Question Number 175077 by rexford last updated on 18/Aug/22
$${Show}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{b}} −{x}^{{a}} }{{lnx}}={ln}\left(\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\right) \\ $$
Answered by Ar Brandon last updated on 18/Aug/22
$${f}\left({b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{b}} }{\mathrm{ln}{x}}{dx}\:\Rightarrow{f}\:'\left({b}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{b}} {dx}=\frac{\mathrm{1}}{{b}+\mathrm{1}} \\ $$$$\Rightarrow{f}\left({b}\right)=\int\frac{\mathrm{1}}{{b}+\mathrm{1}}{dx}=\mathrm{ln}\mid{b}+\mathrm{1}\mid+{C}=\mathrm{ln}\mid{C}_{\mathrm{1}} \left({b}+\mathrm{1}\right)\mid \\ $$$$\Rightarrow{f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} }{\mathrm{ln}{x}}{dx}=\mathrm{ln}\mid{C}_{\mathrm{2}} \left({a}+\mathrm{1}\right)\mid \\ $$$$\Rightarrow{f}\left({b}\right)−{f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{b}} −{x}^{{a}} }{\mathrm{ln}{x}}{dx}=\mathrm{ln}\mid{K}\centerdot\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\mid,\:{K}=\frac{{C}_{\mathrm{1}} }{{C}_{\mathrm{2}} } \\ $$$$\Rightarrow{f}\left({b}\right)−{f}\left({b}\right)=\mathrm{0}=\mathrm{ln}\mid{K}\centerdot\frac{{b}+\mathrm{1}}{{b}+\mathrm{1}}\mid=\mathrm{ln}{K}\Rightarrow{K}=\mathrm{1} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{b}} −{x}^{{a}} }{\mathrm{ln}{x}}{dx}=\mathrm{ln}\mid\frac{{b}+\mathrm{1}}{{a}+\mathrm{1}}\mid \\ $$
Commented by rexford last updated on 24/Aug/22
$${Thank}\:{you} \\ $$