n-integr-natural-calculate-0-dx-x-1-x-2-x-n-

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) \dots \dots (x + n)}

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18

\frac{1}{(x + 1) (x + 2) (x + 3) \dots (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} + \dots + \frac{1}{x + n}} d x

= - {\frac{\frac{e^{t}}{x + 1} + \frac{e^{t}}{x + 2} + \dots + \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 \dots

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}

\dots . . - \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})

\dots . .

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