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

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

Let:

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

Then, by Derivative of Exponential Function:

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

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:

\(\ds \int_0^1 x^m \paren {\ln x}^n \rd x\)

\(=\)

\(\ds \int_\infty^0 \paren {\map \exp {-\frac u {m + 1} } }^m \paren {-\frac u {m + 1} }^n \paren {-\frac 1 {m + 1} \map \exp {-\frac u {m + 1} } \rd u}\)

Integration by Substitution

\(\ds \)

\(=\)

\(\ds \frac 1 {m + 1} \int_0^\infty \paren {\map \exp {-\frac u {m + 1} } }^m \paren {-\frac u {m + 1} }^n \paren {\map \exp {-\frac u {m + 1} } \rd u}\)

Reversal of Limits of Definite Integral

\(\ds \)

\(=\)

\(\ds \frac {\paren {-1}^n} {m + 1} \int_0^\infty \map \exp {-u \paren {\frac m {m + 1} + \frac 1 {m + 1} } } \frac {u^n} {\paren {m + 1}^n} \rd u\)

Exponential of Sum, Power of Power

\(\ds \)

\(=\)

\(\ds \frac {\paren {-1}^n} {\paren {m + 1}^{n + 1} } \int_0^\infty e^{-u} u^{\paren {n + 1} - 1} \rd u\)

\(\ds \)

\(=\)

\(\ds \frac {\paren {-1}^n \map \Gamma {n + 1} } {\paren {m + 1}^{n + 1} }\)

Definition of Gamma Function

$\blacksquare$