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The value of $\int\limits_{0}^{1}(\prod\limits_{r=1}^{n}(x+r))(\sum\limits_{k=1}^{n}\frac{1}{x+k})dx$ is equal to

$(A)n\hspace{1cm}(B)n!\hspace{1cm}(C)(n+1)!\hspace{1cm}(D)n.n!$

I tried:$\int\limits_{0}^{1}(\prod\limits_{r=1}^{n}(x+r))(\sum\limits_{k=1}^{n}\frac{1}{x+k})dx$=$\int\limits_{0}^{1}(x+1)(x+2)...(x+n)(\frac{1}{x+1}+\frac{1}{x+2}+......+\frac{1}{x+n})dx$=$\int\limits_{0}^{1}(x+2)(x+3)...(x+n)+(x+1)(x+3)...(x+n)dx$

I cannot solve it further.Is my approach wrong,I am stuck.What is the right way to solve,Please help...

diya
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3 Answers3

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Consider $$A=\prod_{k=1}^n(x+k)$$ Take logarithms of both sides $$\log(A)=\sum_{k=1}^n \log(x+k)$$ Compute the derivative $$\frac {A'}A=\sum_{k=1}^n \frac 1{x+k}$$

I am sure that you can take from here.

Claude Leibovici
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Differentiate $\prod_{k=1}^n(x+k)$ with respect to $x$, and you get the integrand. So the answer is $(n+1)!-n!=n\cdot n!$.

John Bentin
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First let us make a common denominator for the rationals under the sum sign $$ \left(\prod\limits_{r=1}^n (x+r)\right) \left(\sum\limits_{k=1}^n \frac{1}{x+k} \right)\\ = \left(\prod\limits_{r=1}^n (x+r)\right) \left( \sum\limits_{k=1}^n \frac{\prod\limits_{r=1,r\neq k}^n (x+r)} {\prod\limits_{r=1}^n (x+r)} \right) = \left( \sum\limits_{k=1}^n \prod\limits_{r=1,r\neq k}^n (x+r) \right) = \frac{d}{dx} \left( \prod\limits_{r=1}^n (x+r) \right), $$

where we used the product rule for the last equal sign. Now can you complete your exercise on your own?

andre
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