section_path: Analytic properties/Integral representation of the natural logarithm
section_name: Integral representation of the natural logarithm
paragraph_text: Integral representation of the natural logarithm
  
  The natural logarithm of t can be defined as the definite integral:
  
  ln t = ∫ 1 t 1 x d x. {\displaystyle \ln t=\int _{1}^{t}{\frac {1}{x}}\,dx.} This definition has the advantage that it does not rely on the exponential function or any trigonometric functions; the definition is in terms of an integral of a simple reciprocal. As an integral, ln(t) equals the area between the x -axis and the graph of the function 1/ x, ranging from x = 1 to x = t. This is a consequence of the fundamental theorem of calculus and the fact that the derivative of ln(x) is 1/ x. Product and power logarithm formulas can be derived from this definition. For example, the product formula ln(tu) = ln(t) + ln(u) is deduced as:
  
  ln (t u) = ∫ 1 t u 1 x d x = (1) ∫ 1 t 1 x d x + ∫ t t u 1 x d x = (2) ln (t) + ∫ 1 u 1 w d w = ln (t) + ln (u). {\displaystyle \ln(tu)=\int _{1}^{tu}{\frac {1}{x}}\,dx\ {\stackrel {(1)}{=}}\int _{1}^{t}{\frac {1}{x}}\,dx+\int _{t}^{tu}{\frac {1}{x}}\,dx\ {\stackrel {(2)}{=}}\ln(t)+\int _{1}^{u}{\frac {1}{w}}\,dw=\ln(t)+\ln(u).}
  
  The equality (1) splits the integral into two parts, while the equality (2) is a change of variable (w = x / t). In the illustration below, the splitting corresponds to dividing the area into the yellow and blue parts. Rescaling the left hand blue area vertically by the factor t and shrinking it by the same factor horizontally does not change its size. Moving it appropriately, the area fits the graph of the function f (x) = 1/ x again. Therefore, the left hand blue area, which is the integral of f (x) from t to tu is the same as the integral from 1 to u. This justifies the equality (2) with a more geometric proof.
  
  The power formula ln(t) = r ln(t) may be derived in a similar way:
  
  ln (t r) = ∫ 1 t r 1 x d x = ∫ 1 t 1 w r (r w r − 1 d w) = r ∫ 1 t 1 w d w = r ln (t). {\displaystyle \ln(t^{r})=\int _{1}^{t^{r}}{\frac {1}{x}}dx=\int _{1}^{t}{\frac {1}{w^{r}}}\left(rw^{r-1}\,dw\right)=r\int _{1}^{t}{\frac {1}{w}}\,dw=r\ln(t).} The second equality uses a change of variables (integration by substitution), w = x.
  
  The sum over the reciprocals of natural numbers, 1 + 1 2 + 1 3 + ⋯ + 1 n = ∑ k = 1 n 1 k, {\displaystyle 1+{\frac {1}{2}}+{\frac {1}{3}}+\cdots +{\frac {1}{n}}=\sum _{k=1}^{n}{\frac {1}{k}},} is called the harmonic series. It is closely tied to the natural logarithm: as n tends to infinity, the difference, ∑ k = 1 n 1 k − ln (n), {\displaystyle \sum _{k=1}^{n}{\frac {1}{k}}-\ln(n),} converges (i.e. gets arbitrarily close) to a number known as the Euler–Mascheroni constant γ = 0.5772.... This relation aids in analyzing the performance of algorithms such as quicksort.