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Is it possible to calculate the DC value of signals with undefinable area?

More specificaly, in the case of $x(t)=\frac{1}{t}$. $\int\limits_{-\infty}^\infty\frac1t \, \mathrm dt$ does not converge. Does that mean that its DC can not be determined?

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    $\begingroup$ i'll bet you're studying the Hilbert Transform. one thing you might want to look up is what is called the Principle value of an integral. regarding the Hilbert Transform: $$ \hat{x}(t) \triangleq \mathscr{H}\Big\{ x(t) \Big\} = \frac{1}{\pi} \ \mathbf{pv} \int\limits_{-\infty}^{\infty} \frac{x(\tau)}{t-\tau} \mathrm{d}\tau $$ $\endgroup$ Jan 11, 2019 at 21:08
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    $\begingroup$ If you use the principle value integral, the DC value of $\frac{1}{t}$ over all time is zero. $\endgroup$ Jan 11, 2019 at 21:11
  • $\begingroup$ it turns out that this question has another problem, besides the issues regarding the singularity of $1/t$ at $t=0$, DC is a finite power signal and an infinite energy signal, but $1/t$ is more like an energy signal (but has infinite energy). i think we can all tell that this is coming from the question: "What comes out of a Hilbert transformer with DC going in?" and the answer is "zero". but even if $x(t)$ was unipolar, if it's a finite energy signal, there is no DC in it anyway. $\endgroup$ Jan 12, 2019 at 0:35

2 Answers 2

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You need to be careful with the definition of the DC value of a signal. The actual time average, which is often called DC value is given by

$$\overline{s(t)}=\lim_{T\rightarrow\infty}\frac{1}{T}\int_{-T/2}^{T/2}s(t)dt\tag{1}$$

whereas the value of the signal's Fourier transform at $\omega=0$ is given by

$$S(0)=\int_{-\infty}^{\infty}s(t)dt\tag{2}$$

If $(2)$ is finite, $(1)$ equals zero, and if $(1)$ is finite but non-zero, $(2)$ doesn't exist, and the Fourier transform $S(\omega)$ has a Dirac impulse at $\omega=0$. See also this related answer.

For the given signal $s(t)=1/t$ both integrals in $(1)$ and $(2)$ do not converge in the conventional sense. However, their Cauchy principal value exists and equals zero, as mentioned in a comment by Robert Bristow-Johnson. Hence, the DC-value of $s(t)=1/t$ equals zero, regardless whether you define it by $(1)$ or by $(2)$.

Note that the Fourier transform of $s(t)=1/t$ equals

$$S(\omega)=-j\pi\,\textrm{sgn}(\omega)\tag{3}$$

which is just the frequency response of a Hilbert transformer (scaled by $\pi$).

From $(3)$, $S(0)$ doesn't exist, but the average of the left-sided and right-sided limits $\frac12 (S(0^-)+S(0^+))$ equals zero, and - using the Cauchy principal value according to the definition of the Hilbert transform - the Hilbert transform of a constant is indeed zero.

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try breaking the integral into 2 parts

$$ \int_{-\infty}^{0} f(t) dt + \int_{0}^{\infty} f(t) dt $$

and note that for an odd function, the integrals cancel each other.

edit:

as noted, not a rigorous solution but it satisfies intuition

Matt’s answer should be accepted.

Im not going to delete this answer because sometimes “wrong” answers are instructive.

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    $\begingroup$ That's not how you solve such an improper integral. In general, that integral doesn't converge. Only if you compute the Cauchy principal value does the integral evaluate to zero, otherwise it doesn't because you have to take the limits independently. $\endgroup$
    – Matt L.
    Jan 11, 2019 at 21:31
  • $\begingroup$ Stan, as usual Matt's right about this. $\endgroup$ Jan 11, 2019 at 21:50
  • $\begingroup$ though Matt's answer is formally right, imho the concept of oddity seems sufficient to conclude the result here... afaik we accept $\int_{-\infty}^{\infty} \sin(x)dx = 0$ by just referring to oddity (the limit of the integral in any half does not exist) alone... $\endgroup$
    – Fat32
    Jan 11, 2019 at 23:12
  • $\begingroup$ // ... because sometimes “wrong” answers are instructive.// i agree with you Stan. As I further noted above, there are other issues (besides this Cauchy Principal Value thingie, that makes this question a bit problematic. $\endgroup$ Jan 12, 2019 at 0:39
  • $\begingroup$ i wonder if there is a case where oddity wasn’t sufficient. just curious $\endgroup$
    – user28715
    Jan 12, 2019 at 1:32

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