- #1

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**[**##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.