Integral of Legendre and Chebyshev polynomials. I am trying to expand Legendre polynomials into Chebyshev polynomials, shown as:
$$P_{n}(x)=\sum_{k=0}^{n}a_{k}T_{k}(x), $$
where $P_{n}$ is Legendre polynomials and $T_{k}$ is Chebyshev polynomials, and $a_{k}$ are unknown coefficients.
As the Orthogonality of Chebyshev polynomial is
$$\int_{-1}^{1}T_{j}(x)T_{k}(x)\frac{1}{\sqrt{1-x^2}}dx=\frac{\pi}{2}\delta_{jk},\quad j^{2}+k^{2}\neq0,$$
so
$$a_{k}(n,k)=\frac{2}{\pi}\int_{-1}^{1}P_{n}(x)T_{k}(x)\frac{1}{\sqrt{1-x^2}}dx.$$
But i do not how to do this integral analytically. Сan anyone help? Many thanks!
 A: Basically you're asking about the Fourier expansion of $$P_n(\cos\theta)=\frac{a_{n,0}}{2}+\sum_{k=1}^n a_{n,k} T_k(\cos\theta)=\frac{a_{n,0}}{2}+\sum_{k=1}^n a_{n,k}\cos k\theta.$$
A relatively easy way is to consider the generating function $$\sum_{n=0}^\infty P_n(\cos\theta)t^n=(1-2t\cos\theta+t^2)^{-1/2}=(1-te^{i\theta})^{-1/2}(1-te^{-i\theta})^{-1/2}$$ and use the binomial series $$(1-z)^{-1/2}=\sum_{n=0}^\infty b_n z^n,\qquad b_n=\frac{(2n-1)!!}{(2n)!!}$$ (here $0!!=(-1)!!=1$). Computing the product, we find $$\sum_{n=0}^\infty P_n(\cos\theta)t^n=\sum_{j,k=0}^\infty b_j b_k t^{j+k}e^{i(j-k)\theta}=\sum_{n=0}^\infty t^n\sum_{k=0}^n b_k b_{n-k}e^{i(n-2k)\theta}.$$
Hence, "renaming" the indices, $$a_{n,k}=\begin{cases}\hfill 0,\hfill&n-k\text{ is odd}\\2b_{(n-k)/2}b_{(n+k)/2},&n-k\text{ is even}\end{cases}.$$
A: You can start from the generating function of Legendre polynomials:
\begin{align}
        \sum_{n = 0}^{\infty} P_{n} (\cos\theta) t^n &= \frac{1}{\sqrt{1-2t\cos\theta + t^2}}\\[.2cm]
        &= (1-t\mathrm{e}^{i\theta})^{-\frac{1}{2}}\cdot (1-t\mathrm{e}^{-i\theta})^{-\frac{1}{2}}\\[.2cm]
        &= \left[\sum_{n=0}^{\infty} \binom{n-\frac{1}{2}}{n} t^n \mathrm{e}^{in\theta}\right] \cdot \left[\sum_{n=0}^{\infty} \binom{n-\frac{1}{2}}{n} t^n \mathrm{e}^{-in\theta}\right]\\[.2cm]
        &=\left[\sum_{n=0}^{\infty} \frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma(n+1)}t^{n}\mathrm{e}^{in\theta}\right]\cdot \left[\sum_{n=0}^{\infty} \frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma(n+1)}t^{n}\mathrm{e}^{-in\theta} \right]\\[.2cm]
        &=\sum_{n=0}^{\infty} \left(\sum_{k=0}^{n} \frac{\Gamma\left(k+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma(k+1)}t^{k}\mathrm{e}^{ik\theta}\cdot \frac{\Gamma\left(n-k+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)\Gamma\left(n-k+1\right)}t^{n-k}\mathrm{e}^{-i(n-k)\theta}\right)\\[.2cm]
        &=\sum_{n=0}^{\infty}t^{n} \left(\frac{1}{\pi}\sum_{k=0}^{n}\frac{\Gamma\left(k+\frac{1}{2}\right)}{\Gamma(k+1)}\frac{\Gamma\left(n-k+\frac{1}{2}\right)}{\Gamma(n-k+1)} \mathrm{e}^{i(2k-n)\theta}\right)
\end{align}
Comparing both sides of the equation, you can get:
\begin{equation}
        P_{n}(\cos\theta) = \frac{1}{\pi}\sum_{k=0}^{n}\frac{\Gamma\left(k+\frac{1}{2}\right)}{\Gamma(k+1)}\frac{\Gamma\left(n-k+\frac{1}{2}\right)}{\Gamma(n-k+1)} \mathrm{e}^{i(2k-n)\theta}
\end{equation}
Take the real part of both sides, then
\begin{equation}
 P_{n}(\cos\theta) = \frac{1}{\pi}\sum_{k=0}^{n}\frac{\Gamma\left(k+\frac{1}{2}\right)}{\Gamma(k+1)}\frac{\Gamma\left(n-k+\frac{1}{2}\right)}{\Gamma(n-k+1)} \cos[(n-2k)\theta]\\
\end{equation}
or you can write it as:
\begin{equation}
P_{n}(\cos\theta) =\boxed{\sum_{k=0}^{n} \frac{(2k-1)!!}{(2k)!!} \frac{(2n-2k-1)!!}{(2n-2k)!!} \cos[(n-2k)\theta]}
\end{equation}
Thus, we have obtained the Fourier expansion of Legendre polynomials (expand Legendre polynomials into Chebyshev polynomials).
