I_n =∫_(t=0) ^(+∞) (dt/((t+1)(t+2)...(t+n)))


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Question Number 90135 by Ar Brandon last updated on 21/Apr/20

$I_{n} = \int_{t = 0}^{+ \infty} \frac{dt}{(t + 1) (t + 2) . . . (t + n)}$
	Answered by TANMAY PANACEA. last updated on 21/Apr/20

	$I = \int \frac{d t}{(t + 1) (t + 2) (t + 3) . . (t + n)}$ $\frac{1}{(t + 1) (t + 2) (t + 3) . . (t + n)} = \frac{a_{1}}{t + 1} + \frac{a_{2}}{t + 2} + . . + \frac{a_{n}}{(t + n)}$ $\int_{0}^{\infty} (\frac{a_{1}}{t + 1} + \frac{a_{2}}{t + 2} + . . + \frac{a_{n}}{t + n}) d t$ $=∣ a_{1} l n (t + 1) + a_{2} l n (t + 2) + . . + a_{n} l n (t + n) ∣_{0}^{\infty}$ $n o w c a l c u l a t i o n t o f i n d a_{1}, a_{2} . . . a_{n}$ $1 = a_{1} (t + 2) (t + 3) . . (t + n) + a_{2} (t + 1) (t + 3) . . (t + n) + . . a_{n} (t + 1) (t + 2) . . (t + n - 1)$ $p u t t + 1 = 0$ $1 = a_{1} \times (n - 1)! \to a_{1} = \frac{1}{(n - 1)!}$ $p u t t + 2 = 0$ $a_{2} \times (- 2 + 1) (1.2 . . . n - 2)$ $a_{2} \times (- 1) (n - 2)! = 1 \to a_{2} = \frac{1}{(- 1) (n - 2)!}$ $p u t t + 3 = 0$ $a_{3} (t + 1) (t + 2) (t + 4) . . . (t + n) = 1$ $a_{3} (- 3 + 1) (- 3 + 2) (1.2 . . . . n - 3) = 1$ $a_{3} = \frac{1}{2 \times (n - 3)!}$ $t r i e d t o s o l v e$
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