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Is there a closed form for $$I_n =\int_{0}^{\pi} \frac{\phi}{(1 - r \cos\phi)^n} \,{\rm d}\phi $$ for $\left\vert\,r\,\right\vert < 1$ real and $n > 0$ integer ? The solution to this integral would give a closed form solution for this integral, which describes the interaction energy of vector resonant relaxation in astrophysical dynamics.

Using Mathematica and analytic methods I have derived the following result for $n=\{1,2,3,4,5,6\}$: $$\tag{1} I_n = -\frac{a_n r}{s^{2n-2}} + \frac{b_n}{s^{2n-1}}\left[\chi_2(q) +\left(-\frac{c_n}{b_n}s + {\rm arctanh}(q) \right){\rm arctanh}(r) \right]$$ where $s=\sqrt{1-r^2}$, $q=\sqrt{(1-r)/(1+r)}$, $\chi_2(x)$ is the Legendre chi function, $a_n$, $b_n$, and $c_n$ are constants given by \begin{align} a_1 &= 0, \quad b_1 = 4, \qquad c_1 = 0,\\ a_2 &= 0, \quad b_2 = 4, \qquad c_2 = 2,\\ a_3 &= 1, \quad b_3 = 4 + 2r^2, \qquad c_3 = 3,\\ a_4 &= \frac{7}{3}, \quad b_4 = 4 + 6r^2, \qquad c_4= \frac{11}{3} + \frac{4}{3}r^2,\\ a_5 &= \frac{23}{6}+\frac{11}{12}r^2, \quad b_5 = 4 + 12 r^2 + \frac{3}{2}r^4, \qquad c_5=\frac{25}{6} + \frac{55}{12}r^2,\\ a_6 &=\frac{163}{30} + \frac{47}{12}r^2, \quad b_6 = 4 + 20 r^2 + \frac{15}{2}r^4, \qquad c_6=\frac{137}{30} + \frac{607}{60}r^2 + \frac{16}{15}r^4. \end{align} Is the integral $I_n$ in the general closed form given by (1) for all $n$? If so, what are the constants $a_n$, $b_n$, and $c_n$?

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  • $\begingroup$ Please don't use displaystyle in the title. Thanks. $\endgroup$
    – robjohn
    Commented Nov 26, 2013 at 20:40
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    $\begingroup$ @robjohn The same thing could have been mentioned in a more polite manner. There's nothing wrong in your statement but it came off a bit rude. No offense! $\endgroup$ Commented Nov 26, 2013 at 20:43
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    $\begingroup$ @AhaanRungta: I did not mean it to. I said Please and Thanks. What would have been a more polite way? $\endgroup$
    – robjohn
    Commented Nov 26, 2013 at 20:44
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    $\begingroup$ a.) Please X. Thanks. b.) Could you please x so that Y? Edited this time. Thank you. <- which do you think sounds better? $\endgroup$ Commented Nov 26, 2013 at 20:46
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    $\begingroup$ @CarlMummert: I added the context to the question. $\endgroup$
    – bkocsis
    Commented Nov 27, 2013 at 20:49

3 Answers 3

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First lets make the integral from $-\pi/2$ to $\pi/2$. Set $x = \phi-\pi/2$. We then get $$I = \int_{-\pi/2}^{\pi/2}(1+r\sin(x))^{-n}(x+\pi/2)dx$$ Now recall that $$(1+r\sin(x))^{-n} = \sum_{k=0}^{\infty} (-1)^k \dbinom{n+k}k r^k \sin^k(x)$$ Hence, $$I = \sum_{k=0}^{\infty}(-r)^k \dbinom{n+k}k \int_{-\pi/2}^{\pi/2} (x+\pi/2)\sin^k(x)dx$$ Now from here, we can obtain $$\int_{-\pi/2}^{\pi/2} \sin^k(x) dx \text{ and }\int_{-\pi/2}^{\pi/2} x\sin^k(x) dx$$ Lets call them $I_k$ and $J_k$ respectively. Hence, $$I = \dfrac{\pi}2\sum_{k=0}^{\infty}r^{2k} \dbinom{n+2k}{2k} I_{2k} - \sum_{k=0}^{\infty}r^{2k+1} \dbinom{n+2k+1}{2k+1} J_{2k+1} \tag{$\star$}$$


Expression for $J_{2k+1}$ and convergence of the above summation:

We have $$J_{2k+1} = -\int_{-\pi/2}^{\pi/2}x \sin^{2k}(x) d(\cos(x)) = \int_{-\pi/2}^{\pi/2} \cos(x) \sin^{2k}(x) dx + 2k \int_{-\pi/2}^{\pi/2} x \cos^2(x) \sin^{2k-1}(x) dx$$ $$\int_{-\pi/2}^{\pi/2} \cos(x) \sin^{2k}(x) dx = \int_{-1}^1 t^{2k} dt = \dfrac2{2k+1}$$ $$\int_{-\pi/2}^{\pi/2} x \cos^2(x) \sin^{2k-1}(x) dx = J_{2k-1} - J_{2k+1}$$ Hence, $$(2k+1)J_{2k+1} = \dfrac2{2k+1} + 2k J_{2k-1}$$ Using this recurrence you can obtain $J_{2k+1}$. Also, it is easy to show that $\left \vert J_{2k+1} \right \vert \leq 2$ using induction.

Similarly, $I_{2k} = \dfrac1{4^k} \dbinom{2k}k \pi \leq \pi$.

Now, for $\vert r \vert < 1$, we have $\displaystyle \sum_{k=0}^{\infty} \dbinom{n+2k}{2k} r^{2k}$ and $\displaystyle \sum_{k=0}^{\infty} \dbinom{n+2k+1}{2k+1} r^{2k}$ converges.

This ensures $\star$ makes perfect sense.

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  • $\begingroup$ I was looking for a closed form expression, i.e. a finite sum of known functions. See edited version of the question. $\endgroup$
    – bkocsis
    Commented Dec 1, 2013 at 17:41
  • $\begingroup$ @bkocsis I don't understand what you mean by closed form. The above is a closed form in my opinion. If you get the answer as $\sin(k)$ or $\cos(k)$ or $\Gamma(k)$ or $\arctan(k)$, etc, do you think that is closed form? All these are infinite series and not a closed form. By truncating the above sum, you can get an arbitrary accurate answer. $\endgroup$
    – user17762
    Commented Dec 1, 2013 at 17:48
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    $\begingroup$ @bkocsis Also, can you remove the down-vote. My answer is a perfectly valid answer. $\endgroup$
    – user17762
    Commented Dec 1, 2013 at 17:55
  • $\begingroup$ @bkocsis You are showing a wiki article to mean what a closed form is? I assume that you are interested in calculating the value of the integral for a fixed $r$ and $n$. If so, the infinite series above is a perfectly valid answer. $\endgroup$
    – user17762
    Commented Dec 1, 2013 at 23:17
  • $\begingroup$ The beauty of closed form expressions is that the parameter dependencies become tractable, and that they are often calculable quickly. Most CPUs and GPUs can calculate many transcendental functions extremely efficiently. Closed form results can speed up simulations very significantly. If you can show that your sums converge for $|r|<1$ and give $J_{2k+1}$ specifically, I can remove the down vote. $\endgroup$
    – bkocsis
    Commented Dec 2, 2013 at 0:14
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The integral can be evaluated by realizing that it is a derivative with respect to $R=1/r$, i.e. $$I_n =\int_{0}^{\pi} \frac{\phi}{(1 - r \cos\phi)^n} \,{\rm d}\phi = R^n \int_{0}^{\pi} \frac{\phi}{(R - \cos\phi)^n} \,{\rm d}\phi = \frac{(-1)^{n-1} R^n}{(n-1)!} \frac{d^{n-1}}{dR^{n-1}}\int_{0}^{\pi} \frac{\phi}{R - \cos\phi} \,{\rm d}\phi. $$ This integral can be evaluated with a Weierstrass substitution $t = \tan (\phi/2)$, $$F(R) = \int_{0}^{\pi} \frac{\phi}{R - \cos\phi} \,{\rm d}\phi = \frac{4r}{1+r}\int_0^{\infty} \frac{\arctan t}{q^2 + t^2} dt = \frac{4}{\sqrt{R^2-1}}\int_0^{\pi/2} \arctan \left(q \tan y \right) dy = \frac{4}{\sqrt{R^2-1}}\left[\chi_2(q) - {\rm arctanh}(q) \ln(q) \right] =\frac{4}{\sqrt{R^2-1}}\left[\chi_2(q) +\frac{1}{2} {\rm arccosh}(R) {\rm arctanh}(R) \right],$$ where $q=\sqrt{(1-r)/(1+r)}$ and $\chi_2(x)$ is the Legendre chi function. We can now derive the integral by taking derivatives of this function $$I_n = \frac{(-1)^{n-1} R^n}{(n-1)!} \frac{d^{n-1}}{dR^{n-1}}F(R).$$ Note that ${\rm arccosh\,}R=2{\rm arctanh\,}q$, ${\rm arccoth\,}R=\ln q$, and $$\frac{d}{dx}\chi_2(x) = \frac{{\rm arctanh}(x)}{x} \quad{\rm so\quad} \frac{d}{dR}\chi_2(q) = \frac{{\rm arctanh}(q)}{R^2 - 1} = \frac{1}{2}\frac{{\rm arccosh}(R)}{R^2 - 1} $$ $$\frac{d}{dR} {\rm arccosh}(R) = \frac{1}{\sqrt{R^2 - 1}} \quad{\rm and\quad} \frac{d}{dR} {\rm arccoth}(R) = \frac{-1}{R^2 - 1} = \frac{1}{2}\left(\frac{1}{R+1} - \frac{1}{R-1}\right).$$ The derivatives for $n\geq 3$ simplify to $$I_n = \frac{R^n}{S^{2n-2}}\left\{ \frac{4 A_{n-1}}{S}\left[\chi_2(q) - {\rm arctanh(q)\ln q}\right]\right. \\ \left.+2\sum_{k=0}^{n-2} \frac{A_k A_{n-k-2}}{k+1} \ln q + \sum_{j=0}^{n-3}\sum_{k=0}^{n-3-j}\frac{A_k A_{n-3-j-k}}{n-1-k} \frac{(R-1)^{1+j}-(R+1)^{1+j}}{1+j} \right\} $$ where $S=\sqrt{R^2-1}$ and $A_n$ is a polynomial of $R$ defined as $$A_n = \frac{(-1)^n}{n!}\left(R^2-1\right)^{n+(1/2)}\frac{d^{n}}{dR^{n}} \left(R^2-1\right)^{-1/2}\\ =\left(\frac{R+1}{4}\right)^n\sum_{s=0}^{n}\frac{(2s)!}{(s!)^2}\frac{[2(n-s)]!}{[(n-s)!]^2}q^{2s}.$$ The coefficients are related to Legendre polynomials $P_{2n}(0)=(-1)^n(2n)!/[4^n (n!)^2]$. The result simplifies to $$ I_n=\left(\frac{R}{R-1}\right)^n \frac{q^2}{4^{n-2}}\left\{ a(q) [\chi_2(q) - {\rm arctanh(q)}\ln q] +b(q)\ln q + c(q)\right\}$$ where $$ a(q)=\sum_{K=0}^{n-1}q^{2K-1} \frac{(2K)!}{(K!)^2} \frac{[2(n-1-K)]!}{[(n-1-K)!]^2}\,, $$ $$ b(q)=\sum_{K=0}^{n-2}q^{2K} \sum_{s=0}^{K}\frac{(2s)!}{(s!)^2}\frac{[2(K-s)]!}{[(K-s)!]^2} \sum_{j=K}^{n-2}\frac{[2(j-K)]!}{[(j-K)!])^2}\frac{[2(n-2-j)]!}{[(n-2-j)!]^2}\frac{2}{j-s+1}\,, $$ $$ c(q)=\sum_{K=0}^{n-3}q^{2K+2} \sum_{J=0}^{K} \frac{4^{K-J+1}}{1+K-J} \sum_{s=0}^{J}\frac{(2s)!}{(s!)^2}\frac{[2(J-s)]!}{[(J-s)!]^2}\\ \quad\times \sum_{j=K}^{n-3}\frac{[2(j-K)]!}{[(j-K)!])^2}\frac{[2(n-3-j)]!}{[(n-3-j)!]^2}\frac{1}{n-1-j+s+K-J}\\ \quad+\sum_{K=0}^{n-3}q^{2K} \sum_{J=K}^{L-3} \frac{-4^{J-K+1}}{1+J-K}\sum_{s=0}^{K}\frac{(2s)!}{(s!)^2}\frac{[2(K-s)]!}{[(K-s)!]^2}\\ \quad\times \sum_{j=J}^{n-3}\frac{[2(j-J)]!}{[(j-J)!])^2}\frac{[2(n-3-j)]!}{[(n-3-j)!]^2}\frac{1}{n-1-j+s+J-K}\,.$$

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Not a full answer, but maybe interesting:

$I_{n}(r)$ obeys a recursion relation: $$ I_{n+1}(r)=I_{n}(r)+\frac{r}{n} \frac{\partial}{\partial r}I_{n}(r) $$

A large $n$ approximation is: $$ I_{n}(r)\stackrel{n \rightarrow \infty}{\sim} \int_{0}^{\pi}\phi \exp(n\, r \cos \phi)\mathrm{d}\phi, $$ which shows that for large $n$ the integral depends only on the product $n\,r$. Interestingly, also the approximation seems not to be solvable in closed form.

For the approximation I used: $$ \left(1 - \frac{z}{n}\right)^{n} \stackrel{n \rightarrow \infty}{\sim} e^{- z} $$

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