Difference between revisions of "Logarithm of 1"
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By the definition, | By the definition, | ||
$$\log(z) = \displaystyle\int_1^z \dfrac{1}{\tau} \mathrm{d}\tau.$$ | $$\log(z) = \displaystyle\int_1^z \dfrac{1}{\tau} \mathrm{d}\tau.$$ | ||
| − | Plugging in $z=1$ and the [[integral from a to a]], | + | Plugging in $z=1$ and using the [[integral from a to a]], |
$$\begin{array}{ll} | $$\begin{array}{ll} | ||
\log(1) &= \displaystyle\int_1^1 \dfrac{1}{\tau} \mathrm{d}\tau \\ | \log(1) &= \displaystyle\int_1^1 \dfrac{1}{\tau} \mathrm{d}\tau \\ | ||
Revision as of 12:24, 17 September 2016
Theorem
The following formula holds: $$\log(1)=0,$$ where $\log$ denotes the logarithm.
Proof
By the definition, $$\log(z) = \displaystyle\int_1^z \dfrac{1}{\tau} \mathrm{d}\tau.$$ Plugging in $z=1$ and using the integral from a to a, $$\begin{array}{ll} \log(1) &= \displaystyle\int_1^1 \dfrac{1}{\tau} \mathrm{d}\tau \\ &= 0, \end{array}$$ as was to be shown.
References
- 1964: Milton Abramowitz and Irene A. Stegun: Handbook of mathematical functions ... (previous) ... (next): $4.1.12$
