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?

@MhenniBenghorbal: that technique does not apply at least when $n$ is odd; in that case, one cannot just simply extend the integral over the unit circle. — Ron Gordon, Nov 26 '13 at 05:34

Felix Marin · Answer 1 · 2013-11-26T19:15:39.133

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

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

Can a similar technique work for non-integer $m$? I wonder if you can solve this integral with an appropriate limit for $m\rightarrow 0^+$ http://math.stackexchange.com/questions/582397/closed-form-for-integral-int-0-pi-1-r-cos-phi-n-phi-d-phi — bkocsis, Nov 26 '13 at 20:27
@bkocsis, you may have already computed this, but for reference: for $r = 0$, $I_{11} = \int_0^\pi \sin \phi , d\phi = 2$. For $|r| < 1$ and $r \not= 0$, $u$-substitution (with $u = 1 + r \cos \phi$) yields $I_{11} = \int_0^\pi \frac{\sin \phi}{1 + r \cos \phi} , d\phi = \int_{1+r}^{1-r} \frac{1}{u} , \frac{du}{-r} = \frac{\ln(1+r) - \ln(1-r)}{r}$. — Kyle, Dec 01 '13 at 05:40
@Kyle Thanks. The last one ( the logarithmic one ) is found from the limit $\large n \to 1$ of my result with L'Hopital Rule. Thanks for your clear help. — Felix Marin, Dec 01 '13 at 23:19

score 2 · Accepted Answer · answered Dec 07 '13 at 04:31

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

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

score 1 · Answer 3 · 2013-12-05T15:44:33.997

1

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.

edited Dec 05 '13 at 15:44

answered Nov 26 '13 at 04:09

Felix Martin's solution shows that this simplifies to a finite sum. Can you generate an explicit solution with Chebysev polynomials? – bkocsis Nov 26 '13 at 15:52
I didn't downvote, but shouldn't this be $$ \sum_{k=0}^\infty(-r)^k\binom{k+n-1}{k}\int_0^\pi\sin(mx)\cos^k(x),\mathrm{d}x $$ – robjohn Dec 05 '13 at 14:12
@robjohn Thanks, It should be $\dbinom{n+k}k$. Fixed. – Dec 05 '13 at 14:39
$\binom{-n}{k}=(-1)^k\binom{n+k-1}{k}$ Third formula here, or here. – robjohn Dec 05 '13 at 15:30
@robjohn Oh yes. Thanks. – Dec 05 '13 at 15:44
Heh. I just came across this question which we both answered. – robjohn Dec 05 '13 at 20:53
@robjohn Ha. Yes :-) – Dec 05 '13 at 21:01

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

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