Question

The value of $\int\limits_{0}^{1}{\left( \prod\limits_{r=1}^{n}{(x+r)} \right)\left( \sum\limits_{k=1}^{n}{\dfrac{1}{x+k}} \right)}dx$ is
A)$n$
B)$n!$
C)$\left( n+1 \right)!$
D) $n\text{ }n!$

Answer 1

Hint: Separate the given integral into two parts, evaluate each part separately and then add to get the result. Use the fact that$\dfrac{d}{dx}\left( {}_{n}{{C}^{x+n}} \right)=(x+2)(x+3)...(x+n)+(x+1)(x+3)...(x+n)+...+(x+1)(x+2)...(x+n-1)$

Complete step-by-step answer:
The integral given in the question is $\int\limits_{0}^{1}{\left( \prod\limits_{r=1}^{n}{(x+r)} \right)\left( \sum\limits_{k=1}^{n}{\dfrac{1}{x+k}} \right)}dx$.
We can see that the integral has two parts in it, given by,
$\prod\limits_{r=1}^{n}{(x+r)}$ and $\sum\limits_{k=1}^{n}{\dfrac{1}{x+k}}$
We have to solve these two parts separately and then evaluate the integral.
So, let us first solve the part $\sum\limits_{k=1}^{n}{\dfrac{1}{x+k}}$. Since, it is summation, we can expand it for n number of terms.
Therefore, expanding the above summation we get,
$\sum\limits_{k=1}^{n}{\dfrac{1}{x+k}}=\dfrac{1}{x+1}+\dfrac{1}{x+2}+\dfrac{1}{x+3}+...\dfrac{1}{x+n}$
Now, we can make the denominator common by making every denominator equal to $(x+1)(x+2)(x+3)...(x+n)$. We can do this by multiplying and dividing each term with appropriate terms as shown below,
$\sum\limits_{k=1}^{n}{\dfrac{1}{x+k}}=\dfrac{(x+2)(x+3)...(x+n)}{(x+1)(x+2)...(x+n)}+\dfrac{(x+1)(x+3)...(x+n)}{(x+1)(x+2)...(x+n)}+...+\dfrac{(x+1)(x+2)...(x+n-1)}{(x+1)(x+2)...(x+n)}$
Clubbing the terms, we get the summation as,
$\sum\limits_{k=1}^{n}{\dfrac{1}{x+k}}=\dfrac{(x+2)(x+3)...(x+n)+(x+1)(x+3)...(x+n)+....+(x+1)(x+2)...(x+n-1)}{(x+1)(x+2)(x+3)...(x+n)}$
From the above-obtained result, we can observe that it is of the form given below,
$\dfrac{d}{dx}\left( {}_{n}{{C}^{x+n}} \right)=(x+2)(x+3)...(x+n)+(x+1)(x+3)...(x+n)+...+(x+1)(x+2)...(x+n-1)$
So, we can rewrite the obtained result as,
$\sum\limits_{k=1}^{n}{\dfrac{1}{x+k}}=\dfrac{d}{dx}\left( C_{n}^{x+n} \right)\dfrac{1}{(x+1)(x+2)...(x+n)}$
This is the result of the first part of the integral.
Therefore, the second part can be expanded as,
$\prod\limits_{r=1}^{n}{(x+r)}=(x+1)(x+2)(x+3)...(x+n)$
We have obtained the second part of the integral also.
Therefore, we can substitute the results and the integral becomes,
$\int\limits_{0}^{1}{\left( \prod\limits_{r=1}^{n}{(x+r)} \right)\left( \sum\limits_{k=1}^{n}{\dfrac{1}{x+k}} \right)}dx=\int\limits_{0}^{1}{\dfrac{(x+1)(x+2)...(x+n)}{(x+1)(x+2)...(x+n)}\dfrac{d}{dx}\left( C_{n}^{x+n} \right)}dx$
Cancelling the like terms, we get
$\int\limits_{0}^{1}{\left( \prod\limits_{r=1}^{n}{(x+r)} \right)\left( \sum\limits_{k=1}^{n}{\dfrac{1}{x+k}} \right)}dx=\int\limits_{0}^{1}{\dfrac{d}{dx}\left( C_{n}^{x+n} \right)}dx$
The derivative and integration gets cancelled, so we get
$\int\limits_{0}^{1}{\left( \prod\limits_{r=1}^{n}{(x+r)} \right)\left( \sum\limits_{k=1}^{n}{\dfrac{1}{x+k}} \right)}dx=\left[ C_{n}^{x+n} \right]_{0}^{1}$
Applying the limits, we get
$\int\limits_{0}^{1}{\left( \prod\limits_{r=1}^{n}{(x+r)} \right)\left( \sum\limits_{k=1}^{n}{\dfrac{1}{x+k}} \right)}dx=\left[ C_{n}^{n+1}-C_{n}^{n+0} \right]$
Now we know, $C_{r}^{n}=\dfrac{n!}{r!(n-r)!}$, so the above equation can be written as,

& \int\limits_{0}^{1}{\left( \prod\limits_{r=1}^{n}{(x+r)} \right)\left( \sum\limits_{k=1}^{n}{\dfrac{1}{x+k}} \right)}dx=\dfrac{(n+1)!}{n!(n+1-n)!}-1 \\\ & \Rightarrow \int\limits_{0}^{1}{\left( \prod\limits_{r=1}^{n}{(x+r)} \right)\left( \sum\limits_{k=1}^{n}{\dfrac{1}{x+k}} \right)}dx=\dfrac{(n+1)!}{n!}-1 \\\ & \Rightarrow \int\limits_{0}^{1}{\left( \prod\limits_{r=1}^{n}{(x+r)} \right)\left( \sum\limits_{k=1}^{n}{\dfrac{1}{x+k}} \right)}dx=\dfrac{(n+1)n!}{n!}-1 \\\ \end{aligned}$$ Cancelling the like terms, we get $$\int\limits_{0}^{1}{\left( \prod\limits_{r=1}^{n}{(x+r)} \right)\left( \sum\limits_{k=1}^{n}{\dfrac{1}{x+k}} \right)}dx=n+1-1=n$$ This is our answer, which is option A). Note: It is also very important to make the observation $$\dfrac{d}{dx}\left( {}_{n}{{C}^{x+n}} \right)=(x+2)(x+3)(x+4)...(x+n)+(x+1)(x+3)(x+4)...(x+n)....+(x+1)(x+2)(x+3)...(x+n-1)$$ Another approach is directly solving the integral as it is without separating the summation and production. In this way students can get confused.