How do you prove that $\ln(x) = \int_0^\infty \frac{e^{-t}-e^{-xt}}{t}$?

Question

I got the following result using the technique "Integral Milking":

$$\ln(x) = \int_0^\infty \frac{e^{-t}-e^{-xt}}{t} dt= \lim_{n\to0}\left(\operatorname{Ei}(-xn)-\operatorname{Ei}(-n)\right)$$

for $x > 0$. So, I have a proof of it the result, but now I would like to know how to prove starting with either the integral or the limit. I'm not that familiar with the exponenential integral $\operatorname{Ei}(x)$, so my attempts were pretty bad (I'm not so familiar with the technique, but I'm still going to try to differentiate under the integral sign). Personally I have never seen an integral representation of the natural logarithm like this before, and I can't find it anywhere (e.g. here), but WolframAlpha directly gets the limit right.

Question: How do you prove the the integral (or limit) is equal to $\ln(x)$ by starting with the integral (and not using e.g. integral milking)?

Answer 1

The fastest way to prove this is using Frullani's theorem, as you've seen in the comments, but you can also use a double integral to quickly solve this problem (which is one method how Frullani's theorem is proven). $$I=\int_0^{\infty} \frac{e^{-t}-e^{-tx}}{t} \; dt = \int_0^{\infty} \frac{1}{t}\int_{1}^{x} te^{-tw} \; dw \; dt$$ Where $w$ and $t$ are dummy variables. Now switch the order of integration: $$I=\int_1^x \int_0^{\infty} \frac{1}{t} \cdot te^{-tw} \; dt \; dw= \int_1^x \int_0^{\infty} e^{-tw} \; dw \; dt\int_1^x \frac{1}{w} \; dw = \boxed{\ln{x}}$$

You can generalize this to the $\int_0^{\infty} \frac{f(ax)-f(bx)}{x} \; dx$. Following these steps, you should see that the generalized integral is just $$[f(\infty)-f(0)] \ln{\left(\frac{b}{a}\right)}$$

Answer 2

Well, we have the following integral:

$$\mathcal{I}_\text{n}\left(\alpha,\beta\right):=\int_0^\infty\frac{\exp\left(-\alpha x\right)-\exp\left(-\beta x\right)}{x^\text{n}}\space\text{d}x\tag1$$

Using the lineairy of the integral, we can write:

$$\mathcal{I}_\text{n}\left(\alpha,\beta\right)=\underbrace{\int_0^\infty\frac{\exp\left(-\alpha x\right)}{x^\text{n}}\space\text{d}x}_{=\space\text{I}_\text{n}\left(\alpha\right)}-\int_0^\infty\frac{\exp\left(-\beta x\right)}{x^\text{n}}\space\text{d}x\tag2$$

Now, we look at $\text{I}$. Using rules for exponents:

$$\text{I}_\text{n}\left(\alpha\right)=\int_0^\infty x^{-\text{n}}\exp\left(-\alpha x\right)\space\text{d}x\tag3$$

Now, looking at the definition of the Laplace transform we can observe:

$$\text{I}_\text{n}\left(\alpha\right)=\mathcal{L}_x\left[x^{-\text{n}}\right]_{\left(\alpha\right)}\tag4$$

Using the table of selected Laplace transforms, we can see:

$$\text{I}_\text{n}\left(\alpha\right)=\frac{\Gamma\left(1-\text{n}\right)}{\alpha^{1-\text{n}}}\tag5$$

Now, we have your original integral:

$$\mathcal{I}_\text{n}\left(\alpha,\beta\right)=\text{I}_\text{n}\left(\alpha\right)-\text{I}_\text{n}\left(\beta\right)=\frac{\Gamma\left(1-\text{n}\right)}{\alpha^{1-\text{n}}}-\frac{\Gamma\left(1-\text{n}\right)}{\beta^{1-\text{n}}}=\Gamma\left(1-\text{n}\right)\left(\alpha^{\text{n}-1}-\beta^{\text{n}-1}\right)\tag6$$

I let you prove:

$$\mathcal{I}_1\left(\alpha,\beta\right)=\lim_{\text{n}\to1}\mathcal{I}_\text{n}\left(\alpha,\beta\right)=\ln\left(\beta\right)-\ln\left(\alpha\right)$$