Question

If we have an integral as $I\left( m,n \right)=\int\limits_{0}^{1}{{{t}^{m}}{{\left( 1+t \right)}^{n}}dt}$, then expression for $I\left( m,n \right)$ in terms of $I\left( m+1,n-1 \right)$ is:
A). $\dfrac{{{2}^{n}}}{m+1}-\dfrac{n}{m+1}I\left( m+1,n-1 \right)$
B). $\dfrac{n}{m+1}I\left( m+1,n-1 \right)$
C). $\dfrac{{{2}^{n}}}{m+1}+\dfrac{n}{m+1}I\left( m+1,n-1 \right)$
D). $\dfrac{m}{m+1}I\left( m+1,n-1 \right)$

Answer 1

Hint: At first find out the value of $I\left( m+1,n-1 \right)$ from the given expression. Then try to apply integration by parts in such a way so that we can get back $I\left( m,n \right)$. After this keep $I\left( m+1,n-1 \right)$ on one side and take the rest of the terms on the other side.

Complete step-by-step solution
 The given expression is,
 $I\left( m,n \right)=\int\limits_{0}^{1}{{{t}^{m}}{{\left( 1+t \right)}^{n}}dt}............(1)$
We have to express $I\left( m,n \right)$ in terms of $I\left( m+1,n-1 \right)$.
If we replace m by m+1 and n by n+1 in equation (1) then we will get,
$I\left( m+1,n-1 \right)=\int\limits_{0}^{1}{{{t}^{m+1}}{{\left( 1+t \right)}^{n-1}}dt}$
Let us take $u\left( t \right)={{t}^{m+1}},v\left( t \right)=\dfrac{{{\left( 1+t \right)}^{n}}}{n}$.
If we differentiate v(t) with respect to t, we will get,
$\dfrac{dv}{dt}=\dfrac{n{{\left( 1+t \right)}^{n-1}}}{n}={{\left( 1+t \right)}^{n-1}}$
Therefore,
$I\left( m+1,n-1 \right)=\int\limits_{0}^{1}{u\left( t \right)\dfrac{dv}{dt}dt}$
Now we can apply integration by parts.
$I\left( m+1,n-1 \right)=\left[ u\left( t \right)v\left( t \right) \right]_{t=0}^{t=1}-\int\limits_{0}^{1}{\left( \dfrac{du}{dt} \right)v\left( t \right)dt..........(2)}$
 Now substitute the values of u(t) and v(t) in (2)
$\Rightarrow I\left( m+1,n-1 \right)=\left[ {{t}^{m+1}}\dfrac{{{\left( 1+t \right)}^{n}}}{n} \right]_{t=0}^{t=1}-\int\limits_{0}^{1}{\dfrac{d}{dt}\left( {{t}^{m+1}} \right)\dfrac{{{\left( 1+t \right)}^{n}}}{n}}dt$
Differentiate ${{t}^{m+1}}$ with respect to t,
$\Rightarrow I\left( m+1,n-1 \right)=\left[ {{t}^{m+1}}\dfrac{{{\left( 1+t \right)}^{n}}}{n} \right]_{t=0}^{t=1}-\int\limits_{0}^{1}{\left( m+1 \right){{t}^{m+1-1}}\dfrac{{{\left( 1+t \right)}^{n}}}{n}}dt$
$\Rightarrow I\left( m+1,n-1 \right)=\left( {{\left( 1 \right)}^{m+1}}\dfrac{{{\left( 1+1 \right)}^{n}}}{n}-{{\left( 0 \right)}^{m+1}}\dfrac{{{\left( 1+0 \right)}^{n}}}{n} \right)-\dfrac{m+1}{n}\int\limits_{0}^{1}{{{t}^{m}}{{\left( 1+t \right)}^{n}}}dt$
Here we can substitute the value of equation (1),
$\Rightarrow I\left( m+1,n-1 \right)=\left( \dfrac{{{2}^{n}}}{n}-0 \right)-\dfrac{m+1}{n}I\left( m,n \right)$
Now we have to find out the value of $I\left( m,n \right)$. Therefore we will take $I\left( m,n \right)$ on the left-hand side and all the other terms on the right-hand side.
$\Rightarrow I\left( m+1,n-1 \right)=\dfrac{{{2}^{n}}}{n}-\dfrac{m+1}{n}I\left( m,n \right)$
$\Rightarrow I\left( m+1,n-1 \right)-\dfrac{{{2}^{n}}}{n}=-\dfrac{m+1}{n}I\left( m,n \right)$ , taking $\dfrac{{{2}^{n}}}{n}$ from right hand side to left hand side.
$\Rightarrow \dfrac{{{2}^{n}}}{n}-I\left( m+1,n-1 \right)=\dfrac{m+1}{n}I\left( m,n \right)$, multiplying both the sides of the equation by ‘-‘.
$\Rightarrow I\left( m,n \right)=\dfrac{{{2}^{n}}}{n}\times \dfrac{n}{m+1}-\dfrac{n}{m+1}I\left( m+1,n-1 \right)$ , multiplying both the sides by $\dfrac{n}{m+1}$.
$\Rightarrow I\left( m,n \right)=\dfrac{{{2}^{n}}}{m+1}-\dfrac{n}{m+1}I\left( m+1,n-1 \right)$
Therefore option (a) is correct.

Note: Alternatively we can find out the answer by cross checking the options. Like in option (a) if we put the value of $I\left( m+1,n-1 \right)$ we will get $I\left( m,n \right)$ back. That means option (a) is correct. Be careful while you are applying integration by parts. You have to choose u(t) and v(t) in such a way so that you can get back $I\left( m,n \right)$.