Question-147346

Question Number 147346 by mnjuly1970 last updated on 20/Jul/21

Answered by qaz last updated on 20/Jul/21

a_n =2∫_0 ^(π/2) (1−sin x)^n sin xd(sin x) ⇒a_n =2∫_0 ^1 (1−u)^n udu Σ_(n=1) ^∞ (a_n /n)=Σ_(n=1) ^∞ (2/n)∫_0 ^1 (1−u)^n udu =2∫_0 ^1 uΣ_(n=1) ^∞ (((1−u)^n )/n)du =−2∫_0 ^1 uln(1−(1−u))du =−2∫_0 ^1 ulnudu =−2{(1/2)u^2 lnu∣_0 ^1 −(1/2)∫_0 ^1 udu} =(1/2)

$$\mathrm{a}_{\mathrm{n}} =\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\mathrm{x}\right)^{\mathrm{n}} \mathrm{sin}\:\mathrm{xd}\left(\mathrm{sin}\:\mathrm{x}\right) \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{n}} =\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{u}\right)^{\mathrm{n}} \mathrm{udu} \\ $$$$\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{a}_{\mathrm{n}} }{\mathrm{n}}=\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}}{\mathrm{n}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−\mathrm{u}\right)^{\mathrm{n}} \mathrm{udu} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{u}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{1}−\mathrm{u}\right)^{\mathrm{n}} }{\mathrm{n}}\mathrm{du} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{uln}\left(\mathrm{1}−\left(\mathrm{1}−\mathrm{u}\right)\right)\mathrm{du} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ulnudu} \\ $$$$=−\mathrm{2}\left\{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{u}^{\mathrm{2}} \mathrm{lnu}\mid_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{udu}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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