Suppose the fourier transform is defined as $$\hat{f}(k)=\frac{1}{2}\int_{\infty}^{\infty}f(x)e^{-i2\pi kx}dx$$
and the inverse: $$f(x)=B\int_{-\infty}^{\infty}\hat{f}(k)e^{+i2\pi kx}dk$$ I want to find B, one approach would be to use the definition $\mathcal{F}^{-1}\mathcal{F}(f(x))=f(x)$: $$\mathcal{F}^{-1}\mathcal{F}(f)(t)=\frac{B}{2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(x)e^{-i2\pi kx}e^{+i2\pi kt}dk=\int_{-\infty}^{\infty}f(x)\left( \frac{B}{2}\int_{-\infty}^{\infty}e^{i2\pi(t-x)k}dk\right)dt$$
But for the RHS=$f(t)$, it is required that $$\delta(t-x)=\frac{B}{2}\int_{-\infty}^{\infty}e^{i2\pi(t-x)k}dk$$
and this forms a loop, because $\delta(x)$ is defined using the inverse the fourier transform? How do I find B?
Let the Fourier transform and its inverse be given by $$\begin{align} \mathcal{F}f(\xi) &= A \int_{-\infty}^{\infty} f(x) e^{-iax\xi} \, dx \\ f(x) &= B \int_{-\infty}^{\infty} \mathcal{F}f(\xi) e^{iax\xi} \, d\xi. \\ \end{align}$$ Note that $$ f(x) = \frac{B}{A} \mathcal{F}(\mathcal{F}f)(-x). \label{double-apply}\tag{*} $$
It's easy to show that $$ \mathcal{F}\{ e^{-ax^2/2} \} = A \sqrt{\frac{2\pi}{a}} e^{-a\xi^2/2} $$ which inserted into \eqref{double-apply} gives $$ e^{-ax^2/2} = \frac{B}{A} \cdot A^2 \frac{2\pi}{a} e^{-a(-x)^2/2} = AB \, \frac{2\pi}{a} \, e^{-ax^2/2} $$ so we must have $AB \, \frac{2\pi}{a} = 1,$ i.e. $$ B = \frac{a}{2\pi \, A}. $$
Example 1: With $a=1$ and $A=1$ we get $B=\frac{1}{2\pi}.$
Example 2: With $a=2\pi$ and $A=1$ we get $B=1.$