Integration Questions

Question Number 103871 by Dwaipayan Shikari last updated on 17/Jul/20

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{98}} −\mathrm{99}{x}+\mathrm{98}}{{logx}}{dx} \\$$

Commented by Dwaipayan Shikari last updated on 18/Jul/20

$${I}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} −\mathrm{1}}{{logx}} \\$$$${I}^{'} \left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\partial}{\partial{a}}\left(\frac{{x}^{{a}} −\mathrm{1}}{{logx}}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{{a}} {logx}}{{logx}}=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} =\left[\frac{{x}^{{a}+\mathrm{1}} }{{a}+\mathrm{1}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{{a}+\mathrm{1}} \\$$$$\int{I}^{\:'} \left({a}\right)=\int\frac{\mathrm{1}}{{a}+\mathrm{1}} \\$$$${I}\left({a}\right)={log}\left({a}+\mathrm{1}\right)+{C}=\int\frac{{x}^{{a}} −\mathrm{1}}{{logx}} \\$$$${So}\:, \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{98}} −\mathrm{1}}{{logx}}−\mathrm{99}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}−\mathrm{1}}{{logx}}={log}\mathrm{99}−\mathrm{99}{log}\mathrm{2} \\$$$$\\$$

Commented by Dwaipayan Shikari last updated on 18/Jul/20

$${Kindly}\:{check}\:{my}\:{answer} \\$$

Answered by Ar Brandon last updated on 17/Jul/20

$$\mathrm{Let}\:\mathrm{x}=\mathrm{e}^{\mathrm{u}} \Rightarrow\mathrm{dx}=\mathrm{e}^{\mathrm{u}} \mathrm{du} \\$$$$\Rightarrow\mathcal{I}=\int_{\infty} ^{\mathrm{0}} \frac{\mathrm{e}^{\mathrm{99u}} −\mathrm{99e}^{\mathrm{2u}} +\mathrm{98e}^{\mathrm{u}} }{\mathrm{u}}\mathrm{du} \\$$$$\:\:\:\:\:\:\:\:=−\int_{\mathrm{0}} ^{\infty} \left\{\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{99}^{\mathrm{k}} \mathrm{u}^{\mathrm{k}−\mathrm{1}} }{\mathrm{k}!}−\mathrm{99}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{\mathrm{k}} \mathrm{u}^{\mathrm{k}−\mathrm{1}} }{\mathrm{k}!}+\mathrm{98}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{u}^{\mathrm{k}−\mathrm{1}} }{\mathrm{k}!}\right\}\mathrm{du} \\$$$$\:\:\:\:\:\:\:\:=−\left[\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{99}^{\mathrm{k}} \mathrm{u}^{\mathrm{k}} }{\mathrm{k}\left(\mathrm{k}!\right)}−\mathrm{99}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{2}^{\mathrm{k}} \mathrm{u}^{\mathrm{k}} }{\mathrm{k}\left(\mathrm{k}!\right)}+\mathrm{98}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{u}^{\mathrm{k}} }{\mathrm{k}\left(\mathrm{k}!\right)}\right]_{\mathrm{0}} ^{\infty} \\$$