Definite Integral from 0 to 1 of Power of x by Power of Logarithm of x/Proof 2
Jump to navigation
Jump to search
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 |
$\blacksquare$