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 by Mathematification last updated on 08/Jul/22 | ||
@Mathspace Please check the third and fourth line again sir. | ||