Definite Integral from 0 to 1 of Power of x by Power of Logarithm of x/Proof 2

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Theorem

$\ds \int_0^1 x^m \paren {\ln x}^n \rd x = \frac {\paren {-1}^n \map \Gamma {n + 1} } {\paren {m + 1}^{n + 1} }$


Proof

From Primitive of Power, we have:

$\ds \int_0^1 x^m \rd x = \frac 1 {m + 1}$

We have:

\(\ds \frac {\d^n} {\d m^n} \int_0^1 x^m \rd x\) \(=\) \(\ds \int_0^1 \frac {\partial^n} {\partial m^n} x^m \rd x\) Definite Integral of Partial Derivative
\(\ds \) \(=\) \(\ds \int_0^1 x^m \paren {\ln x}^n \rd x\) Derivative of Power of Constant

So:

\(\ds \int_0^1 x^m \paren {\ln x}^n \rd x\) \(=\) \(\ds \frac {\d^n} {\d m^n} \paren {\frac 1 {m + 1} }\)
\(\ds \) \(=\) \(\ds \frac {\d^{n + 1} } {\d m^{n + 1} } \paren {\map \ln {m + 1} }\) Derivative of Natural Logarithm
\(\ds \) \(=\) \(\ds \frac {\paren {-1}^n \map \Gamma {n + 1} } { \paren {m + 1}^{n + 1} }\) $n$th Derivative of Natural Logarithm

$\blacksquare$