Are there any good examples from other math fields or intuition supporting $\int_0^1\frac1xdx=\int_1^\infty\frac1xdx$?

Question

This question is related to the potential possibilities of classification of divergent integrals more precisely than just "divergent to infinity" and the like. Improper divergent integrals can have many distinguishing properties, like germ at infinity, rate of grow and regularized value. A separate question is whether we can talk about equivalence (in whatever sense) of integrals divergent at infinity and those divergent at a singularity or pole.

For instance, formally applying the Laplace transform we can establish the following formal relations (for $n>0$):

$$\int_0^1 x^{-1+n} dx=\int_1^\infty n!x^{-1-n}dx$$

and

$$\int_1^\infty x^{-1+n} dx=\int_0^1 n!x^{-1-n}dx$$

This works for convergent integrals:

$$\int_0^1 dx=\int_1^\infty \frac1{x^2}dx=1$$

If generalized to divergent integrals, it also seems to work well, the equality

$$\int_1^\infty dx=\int_0^1 \frac1{x^2}dx$$

seems to be supported well by the hyperfunction theory.

But with $n=0$ we have (in the both formulas):

$$\int_0^1 \frac1x dx=\int_1^\infty \frac1x dx$$

This looks counter-intuitive at the first glance, as the figures under the integrals are identical up to a reflection relative the $y=x$ axis, except the first one has an additional $1\times 1$ square.

Of course, since the integrals are divergent, it should not be an obstacle to the identity (as the positions of the figures and the limits of integration are different). The both integrals are regularizable to the value $0$.

So, I wonder, if there is any intuition or example from other fields or mathematics or physics where these integrals demonstrate equivalence in a more strict sense than being both divergent to infinity?

For instance, from geometry or probability theory.

UPDATE

It seems, the equality should not hold, and the more correct one would be $$\int_0^1 \frac1x dx=\gamma+\int_1^\infty \frac1x dx$$

Answer 1

Okay, so these integrals are not equal. Instead, $\int_0^1 \frac1x dx=\gamma+\int_1^\infty \frac1x dx$. Let's see it.

As others pointed out, the integral diverges. Its regularized value is $\gamma$, the Euler-Mascheroni constant.

Let us break down the integral into two parts: $\int_0^\infty \frac1x dx=\int_0^1 \frac1x dx +\int_1^\infty \frac1x dx$.

Now, using the following transformation: $\int_0^\infty f(x)\,dx=\int_0^\infty\mathcal{L}_t[t f(t)](x) \, dx=\int_0^\infty\frac1x\mathcal{L}^{-1}_t[ f(t)](x)\,dx$ involving Laplace transform we can convert each part into equivalent integrals. For the first one we get the following set:

$\int_1^\infty\frac{1}{x}dx=\int_0^\infty\frac{1-e^{-x}}{x}dx=\int_0^\infty\frac{1}{x^2+x}dx=\int_0^\infty-e^x \text{Ei}(-x)dx=\int_0^\infty-\frac{x-x\ln x-1}{(x-1)^2 x}dx$

For the second one we get:

$\int_1^\infty\frac{1}{x}dx=\int_0^\infty\frac{e^{-x}}{x}dx=\int_0^\infty\frac{dx}{x+1}=\int_0^\infty\frac{e^x x \text{Ei}(-x)+1}{x}dx=\int_0^\infty\frac{x-\ln x-1}{(x-1)^2}dx$

We can pick one integral from the first set, (let it be $\int_0^\infty\frac{1-e^{-x}}{x}dx$) and one integral from the second set (let it be $\int_0^\infty\frac{dx}{x+1}$) and find their difference: $\int_0^\infty\frac{1-e^{-x}}{x}dx-\int_0^\infty\frac{dx}{x+1}=\int_0^\infty\left(\frac{1-e^{-x}}{x}-\frac{1}{x+1}\right)dx.$ This difference is equal to Euler-Mascheroni constant $\gamma$, as can be seen from the plot below:

So, $\int_0^1 \frac 1xdx=\int_1^\infty \frac1x dx+\gamma$ and the whole integral is $\int_0^\infty \frac1x dx=2\int_1^\infty \frac1x dx+\gamma$.

We also know that the harmonic series $\sum_1^\infty \frac1x dx$ has regularized value $\gamma$ and the following identity is known: $\sum_{k=1}^\infty \frac1k-\int_1^\infty \frac1x dx=\int_1^\infty\left(\frac1{\lfloor x\rfloor}-\frac1x\right)\,dx=\gamma.$

So, $\sum_{k=1}^\infty \frac1k =\int_0^1 \frac1x dx$, which is not surprising because this series is the Riemann sum for the integral.

Below are the plots of $\int_0^\infty\frac{1-e^{-x}}{x}dx$ (blue), $\int_0^\infty\frac{dx}{x+1}$ (orange) and $\int_0^\infty\frac1{\lfloor x\rfloor}dx$ (green). Tthe last one is equal to the series, shifted left by 1/2:

The areas between green and orange and blue and orange lines both are equal to $\gamma$. Since $\sum_1^\infty \frac1x dx$ has regularized value $\gamma$, we can conclude that $\int_1^\infty \frac1x dx$ has regularized value zero, and so our whole integral $\int_0^\infty\frac1x dx=2\int_1^\infty \frac1x dx+\gamma$ has regularized value $\gamma$.

Answer 2

Your paradox, as I understand it, concerns the correspondence between the divergent integral and the geometric area. Let's be a little more careful and define finite regions which are symmetric about $y=x$ before considering the limit as the regions approach the $x$- and $y$-axes.

There are two such areas we can define. First, there is $A_1$, between $y=1$ and $y=1/x$ on the interval $\epsilon < x< 1$, given by $$A_1 = \int_\epsilon^1 dx\left(\frac1x-1\right) = \int_\epsilon^1 \frac{dx}{x} +\epsilon-1.$$ Here, $0<\epsilon<1$ is a small number.

If we reflect this region about $y=x$, we get $$ A_1 = \int_1^{\epsilon^{-1}} dx\left(\frac1x-\epsilon\right) = \int_1^{\epsilon^{-1}} \frac{dx}{x} + \epsilon-1.$$

A second area can be defined as that between $x=1,$ $x=\epsilon^{-1}$, $y=1/x$, and the $x$-axis: $$ A_2 = \int_1^{\epsilon^{-1}} \frac{dx}{x}.$$

The reflection about $y=x$ is $$ A_2 = \int_0^\epsilon dx (\epsilon^{-1}-1) + \int_\epsilon^1 dx \left(\frac1x-1\right) = \int_\epsilon^1 \frac{dx}{x}.$$

So, in both cases, the geometric areas agree after reflection about $y=x$ if and only if the divergent integrals agree as claimed.

I believe the confusion arises because there are two relevant geometric areas, and which one you get depends on how you take the limit. If you imagine replacing the integral with a finite sum of rectangles, the height of the rectangles may or may not diverge depending on how they are oriented, and this leads to two different answers.