I have already asked a similar question, albeit far more extensive, but it was criticized and closed for being too extensive and promotional. So, here is a greatly truncated and focused version.
Since there are Laplace-based transforms that keep the area under the integral of the function intact, it is reasonable to generalize this to assume that the divergent integrals are kept equal as well under the transform and its inverse:
$$\int_0^\infty f(x)dx=\int_0^\infty\mathcal{L}_t[t f(t)](x)dx=\int_0^\infty\frac1x\mathcal{L}^{-1}_t[ f(t)](x)dx$$
This gives several equivalence classes of divergent integrals that we under the above assumption consider the same. Their regularized values are also preserved. So, it is possible to treat them as various integral representations of infinite "constants".
So here is a short list of the classes, and I wonder, where they can be encountered in areas of math and/or physics:
$\int_0^{\infty } \, dx=\int_0^{\infty } \frac{1}{x^2} \, dx$
$\int_{1/2}^\infty dx=\int_0^\infty \frac{e^{-\frac{x}{2}} (x+2)}{2 x^2}dx=\int_0^2\frac1{x^2} dx$
$\int_{-1/2}^\infty dx=\int_0^\infty \frac{4-e^{-\frac{x}{2}} (x+2)}{2 x^2} dx$
$\int_0^1 \frac1x dx-\gamma=\int_1^\infty \frac1xdx=\int_0^\infty\frac{e^{-x}}{x}dx=\int_0^\infty\frac{dx}{x+1}=\int_0^\infty\frac{e^x x \text{Ei}(-x)+1}{x}dx=\int_0^\infty\frac{x-\ln x-1}{(x-1)^2}dx$
$\int_0^1 \frac1x dx=\int_0^\infty\frac{1-e^{-x}}{x}dx=\int_0^\infty\frac{1}{x^2+x}dx=-\int_0^\infty e^x \text{Ei}(-x)dx=\int_0^\infty\frac{x\ln x-x+1}{(x-1)^2 x}dx$
$\int_1^\infty \sqrt{x^2-1}dx=\int_0^\infty \frac{K_2(x)}{x}dx=\int_0^\infty \left(x-\frac{1}{2 x}\right) dx =\int_0^\infty \left(\frac{2}{x^3}-\frac{1}{2 x}\right)dx$
Are there other notable divergent integrals that appear persistently?
