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See the picture below:

enter image description here

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?

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  • $\begingroup$ The second bracket on the first line should be complex conjugate $\bar{\hat{f}}{{e}^{-i\tau x}}$ so that you also get ${{\left| \hat{f}(t) \right|}^{2}}$ on the last line. In general, the fourier transform will be a complex quantity. $\endgroup$
    – Paul
    Oct 7, 2014 at 10:54
  • $\begingroup$ In your edit you missed the bar (complex conjugation) on top of the $\hat{f}$ in the re-write of line 1. The - sign does not appear due to an inverse transform, it simply arises due to ${{\left| f(x) \right|}^{2}}=f(x)\overline{f(x)}$ and you asre showing that $\int\limits_{-\infty }^{\infty }{{{\left| f(x) \right|}^{2}}dx}=\frac{1}{2\pi }\int\limits_{-\infty }^{\infty }{{{\left| \hat{f}(t) \right|}^{2}}dt}$ $\endgroup$
    – Paul
    Oct 7, 2014 at 12:39

1 Answer 1

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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.

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