For m$$0$$,n$$0$$, let Im,$$n=$$∫$$01$$ $$xm(log$$⁡ $$x)ndx$$, then $$I5$$,$$5$$ is given by

Question

For m>$$0$$,n>$$0$$, let Im,$$n=$$∫$$01$$ $$xm(log$$⁡ $$x)ndx$$, then $$I5$$,$$5$$ is given by

Infinity Learn · Accepted Answer

Integrating by parts, we obtain Im,$$n=xm+1(log$$⁡ $$x)nm+101$$−$$nm+1$$∫$$01$$ $$xm(log$$⁡ $$x)n$$−$$1dxSince$$ limx→$$0+$$ $$xm(log$$⁡ $$x)n=limx$$→$$0+$$ $$xm/nlog$$⁡ $$xn=0So$$, Im, $$n=$$−$$nm+1Im$$, n−$$1Hence$$ $$I5$$, $$5=$$−$$56I5$$, $$4$$ $$=$$−$$56$$−$$46I5$$, $$3=($$−$$1)35.4.363I5$$, $$2=($$−$$1)45.4.3.264I5$$, $$1I5$$, $$1=$$∫$$01$$ $$x5log$$⁡ $$xdx=x6log$$⁡ $$x601$$−$$16$$∫$$01$$ $$x61xdx=$$−$$162So$$ $$I5$$, $$5=($$−$$1)55$$!$$66=$$−$$5$$!$$66$$

For $m > 0, n > 0,$ let $I_{m, n} = \int_{0}^{1} x^{m} (\log x)^{n}$
$d x,$ then $I_{5, 5}$ is given by

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For m>0,n>0, let Im,n=∫01 xm(log⁡ x)ndx, then I5,5 is given by

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For $m > 0, n > 0,$ let $I_{m, n} = \int_{0}^{1} x^{m} (\log x)^{n}$
$d x,$ then $I_{5, 5}$ is given by