^{m+1}}\left[ {{\left( 1+t \right)}^{n-1}} \right]}dt$$ Using eq. (2) in the above equation we get, $$l\left( m,n \right)=\left[ \dfrac{{{2}^{n}}}{m+1} \right]-\dfrac{n}{m+1}.l\left( m+1,n-1 \right)$$ From the above calculation, we have written $ l\left( m,n \right) $ in terms of $ l\left( m+1,n-1 \right) $ as: $$\left[ \dfrac{{{2}^{n}}}{m+1} \right]-\dfrac{n}{m+1}.l\left( m+1,n-1 \right)$$ Hence, the correct option is (a). Note: While solving the integration of $ l\left( m,n \right) $ using integration by parts you might have thought that why we have taken the first term as $ {{\left( 1+t \right)}^{n}} $ and the second term as $ {{t}^{m}} $ rather we might have taken the first term as $ {{t}^{m}} $ and the second term as $ {{\left( 1+t \right)}^{n}} $ . The reason is that if you take the first term as $ {{t}^{m}} $ and the second term as $ {{\left( 1+t \right)}^{n}} $ then you would not get one of the terms as $ \int\limits_{0}^{1}{{{t}^{m+1}}{{\left( 1+t \right)}^{n-1}}dt} $ in the integration of $ l\left( m,n \right) $ instead you will get $ \int\limits_{0}^{1}{{{t}^{m-1}}{{\left( 1+t \right)}^{n+1}}dt} $ which we do not require because it is not in the form of $ l\left( m+1,n-1 \right) $ .