Question Number 207789 by Ghisom last updated on 26/May/24
$$\forall{r}\in\mathbb{R}:\:{H}_{{r}} =\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{t}^{{r}} −\mathrm{1}}{{t}−\mathrm{1}}{dt} \\ $$$${H}_{{r}+\mathrm{2}} −{H}_{{r}} =\mathrm{1} \\ $$$${r}=? \\ $$
Answered by MM42 last updated on 27/May/24
![H_(r+2) −H_r =∫_(0 ) ^1 ((t^(r+2) −t^r )/(t−1))dt =∫_(0 ) ^1 (t^(r+1) +t^r )dt=((t^(r+2) /(r+2))+(t^(r+1) /(r+1)))]_0 ^1 =(1/(r+2))+(1/(r+1))=((2r+3)/(r^2 +3r+2))=1 ⇒r^2 +r−1=0⇒r=((−1+(√5))/2) ✓](https://www.tinkutara.com/question/Q207790.png)
$${H}_{{r}+\mathrm{2}} −{H}_{{r}} =\int_{\mathrm{0}\:} ^{\mathrm{1}} \frac{{t}^{{r}+\mathrm{2}} −{t}^{{r}} }{{t}−\mathrm{1}}{dt} \\ $$$$\left.=\int_{\mathrm{0}\:} ^{\mathrm{1}} \left({t}^{{r}+\mathrm{1}} +{t}^{{r}} \right){dt}=\left(\frac{{t}^{{r}+\mathrm{2}} }{{r}+\mathrm{2}}+\frac{{t}^{{r}+\mathrm{1}} }{{r}+\mathrm{1}}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{r}+\mathrm{2}}+\frac{\mathrm{1}}{{r}+\mathrm{1}}=\frac{\mathrm{2}{r}+\mathrm{3}}{{r}^{\mathrm{2}} +\mathrm{3}{r}+\mathrm{2}}=\mathrm{1} \\ $$$$\Rightarrow{r}^{\mathrm{2}} +{r}−\mathrm{1}=\mathrm{0}\Rightarrow{r}=\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\checkmark\: \\ $$$$ \\ $$
Commented by Ghisom last updated on 26/May/24
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{but}\:\mathrm{I}\:\mathrm{think}\:\mathrm{only}\:\mathrm{the}\:\mathrm{positive} \\ $$$$\mathrm{root}\:\mathrm{is}\:\mathrm{valid} \\ $$
Commented by mr W last updated on 27/May/24
$${why}\:{is}\:{the}\:{negative}\:{root}\:{not}\:{valid}? \\ $$
Commented by MM42 last updated on 27/May/24
$${ok}\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Commented by Frix last updated on 27/May/24
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{t}^{−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}} −\mathrm{1}}{{t}−\mathrm{1}}{dt}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{converge}. \\ $$
Commented by Ghisom last updated on 27/May/24
$$\mathrm{yes} \\ $$