Integrate the Fourier Legendre by parts :$\int_{-1}^{1}\left( x^{2}-1\right) ^{m}\cos \pi x\:dx$

Question

Having difficulty integrating the Fourier Legendre series by parts :

$$\alpha_{m}=\int_{-1}^{1}\left( x^{2}-1\right) ^{m}\cos \pi x\:dx$$

I understand we can use the general formula :

$$uv-\int vdudx$$

To integrate by parts but I keep getting zero, so there must be something I am doing wrong.

Thanks

Answer 1

Hint. A starting point. From $$ \alpha_{m}=\int_{-1}^{1}\left( x^{2}-1\right) ^{m}\cos \pi x\:dx, \quad m=0,1,2\ldots, \tag1 $$ one may integrate by parts twice, for $m\geq2$, $$ \begin{align} &\alpha_{m}=\int_{-1}^{1}\left( x^{2}-1\right) ^{m}\cos \pi x\:dx \\\\&=\left.\left( x^{2}-1\right) ^{m}\frac{\sin \pi x}{\pi} \right|_{-1}^{1}-\frac{2m}{\pi}\int_{-1}^{1}x\left( x^{2}-1\right) ^{m-1}\sin \pi x\:dx \\\\&=0-\frac{2m}{\pi}\left(0+\frac{1}{\pi}\int_{-1}^{1}\left(\left( x^{2}-1\right) ^{m-1}+2(m-1)x^2\left( x^{2}-1\right) ^{m-2}\right)\cos \pi x\:dx\right) \\\\&=-\frac{2m}{\pi}\left(\frac{1}{\pi}\int_{-1}^{1}\left((2m-3)\left( x^{2}-1\right) ^{m-1}+2(m-1)\left(x^{2}-1\right) ^{m-2}\right)\cos \pi x\:dx\right) \\\\&=-\frac{2m(2m-3)}{\pi^2}\alpha_{m-1}-\frac{4m(m-1)}{\pi^2}\alpha_{m-2} \end{align} $$ obtaining

$$ \pi^2\alpha_m+2m(2m-3)\alpha_{m-1}+4m(m-1)\alpha_{m-2}=0, \quad m\geq2 \tag2 $$

with $\alpha_0=0,\,\alpha_1=-4/\pi^2.$