The title of this post is intentionally sensational, but what I am really going to do is to compare the divergent integrals $\int_0^1\frac1xdx$ and $\int_1^\infty\frac1xdx$.
Let's consider the transform $\mathcal{L}_t[t f(t)](x)$. It is notable by the fact that it preserves the area under the curve: $$\int_0^\infty f(x)dx=\int_0^\infty \mathcal{L}_t[t f(t)](x) dx$$
But more interesting, it also works well with divergent integrals, allowing us to define the equivalence classes of divergent integrals. Particularly, by successfully applying this transform to $\int_1^\infty \frac1xdx=\int_0^\infty\frac{\theta (x-1)}{x}dx$, one can obtain the following equivalence class of divergent integrals (the first one and the third one turn out to be similar up to a shift):
$\int_0^\infty\frac{\theta (x-1)}{x}dx=\int_0^\infty\frac{e^{-x}}{x}dx=\int_0^\infty\frac{dx}{x+1}=\int_0^\infty\frac{e^x x \text{Ei}(-x)+1}{x}dx=\int_0^\infty\frac{x-\ln x-1}{(x-1)^2}dx$
On the other hand, applying the transform to $\int_0^1\frac1x dx=\int_0^\infty \frac{\theta (1-x)}{x}dx$ one can obtain another set of equal integrals:
$\int_0^\infty\frac{\theta (1-x)}{x}dx=\int_0^\infty\frac{1-e^{-x}}{x}dx=\int_0^\infty\frac{1}{x^2+x}dx=\int_0^\infty-e^x \text{Ei}(-x)dx=\int_0^\infty-\frac{x-x\ln x-1}{(x-1)^2 x}dx$
Now, having postulated equivalence of divergent integrals in each class, we can pick two integrals, one from each class and compare them. Well, it seems, the integrals in the second class are bigger by an Euler's constant:
$\int_0^{\infty } \left(\frac{1-e^{-x}}{x}-\frac{1}{x+1}\right) \, dx=\gamma$
And thus, we can conclude that $\int_0^1\frac1xdx=\gamma+\int_1^\infty \frac1xdx$. Surprising, is not it, given that one would naively expect $\ln 0=-\ln\infty$?
But we also have an identity $\gamma = \lim_{n\to\infty}\left(\sum_{k=1}^n \frac1{k}-\int_1^n\frac1t dt\right)$ and know the regularized value of harmonic series $\operatorname{reg}\sum_{k=1}^n \frac1{k}=\gamma$. Thus, we obtain $$\int_0^1\frac1xdx=\gamma+\int_1^\infty\frac1xdx=\sum_{k=1}^\infty\frac1k.$$
So, the integrals differ by Euler-Mascheroni constant. Is there any logical explanation for this?
They are equal: $\ln 0 = -\infty$ and $-\ln \infty = -\infty$.
Not at all. $$-\infty = \gamma - \infty.$$ You see, $\infty$ is infinite. And this sort of thing is exactly what makes "infinite" different from "finite".
And per user's comments, it is also true that $-\infty = 1 - \infty$. That too is part of infinity's shtick.
There is no logical inconsistency here. The only problem is that you are trying to enforce intuition developed from finite behavior on infinite behavior where it does not hold.
If infinity didn't behave like this, we would have just added it to the real numbers, just like we added $0$ and negative numbers in the past, and even added a square root of $-1$, though for that one we were at least willing to admit the result deserved its own separate name.
But we don't add $\infty$ to the real numbers, and even tell people that "infinity doesn't exist" for years, as if we couldn't just define it, exactly because it doesn't play well with other numbers. And it is more of pain to deal with all the exceptional behavior than it is to just kick infinity out of the playground.
Only after someone develops the mathematical chops to recognize how it needs to be handled do we admit to them that yes, infinity does exist after all.