Transfrom a Legendre polynomial from $\int_{-1}^{1}\phi_j(x)\phi_k(x)dx $ into $\int_{a}^{b}\phi_j(t)\phi_k(t)dt$ given $t=\dfrac{1}{2}[(b-a)x+(a+b)]$ The Legendre polynomials satisfy
$$\int_{-1}^{1}\phi_j(x)\phi_k(x)dx = \begin{cases}
0 &j\neq k\\\\
\dfrac{2}{2j+1} &j=k
\end{cases}$$
Suppose that the best fit problem is given on the interval $[a,b]$. Show that with the transformation $t=\dfrac{1}{2}[(b-a)x+(a+b)]$ and a slight change of notation, we have
$$\int_{a}^{b}\phi_j(t)\phi_k(t)dt = \begin{cases}
0 &j\neq k\\\\
\dfrac{b-a}{2j+1} &j=k
\end{cases}$$
If $t=\dfrac{1}{2}[(b-a)x+(a+b)]$, then $x=\dfrac{2t-a-b}{b-a}$, and I am stuck at this step, can anyone give me some hints?
 A: Using integration by substitution with the transformation $t=\dfrac{1}{2}\left((b-a)x+(a+b)\right)$ you get from the orthogonality relation on $(-1,1)$
$$
\frac{2}{2j+1} \delta_{j,k} =
\int_{-1}^{1}\phi_j(x)\phi_k(x)dx
=\int_{a}^{b}\frac{2}{b-a}\phi_j\left(\frac{2t-a-b}{b-a}
\right)\phi_k\left(\frac{2t-a-b}{b-a}\right) dt.
$$
Making explicit the slight change of notation by writing the new polynomials on $(a,b)$ with a tilde, i.e.
$$ \tilde{\phi_k}(t) = \phi_k\left(\frac{2t-a-b}{b-a}\right)$$ this gives the relation 
$$\int_{a}^{b}\tilde{\phi_j}(t)\tilde{\phi_k}(t)dt = \frac{b-a}{2} \frac{2}{2j+1} \delta_{j,k} =
\frac{b-a}{2j+1} \delta_{j,k} $$
IMO it is somewhat unfortuate to use the slight change of notation (and $\phi_k$ instead of the standard $P_k$): Take for example the shifted Legendre polynomials on the interval (0,1), denoted normally with $P_k^{*}$. The $P_k$ are obviously  not orthogonal on $(0,1)$ as the example
$$\int_{0}^{1}\phi_1(x)\phi_2(x)dx =\frac{1}{8}$$ shows, but for the new  $\tilde{\phi_k}(x) = P_k^{*}(x) = P_k(2x-1) = \phi_k(2x-1)$ you have
$$\int_{0}^{1}P_1^{*}(x)P_2^{*}(x)dx = \int_{0}^{1}\phi_1(2x-1)\phi_2(2x-1)dx = 0$$
