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In this question product of terms was given so logarithm might help but it does not works.I used the partial fraction but there will be 2021 distinct fractions to be integrated so how to proceed further here? $$\int\frac{1}{\prod_{k=1}^{2021}(x+k)}dx$$ does it involves partial differention?

PHYSION
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  • There will indeed be 2021 fractions if you want to use partial fractions. But is it that hard to compute? Try it! – Trebor Aug 28 '21 at 10:28

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I have solved a few of such integrals on Quora, and I would say the method I am going to present is probably the simplest and also, the only possible method.

First of all we consider the partial fractions of the integrand

$$ \frac{1}{\prod_{k=1}^{2021}(x+k) } = \sum_{k=1}^{2021}\frac{a_k}{x+k} $$ From that, we get

$$ 1 = a_1 (x+2)(x+3)\cdots (x+2021) + a_2 (x+1)(x+3)\cdots (x+2021) \cdots + a_{2021}(x+1)(x+2)\cdots (x+2020) $$

Setting $x=-1,x=-2$ and so on till $x=-2021$, we can deduce a general formula for $k$-th term

$$a_k= \frac{1}{\prod_{1 \le j\le 2021, \\ j \ne k} (j-k) } $$

Using this and the partial fraction expansion, we get

$$\int \frac{\mathrm dx}{\prod_{k=1}^{2021}(x+k)} = \sum_{k=1}^{2021} \frac{\ln(x+k)}{\prod_{1 \le j\le 2021, \\ j \ne k}(j-k) } +C $$

Laxmi Narayan Bhandari
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