Understanding the Complex Conjugate Property in Fourier Transform

[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.

 

[mfb]

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

 

[arpon]

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

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

 

[Svein]

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

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

 

[micromass]

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

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

 

[mfb]

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

 

[arpon]

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

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.

 

[mfb]

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.