Definition of natural logarithm as an integral

[biometrix]

[MATH]\int_{1}^{x} \frac{1}{t} dt = ln(x).[/MATH]
What is the proof of the above equation?

 

[MarkFL]

Hello, and welcome to FMH!

Suppose we let:

[MATH]y=e^x[/MATH]
Then:

[MATH]\d{y}{x}=y[/MATH]
We also know:

[MATH]x=\ln(y)\implies 1=\frac{d}{dx}(\ln(y))\cdot\d{y}{x}\implies \frac{d}{dx}(\ln(y))=\frac{1}{y}[/MATH]
And so, using the anti-derivative form of the FTOC, we may state:

[MATH]\int_1^x \frac{1}{t}\,dt=\ln(x)-\ln(1)=\ln(x)[/MATH]

 

[Steven G]

[MATH]\int_{1}^{x} \frac{1}{t} dt = ln(x).[/MATH]
What is the proof of the above equation?

As you title says it is a definition. There is nothing to prove!

You should know that $\displaystyle \int \dfrac{1}{t}dt = \ln(t)$ and when we evaluate it from t=1 to t=x we get ln(x)-ln(1) = ln(x).

 

[pka]

[MATH]\int_{1}^{x} \frac{1}{t} dt = ln(x).[/MATH]What is the proof of the above equation?

If you can find an early edition of Calculus by Gillman & McDowell this is the way they do the logarithm.
If we define $L(\alpha\beta)=L(\alpha)+L(\beta)$ where $\alpha>0~\&~\beta>0$ as a logarithm function
they then prove that $\displaystyle\log (x) = \int_1^x {\frac{1}{t}dt}$ for $x>0$ has that property.
To prove that look at $\displaystyle\log (\alpha \beta ) = \int_1^{\alpha \beta } {\frac{1}{t}dt} = \log (\alpha ) - \log (\beta )$.
You must find the correct $u$-substitution to accomplish that. [it is a bit tricky]