How to integral legendre polynomial

[Another1]

Question

\(\displaystyle \int_{-1}^{1} cos(x) P_{n}(x)\,dx\)

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my think (maybe incorrect)
\(\displaystyle \int_{-1}^{1} cos(x) P_{n}(x)\,dx\)
\(\displaystyle \frac{1}{2^nn!}\int_{-1}^{1} cos(x) \frac{d^n}{dx^n}(x^2-1)^n\,dx\) This is rodrigues formula
by part n times
\(\displaystyle \frac{1}{2^nn!}(-1)^n\int_{-1}^{1} \frac{d^n}{dx^n}cos(x) \frac{d^{n-n}}{dx^{n-n}}(x^2-1)^n\,dx\)
\(\displaystyle \frac{1}{2^nn!}(-1)^n\int_{-1}^{1} \frac{d^n}{dx^n}cos(x) (x^2-1)^n\,dx\)
in case n = odd number
\(\displaystyle \frac{1}{2^nn!}(-1)^n\int_{-1}^{1} (-1)^{\frac{n+1}{2}}sin(x) (x^2-1)^n\,dx\)
in case n = even number
\(\displaystyle \frac{1}{2^nn!}(-1)^n\int_{-1}^{1} (-1)^{\frac{n}{2}}cos(x) (x^2-1)^n\,dx\)
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how to integral

\(\displaystyle \int_{-1}^{1}sin(x) (x^2-1)^n\,dx\) and \(\displaystyle \int_{-1}^{1}cos(x) (x^2-1)^n\,dx\)

 

[Greg]

Wolfram Alpha as an Online Integrator

 

[Another1]

Wolfram Alpha as an Online Integrator

Thank you. But i don't have pro Wolframalpha.
please. you can show that solution?

 

[Joppy]

Thank you. But i don't have pro Wolframalpha.
please. you can show that solution?

I do. But there's no worked solution for this one anyway.

 

[Another1]

I do. But there's no worked solution for this one anyway.

thank you vary much. now i can solve it

 

[Joppy]

thank you vary much. now i can solve it

Would you mind sharing your solution? I remember there being a trick with these things but I cannot remember!

 

[Another1]

Would you mind sharing your solution? I remember there being a trick with these things but I cannot remember!

Ok

let
\(\displaystyle cos(x) = \sum_{n=0}^{\infty}C_{n}P_{n}(x)\)

orthogonal
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx = \int_{-1}^{1} C_{n}P_{n}(x)P_{n}(x)\,dx\)
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx = C_{n}\int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx\)
see that
\(\displaystyle \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\int_{-1}^{1} \frac{d^n}{dx^n}(x^2-1)^n\frac{d^n}{dx^n}(x^2-1)^n\,dx\)
you can integral by part n time
So..
\(\displaystyle u= \frac{d^n}{dx^n}(x^2-1)^n\) and \(\displaystyle du=\frac{d^{n+1}}{dx^{n+1}}(x^2-1)^n\,dx\)
\(\displaystyle dv= \frac{d^n}{dx^n}(x^2-1)^ndx\) and \(\displaystyle v=\frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n\)

\(\displaystyle udv=uv-vdu\)

\(\displaystyle uv= \frac{d^n}{dx^n}(x^2-1)^n \frac{d^{n-1}}{dx^{n-1}}(x^2-1)^n =0 \) When limit of the integrate from -1 to 1

So when integrate by part n times
\(\displaystyle \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\left[0+0+...+(-1)^n\int_{-1}^{1}\frac{d^{n-n}}{dx^{n-n}}(x^2-1)^n \frac{d^{n+n}}{dx^{n+n}}(x^2-1)^n \,dx\right]\)
\(\displaystyle \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\left[\int_{-1}^{1}(x^2-1)^n \frac{d^{2n}}{dx^{2n}}(1-x^2)^n \,dx\right]\) ; times (-1)^n in (x^n-1)^n

You know \(\displaystyle (1-x^2)^n=\sum_{k=0}^{n}{n \choose k} 1^{n-k}(-1)^n(x^2)^n \)
\(\displaystyle \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\left[\int_{-1}^{1}(x^2-1)^n \frac{d^{2n}}{dx^{2n}}\sum_{k=0}^{n}{n \choose k} 1^{n-k}(-1)^n(x^2)^n \,dx\right]\)

The final term of \(\displaystyle (1-x^2)^n=\sum_{k=0}^{n}{n \choose k} 1^{n-k}(-1)^n(x^2)^n = ...+ {n \choose n }(-1)^n x^{2n}\)

because \(\displaystyle \frac{d^m}{dx^m}x^n = 0\) when n < m

So that
\(\displaystyle \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\left[\int_{-1}^{1}(x^2-1)^n \frac{d^{2n}}{dx^{2n}}(-1)^nx^{2n} \,dx\right]\)
\(\displaystyle \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{1}{2^{2n}(n!)^2}\left[\int_{-1}^{1}(x^2-1)^n (-1)^n (2n)! \,dx\right]\)
\(\displaystyle \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{(2n)!}{2^{2n}(n!)^2}\left[\int_{-1}^{1}(1-x^2)^n \,dx\right]\)

set \(\displaystyle s =\frac{x+1}{2}\)

\(\displaystyle \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{(2n)!}{2^{2n}(n!)^2}\left[\int_{0}^{1} 2\cdot 2^{2n}s^n(1-s)^n\,ds\right]\) beta function
\(\displaystyle \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{(2n)!}{2^{2n}(n!)^2} 2\cdot 2^{2n}\frac{(n!)^2}{(2n+1)!}\)
\(\displaystyle \int_{-1}^{1} P_{n}(x)P_{n}(x)\,dx= \frac{2}{2n+1}\)

see that
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx= \frac{1}{2^{n}n!}\int_{-1}^{1}cos(x)\frac{d^n}{dx^n}(x^2-1)^n\,dx\)

You can integral by part n times.
\(\displaystyle u = con(x)\) and \(\displaystyle dv = \frac{d^n}{dx^n}(x^2-1)^n dx \)
So . . .
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx= \frac{(-1)^n}{2^{n}n!}\int_{-1}^{1}(x^2-1)^n \frac{d^n}{dx^n}cos(x),dx\)

give two solution

in case n = odd number
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^n\int_{-1}^{1} (-1)^{\frac{n+1}{2}}sin(x) (x^2-1)^n\,dx\)
in case n = even number
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^n\int_{-1}^{1} (-1)^{\frac{n}{2}}cos(x) (x^2-1)^n\,dx\)

But \(\displaystyle sin(x) (x^2-1)^n \) are Odd function So \(\displaystyle \frac{1}{2^nn!}(-1)^n\int_{-1}^{1} (-1)^{\frac{n+1}{2}}sin(x) (x^2-1)^n\,dx = 0\)

left only one solution

\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^n\int_{-1}^{1} (-1)^{\frac{n}{2}}cos(x) (x^2-1)^n\,dx\) n = 0,2,4,...
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}} 2 \int_{0}^{1} cos(x) (1-x^2)^n\,dx\) ; even function

Wolfram|Alpha: Computational Intelligence

You know \(\displaystyle cos(x) = \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!}x^{2i} \)
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{2}{2^nn!}(-1)^{\frac{n}{2}} \int_{0}^{1} \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!}x^{2i} (1-x^2)^n\,dx\)
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}} \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!} (2)\int_{0}^{1} x^{2i} (1-x^2)^n\,dx\); Beta function form
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}} \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!} \frac{(i-\frac{1}{2})!n!}{(i-\frac{1}{2}+n+1)!}\)
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}} \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!} \frac{(i-\frac{1}{2})! \Gamma(n+1)}{(i+n+\frac{1}{2})!}\)

from \(\displaystyle (i-\frac{1}{2})! = \frac{1}{i!}(\frac{1}{2})^{2i}\sqrt{\pi}(2i)!\)

\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!} \frac{1}{(i+n+\frac{1}{2})!}\frac{1}{i!}(\frac{1}{2})^{2i}\sqrt{\pi}(2i)!\)
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) \sum_{i=0}^{\infty}\frac{(-1)^i}{(2i)!} \frac{1}{(i+n+\frac{1}{2})!}\frac{1}{i!}(\frac{1}{2})^{2i}\sqrt{\pi}(2i)! \cdot \frac{2^{n+\frac{1}{2}}}{2^{n+\frac{1}{2}}}\)
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{n+\frac{1}{2}}\sqrt{\pi}\sum_{i=0}^{\infty}\frac{(-1)^i}{i!} \frac{1}{(i+n+\frac{1}{2})!}(\frac{1}{2})^{2i} \cdot (\frac{1}{2})^{n+\frac{1}{2}}\)
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{n+\frac{1}{2}}\sqrt{\pi}\sum_{i=0}^{\infty}\frac{(-1)^i}{i!} \frac{1}{(i+n+\frac{1}{2})!}(\frac{1}{2})^{n+\frac{1}{2}+2i}\)
\(\displaystyle \int_{-1}^{1} cos(x)P_{n}(x)\,dx=\frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{n+\frac{1}{2}}\sqrt{\pi}J_{n+\frac{1}{2}}(1)\) ; Bessel polynomialSo
\(\displaystyle C_{n}=\frac{2n+1}{2}\int_{-1}^{1} cos(x)P_{n}(x)\,dx= \frac{2n+1}{2} \frac{1}{2^nn!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{n+\frac{1}{2}}\sqrt{\pi}J_{n+\frac{1}{2}}(1)\)
\(\displaystyle C_{n}=\frac{2n+1}{2}\int_{-1}^{1} cos(x)P_{n}(x)\,dx= \frac{2n+1}{n!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{-\frac{1}{2}}\sqrt{\pi}J_{n+\frac{1}{2}}(1)\) n = 0,2,4,...

and finaly

\(\displaystyle cos(x) = \sum_{n=0}^{\infty}C_{n}P_{n}(x)\)
\(\displaystyle cos(x) = \sum_{n=0}^{\infty} \frac{2n+1}{n!}(-1)^{\frac{n}{2}}\Gamma(n+1) 2^{-\frac{1}{2}}\sqrt{\pi}J_{n+\frac{1}{2}}(1) P_{n}(x)\)

- My solution