Question Number 219552 by Nicholas666 last updated on 28/Apr/25
Answered by mr W last updated on 29/Apr/25
![∫_0 ^1 x_k ^α ln (x_k )dx_k =(1/(α+1))∫_0 ^1 ln (x_k )d(x_k ^(α+1) ) =(1/(α+1)){[ln (x_k )(x_k ^(α+1) )]_0 ^1 −∫_0 ^1 x_k ^(α+1) d(ln (x_k ))} =(1/(α+1)){−∫_0 ^1 x_k ^α dx_k } =−(1/((α+1)^2 )) ∫_0 ^1 ∫_0 ^1 ...∫_0 ^1 ×××dx_1 dx_2 ...dx_n =[−(1/((α+1)^2 ))]^n =(((−1)^n )/((α+1)^(2n) ))](https://www.tinkutara.com/question/Q219584.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}_{{k}} ^{\alpha} \mathrm{ln}\:\left({x}_{{k}} \right){dx}_{{k}} \\ $$$$=\frac{\mathrm{1}}{\alpha+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}\:\left({x}_{{k}} \right){d}\left({x}_{{k}} ^{\alpha+\mathrm{1}} \right) \\ $$$$=\frac{\mathrm{1}}{\alpha+\mathrm{1}}\left\{\left[\mathrm{ln}\:\left({x}_{{k}} \right)\left({x}_{{k}} ^{\alpha+\mathrm{1}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} {x}_{{k}} ^{\alpha+\mathrm{1}} {d}\left(\mathrm{ln}\:\left({x}_{{k}} \right)\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\alpha+\mathrm{1}}\left\{−\int_{\mathrm{0}} ^{\mathrm{1}} {x}_{{k}} ^{\alpha} {dx}_{{k}} \right\} \\ $$$$=−\frac{\mathrm{1}}{\left(\alpha+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} …\int_{\mathrm{0}} ^{\mathrm{1}} ×××{dx}_{\mathrm{1}} {dx}_{\mathrm{2}} …{dx}_{{n}} \\ $$$$=\left[−\frac{\mathrm{1}}{\left(\alpha+\mathrm{1}\right)^{\mathrm{2}} }\right]^{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\alpha+\mathrm{1}\right)^{\mathrm{2}{n}} } \\ $$
Answered by maths2 last updated on 29/Apr/25
![∫_0 ^1 x^a ln(x)dx=(1/(a+1))(x^(a+1) ln(x)]_0 ^1 −(1/(1+a))∫_0 ^1 x^(a+1) .(dx/x) =(((−1))/((a+1)^2 )) ∫_0 ^1 ....∫_0 ^1 (Π_(k=1) ^n x_k )^a Πln(x_k )dx_1 ..dx_n =[∫_0 ^1 x^a ln(x)dx]^n =(((−1)^n )/((a+1)^(2n) ))](https://www.tinkutara.com/question/Q219608.png)
$$\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} {ln}\left({x}\right){dx}=\frac{\mathrm{1}}{{a}+\mathrm{1}}\left({x}^{{a}+\mathrm{1}} {ln}\left({x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{1}+{a}}\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}+\mathrm{1}} .\frac{{dx}}{{x}} \\ $$$$=\frac{\left(−\mathrm{1}\right)}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} ….\int_{\mathrm{0}} ^{\mathrm{1}} \left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{x}_{{k}} \right)^{{a}} \Pi{ln}\left({x}_{{k}} \right){dx}_{\mathrm{1}} ..{dx}_{{n}} \\ $$$$=\left[\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{{a}} {ln}\left({x}\right){dx}\right]^{{n}} =\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({a}+\mathrm{1}\right)^{\mathrm{2}{n}} } \\ $$