When playing with divergent integrals $\int_0^\infty f(x) \, dx$ and their transformations with operators $\int_0^\infty\mathcal{L}_t[t f(t)](x) \, dx$ and $\int_0^\infty\frac1x\mathcal{L}^{-1}_t[ f(t)](x) \, dx$, which preserve the area under the curve (that is, the values of the integrals), and as such, define the equivalence classes of divergent integrals, I have arrived at these interesting results:
$1.\int_0^\infty \frac{1}{x} \, dx=\gamma +2 H$
$2.\int_0^\infty \log (x) \, dx=\tau (H-\gamma-1)$
$3. \int_0^\infty \frac{\log (x)}{x} \, dx=-\frac{\gamma ^2}{2}-\gamma H$
$4. \int_0^\infty \frac{\log (x)}{x^2} \, dx=(2-H) \tau$
where $H=\int_1^\infty \frac1x \, dx$ and $\tau=\int_0^\infty dx$, are simpler divergent integrals, with the regularized value $0$.
It follows thus, that the regularized value of $\int_0^\infty \frac{\log (x)}{x} \, dx$ is $-\frac{\gamma^2}{2}$. On the other hand, the regularized value of $H\tau$ is a mystery for me and I do not know whether one constant can be expressed via the other. As such, the regularized values of integrals $(2)$ and $(4)$ are unknown, except they should be equal by modulus but of the opposite sign.
I tried Borel and other kinds of regularization on the both in Mathematica, with no results so far. I am very curious, what the regularized value could be.
