Solve Integral Equation using Fourier Sine or Cosine Transform $$\frac{1}{a^2+x^2}=\int_0^\infty f(t)\cos(2tx)dt$$
Here is what I've done so far:
$$s=2x, x= \frac{s}{2}$$
$$\frac{1}{a^2+\left(\frac{s}{2}\right)^2}=\int_0^\infty f(t)cos(st)dt$$
$$\sqrt \frac{2}{\pi}\left(\frac{1}{a^2+\left(\frac{s}{2}\right)^2}\right)=\sqrt \frac{2}{\pi}\int_0^\infty f(t)cos(st)dt$$
$$\sqrt \frac{2}{\pi}\left(\frac{1}{a^2+\left(\frac{s}{2}\right)^2}\right)=\mathcal F_c {f(t)} $$
Is there a way for me to find $f(t)$ without using the inverse fourier cosine transform (fct). I believe I can use this: $$ \mathcal F_c (e^{-at})=\sqrt \frac{2}{\pi}\left(\frac{a}{a^2+s^2}\right)$$
Just as my professor on another problem and build off of this for an answer for $f(t)$
Any help would is welcomed. Thanks
Use the scale property: if $\mathcal F_c\{f(t)\}=\hat{F}(s)$ and $\beta>0$, then $$\mathcal F_c\{f(\beta t)\}=\frac{1}{\beta}\hat F_c\left(\frac{s}{\beta}\right)$$
So you'll find for $f(t)=e^{-at}$ with $\mathcal F_c\{f(t)\}=\hat{F}(s)=\sqrt{\frac{2}{\pi}}\frac{a}{a^2+s^2}$ $$\mathcal F_c\{f(2t)\}=\frac{1}{2}\hat F_c\left(\frac{s}{2}\right)=\frac{1}{2}\sqrt{\frac{2}{\pi}}\frac{a}{a^2+(s/2)^2}$$ and then $$\mathcal F^{-1}_c\left\{\sqrt{\frac{2}{\pi}}\frac{a}{a^2+(s/2)^2}\right\}=2f(2t)=2e^{-2at}$$