Fourier transformation of cosine

Question

I am trying to do the Fourier transformation of the expression $x(t) = 10 \cos(50\pi t)$; I have used the Euler representation to get as far as:

$$5\int_{-\infty}^{\infty}e^{-j50\pi t}e^{-j50\pi t}dt$$

Answer 1

Strictly speaking, this function does not have a Fourier transform. However, if we allow for more generalized functions, the following formula should get you to your answer: $$ \mathcal F\left[e^{i\omega_0 t}\right] = \int_{-\infty}^\infty e^{i\omega_0t}e^{-i\omega t}dt = 2\pi\delta(\omega - \omega_0), $$ where $\delta$ is the Dirac delta function.

EDIT: So let me go into more detail about the delta function. The delta function is defined such that $\int_{-\infty}^\infty\delta(x)dx = 1$ and $\delta(x) = 0$ for all $x\ne0$*. Most importantly, the delta function has the property that $\int_{-\infty}^\infty f(x)\delta(x-x_0)dx = f(x_0)$. It is this property that makes the above formula work. If you apply the inverse Fourier transform to $2\pi\delta(\omega - \omega_0)$, you'll see it recovers $e^{i\omega t}$.

*You might notice at this point that no function could have such properties, and you'd be right: the delta function is not a function. But it has many of the same properties of a function so we call it that anyways.