Ackbach said:
You've marked this thread as solved. I'm a bit puzzled by the question myself, though. The two integrals are equal, period, because multiplication is commutative. The equality of those two integrals implies nothing whatsoever. Did you mean, instead, that
$$\lambda^{2}\int_{-1}^{1}y_{n}y_{n}\,dx=\lambda^{2}\int_{-1}^{1}y_{m}y_{m}\,dx?$$
I was looking at a paper about the Legendre polynomials, and I didn't understand what they were doing. I then realized they should have indexed there lambdas.We can rewrite $(1 - x^2)y_{\ell}'' - 2xy_{\ell}' + \left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}$ as
\begin{alignat}{3}
\frac{d}{dx}\left[(1 - x^2)y_{\ell}'\right] + \left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell} & = & 0\notag\\
\frac{d}{dx}\left[(1 - x^2)y_{\ell}'\right] & = & - \left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}\notag\\
y_n\frac{d}{dx}\left[(1 - x^2)y_{\ell}'\right] & = & - \left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_n\notag\\
\int_{-1}^1y_n\left[(1 - x^2)y_{\ell}'\right]'dx & = & - \int_{-1}^1\left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx
\end{alignat}
Integrating (2) by parts where $u = y_n$ and $dv = [(1 - x^2)y_{\ell}']'$, we have
\begin{alignat*}{3}
\underbrace{\left.(1 - x^2)y_n'y_{\ell}\right|_{-1}^1}_{ = 0} - \int_{-1}^1(1 - x^2)y_n'y_{\ell}'dx & = & - \int_{-1}^1\left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx\\
\int_{-1}^1(1 - x^2)y_n'y_{\ell}'dx & = & \int_{-1}^1\left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx
\end{alignat*}
Since we could have started with $y_n$ and the choice of $y_{\ell}$ was arbitrary, we would also have
$$
\int_{-1}^1(1 - x^2)y_n'y_{\ell}'dx = \int_{-1}^1\left(\lambda_{n}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx.
$$
Therefore, we have that
\begin{alignat*}{3}
\int_{-1}^1\left(\lambda_{n}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx & = & \int_{-1}^1\left(\lambda_{\ell}^2 - \frac{m^2}{1 - x^2}\right)y_{\ell}y_ndx\\
\int_{-1}^1\left(\lambda_{n}^2y_{\ell}y_n - \frac{m^2}{1 - x^2}y_{\ell}y_n - \lambda_{\ell}^2y_{\ell}y_n + \frac{m^2}{1 - x^2}y_{\ell}y_n\right)dx & = & 0\\
(\lambda_n^2 - \lambda_{\ell}^2)\int_{-1}^1y_{\ell}yndx & = & 0
\end{alignat*}
When $n\neq\ell$, $\lambda_n^2 - \lambda_{\ell}^2$ is non-zero; therefore, $\int_{-1}^1y_{\ell}yndx = 0$.