# $e^{jwt}$ is an ortho-normal basis proof

I'd like to know how one is supposed to show that the set $$\{ e^{jwt} \}$$, where $$\omega \in \mathbb{R}$$ , is an ortho-normal basis?

So actually, how do I show that for every $$w_1 \neq w_2$$: $$\int_{-\infty}^{\infty}e^{jw_1t}e^{-jw2t}dt=0$$ ?

Kindly

Sammy

• Welcome to Maths SX! Are you sure the bounds are $+\infty$ and $-\infty$? – Bernard Oct 26 at 21:10
• Not entirely actually. How do I know which boundaries to choose in order to show orthogonality for this specific set of functions? I have a feeling that its the [-$\pi$,$\pi$] interval... – Sammy Apsel Oct 26 at 21:13
• Yes, I would compute it over a period. – Bernard Oct 26 at 21:15

Note first that if $$R\in\Bbb R^+$$ and $$\omega\ne0$$,$$\int_{-R}^R\exp(j\omega t)dt=\left[\frac{1}{j\omega}\exp(j\omega t)\right]_{-R}^R=\frac{2\sin(\omega R)}{\omega}=2R\operatorname{sinc}(\omega R).$$The final expression is also valid if $$\omega=0$$. Since $$\frac{1}{\pi}\operatorname{sinc}(\omega)$$ is a nascent delta function,$$\int_{-\infty}^\infty\exp(j\omega t)dt=2\pi\delta(\omega).$$In particular,$$\int_{-\infty}^\infty\overline{\exp(j\omega_1t)}\exp(j\omega_2t)dt=2\pi\delta(\omega_1-\omega_2)$$(in your question, you forgot one factor in the integrand should be complex-conjugated). This is our orthonormality condition (well, if we divide each basis element by $$\sqrt{2\pi}$$).
• Thank you for the answer' but would you be kind enough to explain the right hand side of the following equation: $\int_{-\infty}^\infty\exp(j\omega t)dt=2\pi\delta(\omega)$ ? – Sammy Apsel Oct 26 at 21:17
• @SammyAspel Are you familiar with nascent delta functions? If so, I'm setting $\epsilon=1/R$. – J.G. Oct 26 at 21:18