# Running Integral of sine and cosine functions

In typical signal processing course we were taught that the integral of signal $$x(t)$$ is given by $$y(t) = \int_{-\infty}^{t}x(\tau) d\tau$$ How can we use this definition to evaluate the integrals of eternal sinusoids, $$x(t) = \sin (t)$$ or $$x(t) =\cos(t)$$. I mean how do you evaluate / prove that $$\int_{-\infty}^{t}\cos(\tau)d\tau = \sin(t)$$

Thanks in advance.

Edit: I raised this question because of its (ab)use in frequency modulation(FM) for single tone signal. To give details, any FM wave of single tone signal $$m(t) = \cos(\omega_0 t)$$ is given by \begin{align} \Phi_{FM}(t) & = A\cos\left(\omega_c t + K_f\int_{-\infty}^{t}\cos(\omega_0 \tau)d\tau\right) \end{align} It can be written as \begin{align} \hat{\Phi}_{FM}(t) & = A\: Re\left\{\exp\left(j\omega_c t + jK_f\int_{-\infty}^{t}\cos(\omega_0 \tau)d\tau\right)\right\} \end{align} or \begin{align} \hat{\Phi}_{FM}(t) & = A\:Re\left\{\exp\left(j\omega_c t\right) \exp \left(j K_f\int_{-\infty}^{t}\cos(\omega_0 \tau)d\tau\right)\right\} \end{align} In communication systems textbooks, the argument of exponential term involving cosine term is magically written as $$\dfrac{K_f}{\omega_0} \sin (\omega_0 t)$$ which really fascinates me.

Later they claim this is periodic signal and hence can be represented as Fourier series expansion involving Bessel functions etc.,

As I mentioned it is really magic and am interested in knowing this magic trick. Hence this post. Sorry for being so long....

Thanks to all of you.

• Thanks for the correction. – user5045 Apr 30 '20 at 8:21
• um, in what context does that problem arise? $\lim\limits_{\tau\to-\infty}\cos(\tau)$ is undefined, so in general, you can't. – Marcus Müller Apr 30 '20 at 8:51
• en.wikipedia.org/wiki/Cauchy_principal_value – Jazzmaniac Apr 30 '20 at 9:21
• @Jazzmaniac um, I might be missing something, but $\cos$ has no singularities anywhere, so how does Cauchy's principal value come into play here? – Marcus Müller Apr 30 '20 at 10:05
• Look at first two lines on page 7 personal.utdallas.edu/~saquib/EE4365/FM-Modulation.pdf Hope it helps. – user5045 Apr 30 '20 at 10:49

## 2 Answers

I would not teach integrals this way.

Issues might arise form the definition of terms. In standard calculus, a (note: "a", not "the) primitive function (also called antiderivative or indefinite integral) of a continuous function $$f$$ is a kind of converse of the concept of derivation. SO:

If $$F$$ is "a" differentiable function whose derivative is equal to $$f$$, then $$F$$ is a primitive (function) of $$f$$. There exists many such functions, up to a constant $$c$$, because: $$(F(t)+c)' = (F(t))' =f(t)$$.

Here, you can check that $$t\mapsto \sin(t)+ c$$ is differentiable, with derivative $$t\mapsto \cos(t)$$.

Now (if I remember well), suppose that $$f(x)$$ is a continuous function, AND that it is "integrable" over $$\mathbb{R}$$, then $$F(t) = \int_{-\infty}^t f(\tau)d\tau\;$$ exists, and is a differentiable function. THEN, it is also a primitive of $$f(t)$$ (among others).

But here, it is not possible, namely because $$t\mapsto \cos(t)$$ is not "integrable" over $$\mathbb{R}$$, and thus the integral expression makes no sense.

I would say that one should not define "integrals" the way you wrote it, if one wants mathematical correctness. However, for causal signals, in a lousy engineer perspective, it may be intuitive. But the cosine is not causal.

[Note: there are more general definitions of primitives]

The integration in your question is equivalent to convolution with the unit step function:

$$y(t)=\int_{-\infty}^tx(\tau)d\tau=(x\star u)(t)\tag{1}$$

This means that in the Fourier domain we have

$$Y(j\omega)=X(j\omega)U(j\omega)\tag{2}$$

With

$$U(j\omega)=\frac{1}{j\omega}+\pi\delta(\omega)\tag{3}$$

Eq. $$(2)$$ becomes

$$Y(j\omega)=\frac{X(j\omega)}{j\omega}+\pi X(0)\delta(\omega)\tag{4}$$

For a sinusoidal function $$x(t)=\cos(\omega_0t)$$, $$\omega_0>0$$, we have

$$X(j\omega)=\pi\big[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)\big]\tag{5}$$

and from $$(4)$$ we obtain

\begin{align}Y(j\omega)&=\frac{\pi}{j\omega}\big[\delta(\omega-\omega_0)+\delta(\omega+\omega_0)\big]\\&=\frac{\pi}{j\omega_0}\big[\delta(\omega-\omega_0)-\delta(\omega+\omega_0)\big]\tag{6}\end{align}

because $$X(0)=0$$ and because $$f(x)\delta(x-x_0)=f(x_0)$$ if $$f(x)$$ is continuous at $$x=x_0$$.

From $$(6)$$ we obtain

$$y(t)=\frac{1}{\omega_0}\sin(\omega_0t)\tag{7}$$

We should be able to obtain the result $$(7)$$ also directly from $$(1)$$:

\begin{align}y(t)&=\int_{-\infty}^t\cos(\omega_0\tau)d\tau\\&=\frac{1}{\omega_0}\big[\sin(\omega_0t)-\lim_{t\to-\infty}\sin(\omega_0t)\big]\tag{8}\end{align}

Clearly, we would require $$\lim_{t\to-\infty}\sin(\omega_0t)=0$$ in order to obtain the result $$(7)$$. That limit, however, doesn't exist. More precisely, it doesn't exist in the conventional sense. But since we've used generalized functions (distributions) in the Fourier domain to obtain the result, we also have to be a bit generous in the time domain. As a generalized limit, we have

$$\lim_{t\to\infty}\sin(\omega_0t)=\lim_{t\to\infty}\cos(\omega_0t)=0\tag{9}$$

Eq. $$(9)$$ means that for a well-behaved function $$f(x)$$ ($$L^1$$-integrable, finite) the following holds:

$$\lim_{t\to\infty}\int_{-\infty}^{\infty}f(x)e^{-jx t}dx=0\tag{10}$$

Eq. $$(10)$$ is called the Riemann-Lebesgue-lemma.

In sum, for the results obtained in the time domain and via the Fourier transform to be identical we need to consider generalized limits when computing the improper integral.

• Climbed by the North face of the Fourier "Dirac" Mountain! The hard way:) – Laurent Duval Apr 30 '20 at 10:36
• @LaurentDuval: Well, I think both ways are hard in that case, but it's satisfying that no matter which path we choose, we end up with the same result :) – Matt L. Apr 30 '20 at 10:38
• The converse would have been very unfortunate, for the sake of Science! – Laurent Duval Apr 30 '20 at 10:39
• Can you please relate Eq. (10) with what is obtained in Eq. (9)? – user5045 Apr 30 '20 at 14:17
• @user5045: That is how the limit must be interpreted, namely under an integral (just like a Dirac delta impulse). Note that $e^{-jxt}=\cos(xt)-j\sin(xt)$, so from (10) it also follows that the limits $\int f(x)\cos(xt)dx$ and $\int f(x)\sin(xt)dx$ are zero. And that is exactly how the limits in (9) must be interpreted. – Matt L. Apr 30 '20 at 14:46