# How the Fourier transform of a cosine signal is existed? [duplicate]

This question already has an answer here:

As I know, if an aperiodic continuous-time signal be absolutely integrable, i.e.

$$\int\limits_{-\infty}^\infty \vert x(t) \vert \ dt \ < \ \infty$$

its Fourier transform is existed.

Also, the Fourier transform of $\cos(\omega_0 t)$ is $\pi(\delta(\omega-\omega_0)+\delta(\omega+\omega_0))$.

Now, my question is that: how the Fourier transform of a cosine signal is the above relation, while, this signal is not integrable !?

## marked as duplicate by Matt L., Gilles, MBaz, Laurent Duval, Peter K.♦Aug 23 '16 at 12:51

• Note that integrability is not necessary for the Fourier transform to exist. Take the sinc function as an example: it is not absolutely integrable but its Fourier transform exists (it is a rectangular function). You can take things even further by allowing distributions (such as the Dirac delta impulse); then non-decaying functions (like the step function $u(t)$ and sinusoidal functions) can be transformed. – Matt L. Aug 21 '16 at 21:01
• Thanks a lot for your convincing response. I did not notice that the absolutely integrability is a sufficient condition for existing the Fourier transform. – AllEs Aug 22 '16 at 6:33

The Fourier Transform is defined only for functions that are absolutely integrable or in other words $f\in\mathcal{L}_1$, where $\mathcal{L}_1$ is the set of all absolutely integrable functions.

If you want to be mathematically rigorous, then you should assume that the Fourier transform of the cosine does not exist, as it is not integrable. Apart from it, the Dirac's delta does not fulfill the axioms of a function in modern mathematics, so you cannot say that a Fourier Transform of a function is something that is not a function. Instead of that, the Dirac's delta is defined as a measure that can be used in the Lebesgue integral, that satisfies certain axioms(https://en.wikipedia.org/wiki/Dirac_delta_function). However, the use of the cosine is valid for the common modulation operation, and is mathematically consistent (meaning that its Fourier transform exist):

$\int_{-\infty}^{\infty} |x(t)cos(2\pi t)e^{-i2\pi ft}|dt\leq \int_{-\infty}^{\infty} |x(t)|dt < \infty$

However, if you are just interested in the engineering or physics point of view, you can use the Dirac delta "function" as a trick that works as the measure in order to get the operations you are interested to do. For the sinc function and square integrable functions, a Fourier Transform over the $\mathcal{L}_2$ space, which is the space of square integrable functions, is defined so that you can get the Fourier Transforms for those functions(http://math.mit.edu/~jerison/103/handouts/fourierint1.13.pdf), but if you are just interested in the engineering point of view, it is not important to be so rigorous mathematically.

• Thanks a lot. Your response was very comprehensive. However, I believe that you did not consider that the integrability is not the necessary condition for existing the Fourier transform. It is sufficient condition and it means that a signal could have the Fourier transform even if it does not meet this condition! – AllEs Aug 22 '16 at 10:05
• @AllEs it is a necessary condition for the uniformly convergent Fourier Transform. When a signal is not absolutely but square integrable, then its Fourier transform will be a discontinuous function (which cannnot be the output of a classical Riemann integral transform which is always to be continuous - smooth) such as an ideal low-pass filter frequency response whose convergence is in the mean square error sense. (not convergent at every point of $w$ but the power of error over the whole domain goes to zero) – Fat32 Aug 22 '16 at 10:34

As you say, the cosine signal is not absolutely integrable. However, the product of the cosine signal and the exponential in the definition of the FT is absolutely integrable as you can ensure that this product converges.

you can check the FT step by step in the following link:

FT of cosine step by step

• This is not correct. Unlike the Laplace transform, the exponential in the definition of the Fourier transform has no damping, and, consequently, it doesn't change the integrability of the function. – Matt L. Aug 22 '16 at 9:05