# How does the Fourier Transform work??

I was trying to understand how the Fourier Transform works and wanted to test it on $$\cos(Ax)$$. I know that the FT for $$\cos(Ax)$$ is $$\delta(A)$$ and $$\delta(-A)$$. So I wanted to check if for other values of $$\omega$$ if the FT is zeros. (Please note, I'm still ramping up on creating the math equation. I apologize if something is improper)

The FT of $$\cos(x)$$ (so $$A = 1$$) is: $$F(\omega) = \int_{-\infty}^{\infty}\cos(x)e^{-jwt} dt$$ This implies, I'll have a $$\delta(x)$$ at $$x=1$$ and $$-1$$. For all other cases, it should be 0.  Hence, at $$\omega = 10$$ (10 times the frequency of the original), the FT should be zero...

\begin{align} F(10) &= \int_{-\infty}^{\infty}{(e^{jt}+e^{-jt})e^{-10jt} dt}\\ &= {\frac{1}{2}(e^{-9jt})|_{-\infty}^{\infty}+ \frac{1}{2}(e^{-11jt})|_{-\infty}^{\infty}}\end{align}

• For $\infty$ use \infty. – K.defaoite Jan 13 at 20:25
• To define the Fourier transform of $\cos$, which is not in $L^1(\mathbb R)$, one has to use distribution theory. – md2perpe Jan 13 at 21:08