n integr natural calculate ∫_0 ^∞ (dx/((x+1)(x+2)......(x+n)))


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Question Number 37600 by prof Abdo imad last updated on 15/Jun/18

$n i n t e g r n a t u r a l c a l c u l a t e$ $\int_{0}^{\infty} \frac{d x}{(x + 1) (x + 2) . . . . . . (x + n)}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18
	$(1/((x+1)(x+2)(x+3)...(x+n)))=e^t −{ln(x+1)+ln(x+2)+ln(x+3)..+ln(x+n)= t dt=−{(1/(x+1))+(1/(x+2))+...+(1/(x+n))}dx =−{(((e^t /(x+1))+(e^t /(x+2))+...+(e^t /(x+n)))/((x+1)(x+2)(x+3)..(x+n)))}contd$
	$\frac{1}{(x + 1) (x + 2) (x + 3) . . . (x + n)} = e^{t}$ $- {l n (x + 1) + l n (x + 2) + l n (x + 3) . . + l n (x + n) =$ $t$ $d t = - {\frac{1}{x + 1} + \frac{1}{x + 2} + . . . + \frac{1}{x + n}} d x$ $= - {\frac{\frac{e^{t}}{x + 1} + \frac{e^{t}}{x + 2} + . . . + \frac{e^{t}}{x + n}}{(x + 1) (x + 2) (x + 3) . . (x + n)}} c o n t d$
	Commented by math khazana by abdo last updated on 17/Jun/18

	$y o u a r e r e a l l y a r a i l w a y s . . .$
	Answered by ajfour last updated on 17/Jun/18

	$I = \frac{1}{(n - 1)!} \int_{0}^{\infty} \frac{d x}{x + 1} - \frac{1}{(n - 2)!} \int_{0}^{\infty} \frac{d x}{x + 2}$ $- \frac{1}{2! (n - 3)!} \int_{0}^{\infty} \frac{d x}{x + 3} + \frac{1}{3! (n - 4)!} \int_{0}^{\infty} \frac{d x}{x + 4}$ $. . . . . - \frac{{(- 1)}^{n}}{(n - 1)!} \int_{0}^{\infty} \frac{d x}{x + n}$ $= \lim_{x \to \infty} Σ \frac{{(- 1)}^{r}}{(r - 1)! (n - r)!} \ln (1 + \frac{x}{r})$ $= \frac{1}{(n - 1)!} Σ {(- 1)}^{r} \underset{x \to \infty}{\lim^{n - 1}} C_{r - 1} \ln (1 + \frac{x}{r})$ $. . . . .$
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