Natural logarithm
Integral definition

The natural logarithm of a positive, real number a may be defined as the area under the graph of the hyperbola with equation y = 1/ x between x = 1 and x = a. This is the integral ln a = ∫ 1 a 1 x d x. {\displaystyle \ln a=\int _{1}^{a}{\frac {1}{x}}\,dx.} If a is in (0, 1) {\displaystyle (0,1)}, then the region has negative area, and the logarithm is negative.

This function is a logarithm because it satisfies the fundamental multiplicative property of a logarithm: ln (a b) = ln a + ln b. {\displaystyle \ln(ab)=\ln a+\ln b.}

This can be demonstrated by splitting the integral that defines ln ab into two parts, and then making the variable substitution x = at (so dx = a dt) in the second part, as follows: ln a b = ∫ 1 a b 1 x d x = ∫ 1 a 1 x d x + ∫ a a b 1 x d x = ∫ 1 a 1 x d x + ∫ 1 b 1 a t a d t = ∫ 1 a 1 x d x + ∫ 1 b 1 t d t = ln a + ln b. {\displaystyle {\begin{aligned}\ln ab=\int _{1}^{ab}{\frac {1}{x}}\,dx&=\int _{1}^{a}{\frac {1}{x}}\,dx+\int _{a}^{ab}{\frac {1}{x}}\,dx\\&=\int _{1}^{a}{\frac {1}{x}}\,dx+\int _{1}^{b}{\frac {1}{at}}a\,dt\\&=\int _{1}^{a}{\frac {1}{x}}\,dx+\int _{1}^{b}{\frac {1}{t}}\,dt\\&=\ln a+\ln b.\end{aligned}}}

In elementary terms, this is simply scaling by 1/ a in the horizontal direction and by a in the vertical direction. Area does not change under this transformation, but the region between a and ab is reconfigured. Because the function a /(ax) is equal to the function 1/ x, the resulting area is precisely ln b.

The number e can then be defined to be the unique real number a such that ln a = 1.

The natural logarithm also has an improper integral representation, which can be derived with Fubini's theorem as follows: ln (x) = ∫ 1 x 1 u d u = ∫ 1 x ∫ 0 ∞ e − t u d t d u = ∫ 0 ∞ ∫ 1 x e − t u d u d t = ∫ 0 ∞ e − t − e − t x t d t {\displaystyle \ln \left(x\right)=\int _{1}^{x}{\frac {1}{u}}du=\int _{1}^{x}\int _{0}^{\infty }e^{-tu}\ dt\ du=\int _{0}^{\infty }\int _{1}^{x}e^{-tu}\ du\ dt=\int _{0}^{\infty }{\frac {e^{-t}-e^{-tx}}{t}}dt}
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