MHB Solving wave equation using Fourier Transform

spideyjj1
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I am having trouble with doing the inverse Fourier transform. Although I can find some solutions online, I don't really understand what was going on, especially the part that inverse Fourier transform of cosine function somehow becomes some dirac delta. I've been stuck on it for 2 hrs...
 
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The Fourier transform of f is defined by $F(s)=\int_{-\infty}^{+\infty}f(t)e^{-i2\pi st}dt$. if f(t)=1 let $F_{1}$ be it's Fourier transform for $s\neq0$ you get
$F_{1}(s)=\int_{-\infty}^{+\infty}e^{-i2\pi st}dt=0$ ( odd function ). And for s=0
$F_{1}(0)=\int_{-\infty}^{+\infty}1dt=+\infty$ so
$F_{1}$ is then defined by $F_{1}(s)=0$ if $s\neq0$ and $F_{1}(0)=+\infty$. $F_{1}$ is the Dirac delta "function" ( It's a distribution ). $F_{1}(s)=\delta(s)$
Now if $f(t)= cos(2\pi \omega t)$ then $f(t)=\frac{1}{2}(e^{i2\pi \omega t}+e^{-i2\pi\omega t})$. The Fourier transform of f is then $F(s)= \frac{1}{2}\int_{-\infty}^{+\infty}e^{-i2\pi (s-\omega)t}dt+ \frac{1}{2}\int_{-\infty}^{+\infty}e^{-i2\pi (s+\omega)t}dt= \frac{1}{2}F_{1}(s-\omega)+ \frac{1}{2}F_{1}(s+\omega)= \frac{1}{2}\delta(s-\omega)+ \frac{1}{2}\delta(s+\omega)$
 
There is the following linear Volterra equation of the second kind $$ y(x)+\int_{0}^{x} K(x-s) y(s)\,{\rm d}s = 1 $$ with kernel $$ K(x-s) = 1 - 4 \sum_{n=1}^{\infty} \dfrac{1}{\lambda_n^2} e^{-\beta \lambda_n^2 (x-s)} $$ where $y(0)=1$, $\beta>0$ and $\lambda_n$ is the $n$-th positive root of the equation $J_0(x)=0$ (here $n$ is a natural number that numbers these positive roots in the order of increasing their values), $J_0(x)$ is the Bessel function of the first kind of zero order. I...
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