How to prove the double integral definition of logarithm?

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SUMMARY

The double integral definition of the natural logarithm, expressed as $$\ln n = \int_{0}^{\infty} \int_{1}^{n} e^{-xt} dx dt$$, can be proven by exchanging the order of integration. This method simplifies the calculation, allowing the first integral to be evaluated as $$\int_{0}^{+\infty} e^{-xt} dt = \frac{1}{x}$$, which converges despite being improper. The remaining integral, $$\int_{1}^{n} \frac{1}{x} dx$$, directly yields the result $$\log n$$, confirming the validity of the logarithm's definition.

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td21
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Where does this definition come from: $$\ln n = \int_{0}^{\infty} \int_{1}^{n} e^{-xt} dx dt$$
Thank you very much.
 
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Exchange the order of the integrals, and you can calculate them easily.
 
I don't know but it seems follows from elementary integrals, if you change the order (verify you can do it!) the first is ##\int_{0}^{+\infty}e^{-xt}d\,t=\frac{1}{x}## ( it is improper but converge ) and what remains is simply ##\int_{1}^{n}\frac{1}{x}d\,x## that is ##\log{n}##...
 

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