The terminology comes from the fact that the integral is a type of inner product. We will expand on this in the next section. The theorem has a very short, elegant, and illuminating proof so let us give it here. First, we have the following two equations.

\begin{equation*}
x_1'' + \lambda_1 x_1 = 0
\qquad \text{and} \qquad
x_2'' + \lambda_2 x_2 = 0.
\end{equation*}

Multiply the first by \(x_2\) and the second by \(x_1\) and subtract to get

\begin{equation*}
(\lambda_1 - \lambda_2) x_1 x_2 = x_2'' x_1 - x_2 x_1'' .
\end{equation*}

Now integrate both sides of the equation:

\begin{equation*}
\begin{split}
(\lambda_1 - \lambda_2) \int_a^b x_1 x_2 \,dt
& =
\int_a^b x_2'' x_1 - x_2 x_1'' \,dt \\
& =
\int_a^b \frac{d}{dt} \left( x_2' x_1 - x_2 x_1' \right) \,dt \\
& =
\Bigl[ x_2' x_1 - x_2 x_1' \Bigr]_{t=a}^b
= 0 .
\end{split}
\end{equation*}

The last equality holds because of the boundary conditions. For example, if we consider

(4.1) we have

\(x_1(a) = x_1(b) = x_2(a) = x_2(b) = 0\) and so

\(x_2' x_1 - x_2 x_1'\) is zero at both

\(a\) and

\(b\text{.}\) As

\(\lambda_1 \not= \lambda_2\text{,}\) the theorem follows.

The function \(\sin (n t)\) is an eigenfunction for the problem \(x''+\lambda x = 0\text{,}\) \(x(0) = 0\text{,}\) \(x(\pi) = 0\text{.}\) Hence for positive integers \(n\) and \(m\) we have the integrals

\begin{equation*}
\int_{0}^\pi \sin (mt) \sin (nt) \,dt = 0 ,
\quad
\text{when } m \not = n.
\end{equation*}

Similarly,

\begin{equation*}
\int_{0}^\pi \cos (mt) \cos (nt) \,dt = 0 ,
\quad
\text{when } m \not = n,
\qquad \text{and} \qquad
\int_{0}^\pi \cos (nt) \,dt = 0 .
\end{equation*}

And finally we also get

\begin{equation*}
\int_{-\pi}^\pi \sin (mt) \sin (nt) \,dt = 0 ,
\quad
\text{when } m \not = n,
\qquad \text{and} \qquad
\int_{-\pi}^\pi \sin (nt) \,dt = 0 ,
\end{equation*}

\begin{equation*}
\int_{-\pi}^\pi \cos (mt) \cos (nt) \,dt = 0 ,
\quad
\text{when } m \not = n,
\qquad \text{and} \qquad
\int_{-\pi}^\pi \cos (nt) \,dt = 0 ,
\end{equation*}

and

\begin{equation*}
\int_{-\pi}^\pi \cos (mt) \sin (nt) \,dt = 0
\qquad \text{(even if $m=n$).}
\end{equation*}