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Question Number 100054 by  M±th+et+s last updated on 24/Jun/20

$$I_{n,m}={\int }_0^1{\int }_0^1\frac{{\left(ln\left(x\right)\right)}^n{\left(ln\left(y\right)\right)}^m}{1-xy}dxdy$$

Answered by maths mind last updated on 24/Jun/20

$$since\mid xy\mid <1$$$$\frac{1}{1-xy}=\sum _{k⩾0}{\left(xy\right)}^k$$$$I=\int \int \frac{{ln}^n\left(x\right){ln}^m\left(y\right)}{1-xy}dxdy={\int }_0^1{\int }_0^1\sum _{k⩾0}\left(x^k{y}^k\right){ln}^n\left(x\right){ln}^m\left(y\right)dxdy$$$$=\sum _{k⩾0}{\int }_0^1{\int }_0^1\left(x^k{ln}^n\left(x\right)dx\right)y^k{ln}^m\left(y\right)dy$$$$f\left(s\right)={\int }_0^1{x}^k{ln}^s\left(x\right)dx={\int }_0^{+\mathrm{\infty }}{(-t)}^s{e}^{-(k+1)t}dt$$$$\mathrm{t}\to (k+1)t\iff$$$$f\left(s\right)={(-1)}^s{\int }_0^{+\mathrm{\infty }}\frac{t^s{e}^{-t}}{{(k+1)}^{s+1}}dt={(-1)}^s\frac{\mathrm{\Gamma }(s+1)}{{(k+1)}^{\mathrm{s}+1}}$$$$I=\sum _{k⩾0}{\int }_0^1{x}^k{ln}^n\left(x\right)dx.{\int }_0^1{y}^k{ln}^m\left(x\right)dx$$$$=\sum _{k⩾0}\frac{{(-1)}^n\mathrm{\Gamma }(n+1)}{{(k+1)}^{n+1}}.{(-1)}^m\frac{\mathrm{\Gamma }(m+1)}{{(k+1)}^{m+1}}$$$$={(-1)}^{n+m}\mathrm{\Gamma }(n+1)\mathrm{\Gamma }(m+1).\sum _{k⩾0}\frac{1}{{(1+k)}^{n+m+2}}$$$$={(-1)}^{n+m}n!.m!.\zeta (n+m+2)$$$$\iff I_{n,m}={\int }_0^1{\int }_0^1\frac{{ln}^n\left(x\right){ln}^m\left(y\right)}{1-xy}dxdy={(-1)}^{n+m}n!.m!\zeta (n+m+2)$$$$$$

Commented by  M±th+et+s last updated on 24/Jun/20

$$coolsir.thankyou$$

Commented by maths mind last updated on 24/Jun/20

$$withepleasur$$

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