# DC value of $x(t)=\frac{1}{t}$

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?

• 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: $$\hat{x}(t) \triangleq \mathscr{H}\Big\{ x(t) \Big\} = \frac{1}{\pi} \ \mathbf{pv} \int\limits_{-\infty}^{\infty} \frac{x(\tau)}{t-\tau} \mathrm{d}\tau$$ Jan 11, 2019 at 21:08
• If you use the principle value integral, the DC value of $\frac{1}{t}$ over all time is zero. Jan 11, 2019 at 21:11
• 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. Jan 12, 2019 at 0:35

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.

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

• 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... Jan 11, 2019 at 23:12