# I Problem -- Fourier transform

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1. Dec 25, 2016

### arpon

[$f^*$ represents complex conjugate of $f$. ]

[$\widetilde{f}(k)$ represents fourier transform of the function $f(x)$.]

\begin{align} \int_{-\infty}^{\infty}f^*(x)e^{ikx}\,dx&=\int_{-\infty}^{\infty}f^*(x)\left(e^{-ikx}\right)^*\,dx\\ &=\int_{-\infty}^{\infty}\left(f(x)e^{-ikx}\right)^*\,dx\\ &=\left(\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx\right)^*\\ &=\left[\widetilde{f}(k)\right]^*\\ \end{align}
Now, let
$$f(x)=u(x)+iv(x)$$
where $u(x)$ and $v(x)$ are the real and imaginary parts of $f(x)$.
Again, we have,
\begin{align} \int_{-\infty}^{\infty}f^*(x)e^{ikx}\,dx&=\int_{-\infty}^{\infty}\left(u(x)-iv(x)\right)e^{ikx}\,dx\\ &=\int_{-\infty}^{\infty}u(x)e^{-i(-k)x}\,dx-i\int_{-\infty}^{\infty}v(x)e^{-i(-k)x}\,dx\\ &=\widetilde{u}(-k)-i\widetilde{v}(-k)\\ &=\left[\widetilde{u}(-k)+i\widetilde{v}(-k)\right]^*\\ &=\left[\widetilde{f}(-k)\right]^* \text{ [using linearity property of fourier transform]} \end{align}
So, I am getting different results. What is wrong with this calculation.

2. Dec 25, 2016

### Staff: Mentor

There is a sign error between (3) and (4).

3. Dec 25, 2016

### arpon

I used the defination,
$$\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx=\widetilde{f}(k)$$

4. Dec 25, 2016

### Svein

Well, not quite. The definition is $\widetilde f(k)= \frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-ikx}dx$

5. Dec 25, 2016

### micromass

Staff Emeritus
There are actually many non-equivalent definitions. Most just differ in the constant in front.

6. Dec 25, 2016

### Staff: Mentor

Well, it doesn't matter which definition you use, but keep it consistent. Either eikx or e-ikx. Using the former, (3) is the Fourier transform for -k instead of k.

7. Dec 25, 2016

### arpon

I did not understand why (3) is the Fourier transform of -k instead of k. Look, I used the defination,
$$\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx=\widetilde{f}(k)\\$$
Taking complex conjugate on both sides,
$$\left(\int_{-\infty}^{\infty}f(x)e^{-ikx}\,dx\right)^*=\left[\widetilde{f}(k)\right]^*\\$$
So (3) and (4) are justified.

8. Dec 26, 2016

### Staff: Mentor

Ah, I misread the signs, sorry.

There is a mistake between (7) and (8). u and v are real, but their Fourier transformations in general won't be real. You can use $\widetilde{u}(-k) = \widetilde{u}(k)^*$:

\begin{align} \widetilde{u}(-k)-i\widetilde{v}(-k) &= \widetilde{u}(k)^* -i \widetilde{v}(k)^*\\ \end{align}

Not sure how to simplify that.