See the picture below:
I know if the sign is not '-', the following derivation can not continue，but I really want to know why $$e^{itx}\cdot e^{i\tau x}=e^{i(t-\tau)x}$$ How it can be that? I really want to know.
plus $\hat{f}(t)$ is the standard fourier transform. It means: $$\hat{f}(t)=\int_{-\infty}^{+\infty}f(x)e^{-itx}\,dx$$
And with the @Paul's answer he says:
"The second bracket on the first line should be complex conjugate $\hat{f}(\tau)e^{-i\tau x}$"
I am sure he means that the first line should be as follows: $$\int_{-\infty}^{\infty}f(x)^2\,dx=(\frac{1}{2\pi}\int_{-\infty}^{+\infty}\hat{f}(t)e^{itx}\,dt)(\frac{1}{2\pi}\int_{-\infty}^{+\infty}\hat{f}(\tau)e^{-i\tau x}\,d\tau)$$
The difference is the sign before $i\tau x$.the sign in the very beginning example is positive and in this example is negative.
And now I want to know why the inverse fourier transform could be $$f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\hat{f}(\tau)e^{-i\tau x}\,d\tau$$
Is that should be as follows.The sign before $i\tau x$ is positive instead of negative? $$f(x)=\frac{1}{2\pi}\int_{-\infty}^{+\infty}\hat{f}(\tau)e^{i\tau x}\,d\tau$$
I mean the sign before $i\tau x$ should be always positive in the inverse fourier transform, and can't be negative ,right?
Thaks to Paul，I know what I was confusing:$|f(x)|^2=f(x)\overline{f(x)}$So,the second bracket on the first line should have the negative sign before $i\tau x$ that be $-i\tau x$.
To describe it simply.That is:due to $e^{i\tau x}$ being the complex number,so taking the conjugate make the $e^{i\tau x}$ become $e^{-i\tau x}$.I understand it totally,very very thanks to @Paul.