# Why is X(jw) real and even if x(t) is real and even

So I'm going through Oppenheim's (really excellent) textbook "Signals and Systems" (2nd ed.) and the associated MIT course and I'm having an issue with the properties of the Fourier transform. On page 308 (Sec 4.3, Properties of Continuous-time fourier transform), he makes the claim that if x(t) is real and even, X(jw) is real and even. These are the steps he goes through to prove it:

First, the definition of the fourier transform $$X(j\omega) = \int_ {-\infty}^{\infty} x(t)e^{-j\omega t} dt$$ replacing the argument with its negative, $$X(-j\omega) = \int_ {-\infty}^{\infty} x(t)e^{j\omega t} dt$$ then making the substitution $$\tau = -t$$ he claims that $$X(-j\omega) = \int_ {-\infty}^{\infty} x(-\tau)e^{-j\omega \tau} d\tau$$ and so, since we are assuming x(t) is real and even, we can replace this with

$$X(-j\omega) = \int_ {-\infty}^{\infty} x(\tau)e^{-j\omega t} d\tau$$ which shows that $$X(-j\omega) = X(j\omega)$$ and so X(jw) is even. But he seems to have ignored the differential substitution, because when making the variable switch, since $$d\tau=-dt$$ The whole term on the left would become negative, and we would find the opposite of what the author intended to prove, that for a real even x(t) X(jw) is in fact odd. Am I missing something, or is there a more rigorous proof that the author doesn't go over here?

• Just pointing out that questions on signals and systems might be a better fit over at DSP.SE. Jan 29, 2017 at 20:59
• You're absolutely correct, I will post there regarding pure signal analysis in the future. Jan 30, 2017 at 21:32

The initial limits were from $t=-\infty$ to $t=\infty$. When substituting you would first end up with an integral from $\tau=\infty$ to $\tau=-\infty$. Reversing the order of the limits to get back to an integral from $-\infty$ to $+\infty$ cancels the negative sign on the differential element.