r-R-H-r-0-1-t-r-1-t-1-dt-H-r-2-H-r-1-r-

Question Number 207789 by Ghisom last updated on 26/May/24

\forall r \in R : H_{r} = \underset{0}{\int^{1}} \frac{t^{r} - 1}{t - 1} d t

H_{r + 2} - H_{r} = 1

r = ?

Answered by MM42 last updated on 27/May/24

H_{r + 2} - H_{r} = \int_{0}^{1} \frac{t^{r + 2} - t^{r}}{t - 1} d t

{= \int_{0}^{1} (t^{r + 1} + t^{r}) d t = (\frac{t^{r + 2}}{r + 2} + \frac{t^{r + 1}}{r + 1})]}_{0}^{1}

= \frac{1}{r + 2} + \frac{1}{r + 1} = \frac{2 r + 3}{r^{2} + 3 r + 2} = 1

\Rightarrow r^{2} + r - 1 = 0 \Rightarrow r = \frac{- 1 + \sqrt{5}}{2} ✓

Commented by Ghisom last updated on 26/May/24

thank you but I think only the positive

root is valid

Commented by mr W last updated on 27/May/24

w h y i s t h e n e g a t i v e r o o t n o t v a l i d ?

Commented by MM42 last updated on 27/May/24

o k \underset{―}{\underset{⏟}{⋚}}

Commented by Frix last updated on 27/May/24

\underset{0}{\int^{1}} \frac{t^{- \frac{1 + \sqrt{5}}{2}} - 1}{t - 1} d t {doesn}^{'} t converge .

Commented by Ghisom last updated on 27/May/24

yes

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