Editing Logarithm (section)

===Integral representation of the natural logarithm===
[[File:Natural logarithm integral.svg|right|thumb|The [[natural logarithm]] of ''{{Mvar|t}}'' is the shaded area underneath the graph of the function {{math|1=''f''(''x'') = 1/''x''}}.|alt=A hyperbola with part of the area underneath shaded in grey.]]
The [[natural logarithm]] of {{Mvar|t}} can be defined as the [[definite integral]]:

<math display="block">\ln t = \int_1^t \frac{1}{x} \, dx.</math>
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, {{math|ln(''t'')}} equals the area between the {{mvar|x}}-axis and the graph of the function {{math|1/''x''}}, ranging from {{math|1=''x'' = 1}} to {{math|1=''x'' = ''t''}}. This is a consequence of the [[fundamental theorem of calculus]] and the fact that the derivative of {{math|ln(''x'')}} is {{math|1/''x''}}. Product and power logarithm formulas can be derived from this definition.<ref>{{Citation|last1=Courant|first1=Richard|title=Differential and integral calculus. Vol. I|publisher=[[John Wiley & Sons]]|location=New York|series=Wiley Classics Library|isbn=978-0-471-60842-4|mr=1009558|year=1988}}, section III.6</ref> For example, the product formula {{math|1=ln(''tu'') = ln(''t'') + ln(''u'')}} is deduced as:

<math display="block">\begin{align}
\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).
\end{align}</math>

The equality&nbsp;(1) splits the integral into two parts, while the equality&nbsp;(2) is a change of variable ({{math|1=''w'' = {{mvar|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&nbsp;{{Mvar|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 {{math|1=''f''(''x'') = 1/''x''}} again. Therefore, the left hand blue area, which is the integral of {{math|''f''(''x'')}} from {{Mvar|t}} to {{Mvar|tu}} is the same as the integral from 1 to {{Mvar|u}}. This justifies the equality&nbsp;(2) with a more geometric proof.

[[File:Natural logarithm product formula proven geometrically.svg|thumb|center|A visual proof of the product formula of the natural logarithm|alt=The hyperbola depicted twice. The area underneath is split into different parts.]]

The power formula {{math|1=ln(''t''<sup>''r''</sup>) = ''r'' ln(''t'')}} may be derived in a similar way:

<math display="block">\begin{align}
\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).
\end{align}</math>
The second equality uses a change of variables ([[integration by substitution]]), {{math|1=''w'' = {{mvar|x}}<sup>1/''r''</sup>}}.

The sum over the reciprocals of natural numbers,
<math display="block">1 + \frac 1 2 + \frac 1 3 + \cdots + \frac 1 n = \sum_{k=1}^n \frac{1}{k},</math>
is called the [[harmonic series (mathematics)|harmonic series]]. It is closely tied to the [[natural logarithm]]: as {{Mvar|n}} tends to [[infinity]], the difference,
<math display="block">\sum_{k=1}^n \frac{1}{k} - \ln(n),</math>
[[limit of a sequence|converges]] (i.e. gets arbitrarily close) to a number known as the [[Euler–Mascheroni constant]] {{math|1 = ''γ'' = 0.5772...}}. This relation aids in analyzing the performance of algorithms such as [[quicksort]].<ref>{{Citation|last1=Havil|first1=Julian|title=Gamma: Exploring Euler's Constant|publisher=[[Princeton University Press]]|isbn=978-0-691-09983-5|year=2003}}, sections 11.5 and 13.8</ref>