Fourier transform with initial limit starting from 0

Question

For a function $f(p)$, the Fourier transform is $\hat{f}(p)=\int_{-\infty} ^{\infty}e^{-i xp}f(x)dx$. What are the conditions that i write it as; $\hat{f}(p)=2\int_{0} ^{\infty}e^{-ixp}f(x)dx$? I think if $f(x)$ is even then I can write it. Or is it even possible to write such that initial limit starts from 0?

Answer 1

You can conclude that $\int_{-\infty}^\infty g(x)\,dx =2 \int_0^\infty g(x)\,dx$ if you know that $g$ is even. Since here $g(x) = e^{-ixp} f(x)$, the property you would need for this conclusion is $$ e^{-ixp}f(x) = e^{ixp} f(-x) $$ which unfortunately doesn't hold unless $f\equiv 0$ (considering that $p$ is arbitrary).

However, if $f(-x) = \overline{f(x)}$, it follows that $$ \overline{e^{-ixp}f(x)} = e^{ixp} f(-x) $$ and therefore $$ \operatorname{Re}\hat f(p) = 2\operatorname{Re}\int_0^\infty e^{-ixp} f(x)\,dx $$ This is particularly useful when $\hat f(p)$ is known to be real; a typical scenario in which this happens is taking the inverse Fourier transform related to the Fourier transform of a real function.