Confusion on Parseval's Theorem

Question

Parseval's Theorem says that:

$$\int_{-\infty}^{\infty}g(t)f(t)^\ast dt = \frac{1}{2\pi} \int_{-\infty}^{\infty}G(\omega)F(\omega)^\ast d\omega$$

Although I know how to prove it, it's difficult to imagine how the two integrals can be equal. If we define $g(t)$ and $f(t)$ as time-domain real-number signals (so $f(t)$'s conjugate is also a real-number signal), then the left side is also real numbered. But the $G(\omega)$ and $F(\omega)^\ast$ can be complex numbered, so the integral on the right side is also a complex number.

Then how could a real number equal to a complex number? There must be something wrong with my above understanding. Could anyone help me to point it out?

Answer 1

Assume that $f(t)$ and $g(t)$ are real valued functions. Then, we have

$$\begin{align} F(\omega)&=\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt\\\\ &=\underbrace{\int_{-\infty}^\infty f(t) \cos(\omega t)\,dt}_{\text{Even in}\,\omega}+i\underbrace{\int_{-\infty}^\infty f(t) \sin(\omega t)\,dt}_{\text{Odd in }\,\omega} \end{align}$$

$$\begin{align} G(\omega)&=\int_{-\infty}^\infty f(t)e^{i\omega t}\,dt\\\\ &=\underbrace{\int_{-\infty}^\infty g(t) \cos(\omega t)\,dt}_{\text{Even in}\,\omega}+i\underbrace{\int_{-\infty}^\infty g(t) \sin(\omega t)\,dt}_{\text{Odd in }\,\omega} \end{align}$$

Now, note that the product of $F$ and the complex conjugate of $G$ is given by

$$\begin{align} F(\omega)G^*(\omega)&=\left(\text{Re}(F(\omega))\text{Re}(G(\omega))+\text{Im}(F(\omega))\text{Im}(G(\omega))\right)\\\\ &-i\left(\text{Re}(F(\omega))\text{Im}(G(\omega))-\text{Im}(F(\omega))\text{Re}(G(\omega))\right) \end{align}$$

Therefore, the imaginary part of $F(\omega)G^*(\omega)$ is an odd function. Since convergent integrals are equal to their Cauchy Principal Values, the integral of any odd convergent function over symmetric limits vanishes. And we are done!

Answer 2

Here's a way to understand it using essentially only one fact: the integral of the product of an even function ($f(x)=f(-x)$) and an odd function ($f(x)=-f(-x)$) is zero:

If $f$ is even and $g$ is odd, then $$ \int_{-\infty}^{\infty} f(x)g(x) \, dx = 0. $$

Any function $f$ can be written as the sum of an even function and an odd function: $$ f(x) = \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2}. $$ Call the former fraction $f_e$, the latter $f_o$.

How does the Fourier transform work on the even and odd part? We can split $e^{ikx}=\cos{kx}+i\sin{kx}$, and the cosine is even, the sine odd. Therefore, if we write $F(k) = \int_{-\infty}^{\infty} e^{ikx} f(x) \, dx$, $$ F(k) = \int_{-\infty}^{\infty} (\cos{kx}+i\sin{kx})(f_e(x)+f_o(x)) \, dx \\ = \int_{-\infty}^{\infty} (\cos{kx}) f_e(x) \, dx +i\int_{-\infty}^{\infty} (\sin{kx}) f_o(x) \, dx, $$ since the other integrals vanish. But both these integrands are even, so they are equal to twice the integral over $(0,\infty)$: $$ F(k) = 2\int_{0}^{\infty} (\cos{kx}) f_e(x) \, dx +2i\int_{0}^{\infty} (\sin{kx}) f_o(x) \, dx = F_c(k) + iF_o(k), $$ say. But the evenness of cosine means that $F_c$ is even, and in the same way $F_o$ is odd. Moreover, both are obviously real. But then, considering the right-hand side of Parseval, $$ \int_{-\infty}^{\infty} F(k) G(k)^* \, dk = \int_{-\infty}^{\infty} (F_c(k)+F_s(k)) (G_c(k)-G_s(k)) \, dk = \int_{-\infty}^{\infty} F_c(k) G_c(k) \, dk + \int_{-\infty}^{\infty} F_s(k) G_s(k) \, dk; $$ the other two terms vanish because $F_c,G_c$ are even and $F_s,F_s$ are odd.