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Question Number 32939 by abdo imad last updated on 06/Apr/18

$$1)studytheconvergenceof{\int }_0^1\frac{x^p}{1+x}dx$$$$2)find{lim}_{p\to \mathrm{\infty }}{\int }_0^1\frac{x^p}{1+x}dx.$$

Commented by abdo imad last updated on 08/Apr/18

$$letputI={\int }_0^1\frac{x^p}{1+x}dx.ch.x=\frac{1}{t}give$$$$I={\int }_1^{+\mathrm{\infty }}\frac{1}{t^p(1+\frac{1}{t})}\frac{dt}{t^2}={\int }_1^{+\mathrm{\infty }}\frac{dt}{t^{p+2}+t^{p+1}}$$$$={\int }_1^{+\mathrm{\infty }}\frac{dt}{t^{p+2}(1+\frac{1}{t})}.andatV\left(\mathrm{\infty }\right)\frac{1}{t^{p+2}(1+\frac{1}{t})}\sim \frac{1}{t^{p+2}}$$$${\int }_1^{+\mathrm{\infty }}\frac{dt}{t^{p+2}}converges\iff p+2>1\iff p>-1.so$$$$Iconverges\iff p>-1.$$$$2)\mathrm{\forall }x\in [0,1]\frac{x^p}{1+x}⩽x^p\Rightarrow {\int }_0^1\frac{x^p}{1+x}dx⩽{\int }_0^1{x}^{p}dx=\frac{1}{p+1}\Rightarrow$$$${lim}_{p\to \mathrm{\infty }}{\int }_0^1\frac{x^p}{1+x}dx=0.$$

Answered by JDamian last updated on 08/Apr/18

$$$$$$1)\frac{x^p}{1+x}=x^{p-1}-x^{p-2}+\cdot \cdot \cdot {(-1)}^{p+1}+\frac{{(-1)}^p}{x+1}=\underset{k=0}{\sum ^{p-1}}{(-1)}^{p+k+1}{x}^k$$$${\int }_0^1\frac{x^p}{1+x}={[\frac{x^p}{p}-\frac{x^{p-1}}{p-1}+\cdot \cdot \cdot +{(-1)}^{p+1}+{(-1)}^{p}ln\mid x+1\mid ]}_0^1=$$$$={[\underset{k=1}{\sum ^p}{(-1)}^{p+k}\frac{x^k}{k}+{(-1)}^{p}ln\mid x+1\mid ]}_0^1=\underset{k=1}{\sum ^p}\frac{{(-1)}^{p+k}}{k}+{(-1)}^{p}ln2=$$$$=\frac{1}{p}-\frac{1}{p-1}+\cdot \cdot \cdot +{(-1)}^{p+1}+{(-1)}^{p}ln2$$

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