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Integration Questions

Question Number 172764 by Mathematification last updated on 01/Jul/22

$$\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)}{\mathrm{x}}\right)^{\mathrm{2}} \:\mathrm{dx}=\:? \\$$

Answered by Mathspace last updated on 01/Jul/22

$$\Upsilon=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{{x}^{\mathrm{2}} }{dx}\left({u}^{'} =\frac{\mathrm{1}}{{x}^{\mathrm{2}} }{andv}={ln}^{\mathrm{2}} \left(\mathrm{1}+{x}\right)\right) \\$$ $$\Upsilon=\left[−\frac{\mathrm{1}}{{x}}{ln}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)\right]_{\mathrm{0}} ^{+\infty} \\$$ $$+\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}}\left(−\mathrm{2}\right){ln}\left(\mathrm{1}−{x}\right){dx} \\$$ $$=−\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}{dx} \\$$ $${ln}^{'} \left(\mathrm{1}−{x}\right)=−\frac{\mathrm{1}}{\mathrm{1}−{x}}=−\sum_{{n}=\mathrm{0}} ^{\infty} {x}^{{n}} \Rightarrow \\$$ $${ln}\left(\mathrm{1}−{x}\right)=−\sum_{{n}=\mathrm{0}} ^{\infty} \frac{{u}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:+{c}\left({c}=\mathrm{0}\right) \\$$ $$=−\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{x}^{{n}} }{{n}}\:\Rightarrow−\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}} \\$$ $$=\sum_{{n}=\mathrm{1}} ^{\infty} \frac{{x}^{{n}−\mathrm{1}} }{{n}}\:\Rightarrow \\$$ $$\Upsilon=\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{n}−\mathrm{1}} {dx} \\$$ $$=\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\mathrm{2}.\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\frac{\pi^{\mathrm{2}} }{\mathrm{3}} \\$$ $$\Upsilon=\frac{\pi^{\mathrm{2}} }{\mathrm{3}} \\$$

Commented byMathematification last updated on 08/Jul/22

@Mathspace Please check the third and fourth line again sir.

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