Closed form for integral $ \int_0^{\pi} \frac{\sin (m \phi)}{(1 + r \cos \phi)^n} d\phi$

Question

Is there a closed form for $n>0$ integer, $m\neq 0$ integer, and $|r|<1$ real?

Answer 1

$\int_0^\pi\dfrac{\sin m\phi}{(1+r\cos\phi)^n}d\phi$

$=\int_0^\pi\dfrac{\sin\phi~U_{m-1}(\cos\phi)}{(1+r\cos\phi)^n}d\phi$ (according to http://mathworld.wolfram.com/Multiple-AngleFormulas.html)

$=-\int_0^\pi\dfrac{U_{m-1}(\cos\phi)}{(1+r\cos\phi)^n}d(\cos\phi)$

$=\int_{-1}^1\dfrac{U_{m-1}(x)}{(1+rx)^n}dx$

Can you take it from here?

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\ds{{\cal I}_{mn}\pars{r} \equiv \int_{0}^{\pi} {\sin\pars{m\phi} \over \bracks{1 + r\cos\pars{\phi}}^{n}}\,\dd\phi\,,\qquad \mbox{Notice that}\ {\cal I}_{mn}\pars{r} = -{\cal I}_{-m,n}\pars{r}}$

\begin{align} {\cal I}_{mn}\pars{r} &= \left.{-\cos\pars{m\phi}/m \over \bracks{1 +r\cos\pars{\phi}}^{n}}\right\vert_{0}^{\pi} - \int_{0}^{\pi}\bracks{-\,{\cos\pars{m\phi} \over m}}\braces{% -n\,{-r\sin\pars{\phi} \over \bracks{1 + r\cos\pars{\phi}}^{n + 1}}}\,\dd\phi \\[3mm]&= {1 \over m}\bracks{% {\pars{-1}^{m + 1} \over \pars{1 - r}^{n}} + {1 \over \pars{1 + r}^{n}}} + {nr \over m}\int_{0}^{\pi} {\cos\pars{m\phi}\sin\pars{\phi} \over \bracks{1 + r\cos\pars{\phi}}^{n + 1}}\,\dd\phi \\[3mm]&= {1 \over m}\bracks{% {\pars{-1}^{m + 1} \over \pars{1 - r}^{n}} + {1 \over \pars{1 + r}^{n}}} + {nr \over m}\int_{0}^{\pi} {\braces{\sin\pars{\bracks{m + 1}\phi} - \sin\pars{\bracks{m - 1}\phi}}/2 \over \bracks{1 + r\cos\pars{\phi}}^{n + 1}}\,\dd\phi \\[3mm]&= {1 \over m}\bracks{% {\pars{-1}^{m + 1} \over \pars{1 - r}^{n}} + {1 \over \pars{1 + r}^{n}}} + {nr \over 2m}\bracks{{\cal I}_{m + 1,n + 1}\pars{r} - {\cal I}_{m - 1,n + 1}\pars{r}} \end{align}

$$ {\cal I}_{m + 1,n + 1}\pars{r} = -\,{2 \over nr}\bracks{% {\pars{-1}^{m + 1} \over \pars{1 - r}^{n}} + {1 \over \pars{1 + r}^{n}}} + {2m \over nr}\,{\cal I}_{mn}\pars{r} + {\cal I}_{m - 1,n + 1}\pars{r}\,,\qquad m \geq 1 $$ with $$ {\cal I}_{0n}\pars{r} = 0 \qquad\mbox{and}\qquad {\cal I}_{1n}\pars{r} = {\pars{1 - r}^{-\pars{n - 1}} - \pars{1 + r}^{-\pars{n - 1}} \over r\pars{n - 1}}\,, \quad {\cal I}_{-m,n}\pars{r} = -{\cal I}_{mn}\pars{r} $$

Answer 3

Let $\displaystyle \int_0^{\pi} \sin(m x) \cos^k(x)dx = I(m,k)$. We have $$I = \int_0^{\pi} \dfrac{\sin(m x)dx}{(1+r \cos(x))^n} = \sum_{k=0}^{\infty} (-r)^k \dbinom{n+k-1}k\int_0^{\pi} \sin(mx) \cos^k(x)dx = \sum_{k=0}^{\infty} (-r)^k \dbinom{n+k-1}k I(m,k)$$

Note that we also have

$$I(m,k) = \int_0^{\pi}\sin(mx) \cos^k(x) dx = \int_0^{\pi}U_{m-1}(\cos(x)) \sin(x) \cos^k(x) dx = \int_{-1}^1 t^k U_{m-1}(t)dt$$ where $U_{m-1}$ is the $(m-1)^{th}$ Chebyshev polynomial of the second kind. Hence, $I(m,k)$ is the $k^{th}$ moment of the $(m-1)^{th}$ Chebyshev polynomial of the second kind.