Let ( 0 , 1 ) and = log e ( 1 - ) . Let P n ( x ) = x + x 2 2 + x 3 3 + … + x n n , x ( 0 , 1 ) . Then the integral 0 t 50 1 - t d t is equal to [2023]

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Solution

Q.
Let $α \in (0, 1)$ and $β = \log_{e} (1 - α)$ . Let $P_{n} (x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots + \frac{x^{n}}{n}, x \in (0, 1) .$ Then the integral $\int_{0}^{α} \frac{t^{50}}{1 - t} d t$ is equal to [2023]

1 $P_{50} (α) - β$

2 $- (β + P_{50} (α))$

3 $β + P_{50} (α)$

4 $β - P_{50} (α)$

Ans.
(2)

We have, $P_{n} (x) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \dots + \frac{x^{n}}{n}, x \in (0, 1)$ ...(i)

Let $I = \int_{0}^{α} \frac{t^{50}}{1 - t} d t = \int_{0}^{α} \frac{t^{50} + 1 - 1}{1 - t} d t = \int_{0}^{α} (\frac{- (1 - t^{50})}{1 - t} + \frac{1}{1 - t}) d t$

$= \int_{0}^{α} - (1 + t + t^{2} + \dots + t^{49}) d t + \int_{0}^{α} \frac{1}{1 - t} d t$

$= - {[t + \frac{t^{2}}{2} + \frac{t^{3}}{3} + \dots + \frac{t^{50}}{50}]}_{0}^{α} - {[\log (1 - t)]}_{0}^{α}$

$= - \log (1 - α) - [α + \frac{α^{2}}{2} + \frac{α^{3}}{3} + \dots + \frac{α^{50}}{50}]$

$= - β - P_{50} (α)$ $[∵ β = \log_{e} (1 - α) and from (i)]$

$= - (β + P_{50} (α))$

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