inner product zero?

I am studying about Fourier series from book"Signals and Systems Laboratory with MATLAB"

I came across topic "Orthogonality of Complex Exponential Signals"

I am confused in case when m=k, will the two signals still orthogonal in this case ?(highlighted yellow)

• When m=k, the two signals are the same signal. They cannot be orthogonal to themselves. – Juancho May 7 at 18:16

Well, when $$m=k$$ the integral is: $$\int_0^T e^{j(m-k)\Omega_0t} dt = \int_0^T e^{j \cdot 0 \cdot\Omega_0t} dt = \int_0^T dt = T$$

So as Juancho says in the comments, it's the same signal and so can't be orthogonal to itself.

The text is a bit cumbersome, in that it states "things" before defining them. And the zero-signal (or vector) is considered orthogonal to every other vector. This is why the sentence "If $$I=0$$ for..." seems unnecessarily complicated to me. And the $$m$$ and $$k$$ indices introduced at the beginning come out of nowhere until we understand they are integer complex exponential indices. Moreover, introducing $$x$$ and $$y$$ at the very end is quite inconsistent with the previous $$x_m$$ and $$x_k$$ notations.

In a mood for conciseness and precision, let me rewrite the text (with missing hypotheses, like $$\Omega_{0}\neq 0$$), hoping it will be more direct.

Let us consider in the following complex-valued continuous-time periodic signals with period $$T>0$$. The inner product of $$x_1$$ and $$x_2$$ is defined as:

$$I(x_1,x_2)=\int_{t_{0}}^{t_{0}+T} x_1(t) x_2^{*}(t) d t$$

where $$x^{*}$$ denotes the complex conjugate of $$x$$. Two signals are orthogonal if their inner product is zero. Suppose now that $$(m,k)\in \mathbb{Z}^2$$, $$\Omega_{0}\neq 0$$ and define $$x_{1}(t)=e^{j m \Omega_{0} t}$$ and $$x_{2}(t)=e^{j k \Omega_{0} t}$$. Since: $$I(x_1,x_2)=\int_{0}^{T} e^{j m \Omega_{0} t_{e}-j k \Omega_{0} t} d t=\int_{0}^{T} e^{j(m-k) \Omega_{0} t} d t=\left\{\begin{array}{ll}{T,} & {k=m} \\ {0,} & {k \neq m}\end{array}\right.\,,$$ then the complex exponentials $$e^{j m \Omega_{0} t}$$ and $$e^{j k \Omega_{0} t}$$ are orthogonal if $$m\neq k$$.