# 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}

But I'm stuck here and it looks like the answer is non-zero. From this page, I'm getting a non-zero result. Could someone please help in solving this integral? Also, please correct me if my assumption is totally wrong.

• 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
• @K.defaoite will do. @ md2perpe, the link was enough for me to give up on this apparent rabbit hole immediately. Any page which concentrates on the cosine directly? – jay Jan 14 at 6:53
• Delta function makes sense as an integral with other functions. This is its meaning - to "cut out" one point in a smooth function when integrating the product of a delta function and a smooth function. If you are trying to get an "explicit" delta function - it can be a function that does not necessarily equal zero at other points. For example, it can be a highly oscillating function - like the expressions you got. Integral from these strongly oscillating functions with a smooth function will give zero over any interval not including the singularity point (i.e., the zero point in your case). – Svyatoslav Jan 14 at 10:56