Question Number 28169 by abdo imad last updated on 21/Jan/18
$${prove}\:{that}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\left({lnx}\right)^{{k}} }{\mathrm{1}−{x}}{dx}=\left(−\mathrm{1}\right)^{{k}} \:\left({k}!\right)\xi\left({k}+\mathrm{1}\right)\:\:. \\ $$
Commented by abdo imad last updated on 23/Jan/18
![we have ∫_0 ^1 (((ln(x))^k )/(1−x))dx= ∫_0 ^1 (lnx)^k (Σ_(p=0) ^(+∞) x^p )dx = Σ_(p=0) ^(+∞) ∫_0 ^1 x^p (lnx)^k dx let put I_(p,k) =∫_0 ^1 x^p (lnx)^k dx by parts we get I_(p,k) = [ (1/(p+1)) x^(p+1) (lnx)^k ]_0 ^1 − ∫_0 ^1 (1/(p+1)) x^(p+1) (k/x) (lnx)^(k−1) dx =((−k)/(p+1)) ∫_0 ^1 x^p (lnx)^(k−1) dx= ((−k)/(p+1)) I_(p,k−1 ) =(((−1)^2 k(k−1))/((p+1)^2 )) I_(p,k−2) =(((−1)^k k!)/((p+1)^k )) I_(p,0) =(((−1)^k k!)/((p+1)^(k+1) )) so ∫_0 ^1 (((lnx)^k )/(1−x))dx = Σ_(p=0) ^(+∞) (((−1)^k k!)/((p+1)^(k+1) )) = (−1)^k k! Σ_(p=0) ^(+∞) (1/((p+1)^(k+1) )) =(−1)^k k! Σ_(p=1) ^(+∞) (1/p^(k+1) )= (−1)^k k! ξ(k+1) .](https://www.tinkutara.com/question/Q28287.png)
$${we}\:{have}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\left({ln}\left({x}\right)\right)^{{k}} }{\mathrm{1}−{x}}{dx}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({lnx}\right)^{{k}} \left(\sum_{{p}=\mathrm{0}} ^{+\infty} \:{x}^{{p}} \right){dx} \\ $$$$=\:\sum_{{p}=\mathrm{0}} ^{+\infty} \:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{p}} \:\left({lnx}\right)^{{k}} {dx}\:\:{let}\:{put}\: \\ $$$${I}_{{p},{k}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{p}} \left({lnx}\right)^{{k}} {dx}\:\:{by}\:{parts}\:{we}\:{get} \\ $$$${I}_{{p},{k}} \:\:=\:\left[\:\frac{\mathrm{1}}{{p}+\mathrm{1}}\:{x}^{{p}+\mathrm{1}} \:\left({lnx}\right)^{{k}} \right]_{\mathrm{0}} ^{\mathrm{1}} \:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{{p}+\mathrm{1}}\:{x}^{{p}+\mathrm{1}} \:\frac{{k}}{{x}}\:\left({lnx}\right)^{{k}−\mathrm{1}} \:{dx} \\ $$$$=\frac{−{k}}{{p}+\mathrm{1}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{x}^{{p}} \:\left({lnx}\right)^{{k}−\mathrm{1}} {dx}=\:\frac{−{k}}{{p}+\mathrm{1}}\:{I}_{{p},{k}−\mathrm{1}\:} \:=\frac{\left(−\mathrm{1}\right)^{\mathrm{2}} {k}\left({k}−\mathrm{1}\right)}{\left({p}+\mathrm{1}\right)^{\mathrm{2}} }\:{I}_{{p},{k}−\mathrm{2}} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{k}} \:{k}!}{\left({p}+\mathrm{1}\right)^{{k}} }\:{I}_{{p},\mathrm{0}} =\frac{\left(−\mathrm{1}\right)^{{k}} \:{k}!}{\left({p}+\mathrm{1}\right)^{{k}+\mathrm{1}} }\:\:{so} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\left({lnx}\right)^{{k}} }{\mathrm{1}−{x}}{dx}\:\:=\:\sum_{{p}=\mathrm{0}} ^{+\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} \:{k}!}{\left({p}+\mathrm{1}\right)^{{k}+\mathrm{1}} }\:=\:\left(−\mathrm{1}\right)^{{k}} \:{k}!\:\sum_{{p}=\mathrm{0}} ^{+\infty} \:\:\:\frac{\mathrm{1}}{\left({p}+\mathrm{1}\right)^{{k}+\mathrm{1}} } \\ $$$$=\left(−\mathrm{1}\right)^{{k}} \:{k}!\:\sum_{{p}=\mathrm{1}} ^{+\infty} \:\:\frac{\mathrm{1}}{{p}^{{k}+\mathrm{1}} }=\:\left(−\mathrm{1}\right)^{{k}} \:{k}!\:\xi\left({k}+\mathrm{1}\right)\:\:. \\ $$