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Question Number 107596 by Rio Michael last updated on 11/Aug/20
Given    I_(m,n)  = ∫_1 ^e x^m  (ln x)^n  dx where m,n ∈ N^∗   Show that (1 + m)I_(m,n)  = e^(m+1) −nI_(m,n−1)  for m >0 and n>0  also, evaluate I_(2,3)
$$\mathrm{Given}\: \\ $$$$\:{I}_{{m},{n}} \:=\:\underset{\mathrm{1}} {\overset{{e}} {\int}}{x}^{{m}} \:\left(\mathrm{ln}\:{x}\right)^{{n}} \:{dx}\:\mathrm{where}\:{m},{n}\:\in\:\mathbb{N}^{\ast} \\ $$$$\mathrm{Show}\:\mathrm{that}\:\left(\mathrm{1}\:+\:{m}\right){I}_{{m},{n}} \:=\:{e}^{{m}+\mathrm{1}} −{nI}_{{m},{n}−\mathrm{1}} \:\mathrm{for}\:{m}\:>\mathrm{0}\:\mathrm{and}\:{n}>\mathrm{0} \\ $$$$\mathrm{also},\:\mathrm{evaluate}\:{I}_{\mathrm{2},\mathrm{3}} \\ $$
Answered by Ar Brandon last updated on 11/Aug/20
I= ∫_1 ^e x^m  (ln x)^n  dx    =[(lnx)^n ∫x^m −∫{((d(lnx)^n )/dx)∙∫x^m dx}dx]_1 ^e     =[(((lnx)^n x^(m+1) )/(m+1))]_1 ^e −(n/(m+1))∫_1 ^e x^m (lnx)^(n−1) dx    =(e^(m+1) /(m+1))−(n/(m+1))I_((m,n−1))   ⇒(m+1)I_(m,n) =e^(m+1) −nI_((m,n−1))
$$\mathrm{I}=\:\underset{\mathrm{1}} {\overset{{e}} {\int}}{x}^{{m}} \:\left(\mathrm{ln}\:{x}\right)^{{n}} \:{dx} \\ $$$$\:\:=\left[\left(\mathrm{lnx}\right)^{\mathrm{n}} \int\mathrm{x}^{\mathrm{m}} −\int\left\{\frac{\mathrm{d}\left(\mathrm{lnx}\right)^{\mathrm{n}} }{\mathrm{dx}}\centerdot\int\mathrm{x}^{\mathrm{m}} \mathrm{dx}\right\}\mathrm{dx}\right]_{\mathrm{1}} ^{\mathrm{e}} \\ $$$$\:\:=\left[\frac{\left(\mathrm{lnx}\right)^{\mathrm{n}} \mathrm{x}^{\mathrm{m}+\mathrm{1}} }{\mathrm{m}+\mathrm{1}}\right]_{\mathrm{1}} ^{\mathrm{e}} −\frac{\mathrm{n}}{\mathrm{m}+\mathrm{1}}\int_{\mathrm{1}} ^{\mathrm{e}} \mathrm{x}^{\mathrm{m}} \left(\mathrm{lnx}\right)^{\mathrm{n}−\mathrm{1}} \mathrm{dx} \\ $$$$\:\:=\frac{\mathrm{e}^{\mathrm{m}+\mathrm{1}} }{\mathrm{m}+\mathrm{1}}−\frac{\mathrm{n}}{\mathrm{m}+\mathrm{1}}\mathrm{I}_{\left(\mathrm{m},\mathrm{n}−\mathrm{1}\right)} \\ $$$$\Rightarrow\left(\mathrm{m}+\mathrm{1}\right)\mathrm{I}_{\mathrm{m},\mathrm{n}} =\mathrm{e}^{\mathrm{m}+\mathrm{1}} −\mathrm{nI}_{\left(\mathrm{m},\mathrm{n}−\mathrm{1}\right)} \\ $$
Answered by hgrocks last updated on 11/Aug/20

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