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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.

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  • $\begingroup$ For $\infty$ use \infty. $\endgroup$ – K.defaoite Jan 13 at 20:25
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    $\begingroup$ To define the Fourier transform of $\cos$, which is not in $L^1(\mathbb R)$, one has to use distribution theory. $\endgroup$ – md2perpe Jan 13 at 21:08
  • $\begingroup$ @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? $\endgroup$ – jay Jan 14 at 6:53
  • $\begingroup$ 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). $\endgroup$ – Svyatoslav Jan 14 at 10:56

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