I have a doubt in the attached solved example (The underlined part)
If I compare following two equations as given in the attachment
$$e^{-a}=\int_0^\infty f(x)\cos(ax)\,dx$$ and
$$g(x)=\frac 2 \pi \int_0^\infty A(\alpha)\cos(\alpha x)\,dx$$
I see following three comparisons:
$$f(x)=A(\alpha)$$
$$x=\alpha$$
$$a=\alpha$$
So I am unable to understand the comparisons given in the attached text. Also last two comparisons of $(x=\alpha)$ and ($x=a$) cannot hold good at the same time. So I request clarification on how these comparisons are done
The Fourier cosine transform and inversion together give you $$ g(y) = \frac{2}{\pi}\int_{0}^{\infty}\left(\int_{0}^{\infty}g(x)\cos(sx)dx\right)\cos(sy)ds $$ And you want to solve for $f$ such that $$ e^{-a}=\int_{0}^{\infty}f(s)\cos(sa)ds $$ Therefore, $$ f(s) = \frac{2}{\pi}\int_{0}^{\infty}e^{-x}\cos(sx)dx. $$