Primitive of Logarithm of x/Proof 2
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Theorem
- $\ds \int \ln x \rd x = x \ln x - x + C$
Proof
Note that we have:
| \(\ds \int_0^1 \ln x \rd x\) | \(=\) | \(\ds \int_0^1 x^0 \paren {\ln x}^1 \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {\paren {-1}^1 \map \Gamma 2} {1^2}\) | Definite Integral from $0$ to $1$ of $x^m \paren {\ln x}^n$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -1\) | Gamma Function Extends Factorial |
We therefore have:
| \(\ds -1\) | \(=\) | \(\ds \int_0^1 \ln x \rd x\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 a \int_0^a \map \ln {\frac u a} \rd u\) | substituting $x = \dfrac u a$ where $a > 0$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 a \int_0^a \paren {\ln u - \ln a} \rd u\) | Difference of Logarithms | |||||||||||
| \(\ds \) | \(=\) | \(\ds \frac 1 a \int_0^a \ln u \rd u - \ln a\) | Primitive of Constant, Fundamental Theorem of Calculus |
giving:
- $\ds \frac 1 a \int_0^a \ln u \rd u = \ln a - 1$
so:
- $\ds \int_0^a \ln u \rd u = a \ln a - a$
for all real $a > 0$.
By Fundamental Theorem of Calculus: First Part, we have that:
- $x \ln x - x$ is a primitive of $\ln x$ on $x > 0$.
We therefore conclude that, by Primitives which Differ by Constant:
- $\ds \int \ln x \rd x = x \ln x - x + C$
for $x > 0$.
$\blacksquare$