Definite Integral from 0 to 1 of Power of x by Power of Logarithm of x

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Theorem

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

where:

$n$ is a non-negative integer
$m$ is a real number with $m > -1$.


Proof 1

Let:

$\displaystyle x = \map \exp {-\frac u {m + 1} }$

Then, by Derivative of Exponential Function:

$\displaystyle \frac {\d x} {\d u} = -\frac 1 {m + 1} \map \exp {-\frac u {m + 1} }$

We have by Exponential of Zero:

as $x \to 1$, $u \to 0$

We also have, by Exponential Tends to Zero and Infinity:

as $x \to 0$, $u \to \infty$

So:

\(\displaystyle \int_0^1 x^m \paren {\ln x}^n \rd x\) \(=\) \(\displaystyle -\frac 1 {m + 1} \int_\infty^0 \map \exp {-\frac u {m + 1} } \map \exp {-\frac {u m} {m + 1} } \paren {-\frac u {m + 1} }^n \rd x\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {-1}^n} {m + 1} \int_0^\infty \map \exp {-u \paren {\frac m {m + 1} + \frac 1 {m + 1} } } \frac {u^n} {\paren {m + 1}^n} \rd x\) Reversal of Limits of Definite Integral, Exponential of Sum
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {-1}^n} {\paren {m + 1}^{n + 1} } \int_0^\infty e^{-u} u^{\paren {n + 1} - 1} \rd u\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\paren {-1}^n \map \Gamma {n + 1} } {\paren {m + 1}^{n + 1} }\) Definition of Gamma Function

$\blacksquare$


Proof 2

From Primitive of Power, we have:

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

We have:

\(\displaystyle \frac {\d^n} {\d m^n} \int_0^1 x^m \rd x\) \(=\) \(\displaystyle \int_0^1 \frac {\partial^n} {\partial m^n} x^m \rd x\) Definite Integral of Partial Derivative
\(\displaystyle \) \(=\) \(\displaystyle \int_0^1 x^m \paren {\ln x}^n \rd x\) Derivative of Exponential Function: Corollary 2

So:

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

$\blacksquare$


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