What is this function? $f(x)=\frac12\int_{-\infty}^\infty \frac{{\rm sgn} (x-t)}t dt$?

Question

What is $f(x)=\frac12\int_{-\infty}^\infty \frac{\operatorname{sgn} (x-t)}t dt$ in closed form?

The integral should be understood as the Cauchy principal value and regularized values when diverges.

The question arose in the context of natural integral.

It seems to me, the correct answer is $f(x)=\ln |x|$, but the Fourier transform method using Mathematica gives $f(x)=\ln |x| - \gamma$.

Answer 1

if we split the integral we get: $$\int_{-\infty}^x\frac{1}{t}dt - \int_{x}^{\infty}\frac{1}{t}dt$$ using the Cauchy principal we get: $$\lim\limits_{R \to \infty}\int_{-R}^x\frac{1}{t}dt - \int_{x}^{R}\frac{1}{t}dt$$ $$\lim\limits_{R \to \infty}(ln(|x|)-ln(|-R|))-(ln(|R|)-ln (x))$$ R cancels out and we get: $$\frac{1}{2}\int_{-\infty}^{\infty}\frac{sgn(x-t)}{t}dt=ln(|x|)$$

Answer 2

As noted bt razivo, this integral is $$ \int_{-\infty}^x\frac{1}{t}dt - \int_{x}^{\infty}\frac{1}{t}dt $$ It is divergent. It diverges at $-\infty$ since it behaves like $1/t$ there. It diverges at $+\infty$ since it behaves like $-1/t$ there. It diverges at $0$ since it behaves like $1/t$ or $-1/t$ (or both, one on each side) there.

The correct answer to the question is: the integral diverges.

Now solver razivo proposes to find a "principal value" for it. His method attempts to take care of the divergence at $\pm\infty$, but ignores the divergence at $0$.

Here is the integrand, when $x=1$:

There is a jump discontinuity at $x$. A "principal value" computation near $t=0$ will cancel the two sides, giving us a value. But a "principal value" computation at $\pm \infty$ will not cancel. Both sides have integral $-\infty$. So: $$ \mathrm{P.V.}\int_{-\infty}^{\infty}\frac{\operatorname{sgn}(x-t)}{t}dt=-\infty $$ This calculation is for the case $x \ne 0$.

Here is the integrand when $x=0$.

This is even worse. Not only does the integral diverge to $-\infty$ at $\pm\infty$, preventing the cancellation needed for a principal value, it also diverges to $-\infty$ at $0$, preventing the cancellation there.

But again: $$ \mathrm{P.V.}\int_{-\infty}^{\infty}\frac{\operatorname{sgn}(0-t)}{t}dt=-\infty $$

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Regularized? I'm guessing this is what Anixx means (but there are many other possible meanings).

Assume $x>0$. First, $$ \mathrm{P.V.} \int_{-1}^1 \frac{dt}{t} = 0, \tag1$$ Next, compute for $s>1$, $$ \int_{-\infty}^{-1} \frac{-dt}{(-t)^s} = \frac{1}{s-1} $$ The Laurent series for this in powers of $s-1$ has constant term $0$, so we substitute $s=1$ and say $$ \mathrm{Reg}\int_{-\infty}^{-1} \frac{dt}{t} = 0 \tag2$$

Next, $$ \int_1^\infty \frac{\operatorname{sgn}(x-t)}{t^s}\;dt =\int_{1}^x\frac{1}{t^s}dt - \int_{x}^{\infty}\frac{-1}{t^s}dt = \frac{1-2x^{1-s}}{s-1} $$ The Laurent series for this in powers of $s-1$ is $$ \frac{-1}{s-1} + 2\log x + O(s-1) . $$ So we substitute $s=1$ and say $$ \operatorname{Reg}\int_1^{+\infty}\frac{\operatorname{sgn}(x-t)}{t^s}\;dt = 2\log x. \tag3$$

Now, adding $(1), (2), (3)$ we have $$ \operatorname{Reg\ P.V.}\int_{-\infty}^{+\infty}\frac{\operatorname{sgn}(x-t)}{t}\;dt = 2\log x. $$

Similarly, for $x<0$ we get $$ \operatorname{Reg\ P.V.}\int_{-\infty}^{+\infty}\frac{\operatorname{sgn}(x-t)}{t}\;dt = 2\log (-x). $$

And for $x=0$ we have to regularize also at $t=0$. I guess we now take $s<1$ to get convergence of $$ \int_{-1}^1\frac{-1}{|t|^s}\;dt = \frac{2}{s-1} $$ with zero constant term, so $$ \mathrm{Reg}\int_{-\infty}^{+\infty}\frac{-1}{|t|}\;dt = 0 $$ However, it does seem strange to regularize with $s>1$ at $\pm \infty$ and with $s<1$ at $0$.