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-gt-0-and-n-gt-0-also-evaluate-I-2-3-

Question Number 107596 by Rio Michael last updated on 11/Aug/20

Given

I_{m, n} = \underset{1}{\int^{e}} x^{m} {(\ln x)}^{n} d x where m, n \in N^{*}

Show that (1 + m) I_{m, n} = e^{m + 1} - {n I}_{m, n - 1} for m > 0 and n > 0

also, evaluate I_{2, 3}

Answered by Ar Brandon last updated on 11/Aug/20

I = \underset{1}{\int^{e}} x^{m} {(\ln x)}^{n} d x

= {[{(lnx)}^{n} \int x^{m} - \int {\frac{d {(lnx)}^{n}}{dx} \cdot \int x^{m} dx} dx]}_{1}^{e}

= {[\frac{{(lnx)}^{n} x^{m + 1}}{m + 1}]}_{1}^{e} - \frac{n}{m + 1} \int_{1}^{e} x^{m} {(lnx)}^{n - 1} dx

= \frac{e^{m + 1}}{m + 1} - \frac{n}{m + 1} I_{(m, n - 1)}

\Rightarrow (m + 1) I_{m, n} = e^{m + 1} - {nI}_{(m, n - 1)}

Answered by hgrocks last updated on 11/Aug/20

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