# Integral in Polar Co-ordinates: Can you help evaluate it?

I have

$$\int_{0}^{r_{0}}\int_{a}^{b}r_{1}e^{-\beta(r_{1}^{2}+r_{2}^{2}-2r_{1}r_{2}\cos(\theta))}d\theta dr_{1}$$

Can anyone help me break it down for general $a$ and $b$?

Alex

• Can $a.b$ be functions of $r_1$? Jul 25, 2013 at 21:05
• possibly, in my case they aren't - its easier if they aren't I think
– apg
Jul 25, 2013 at 21:09
• perhaps any help if a and b are just constants?
– apg
Jul 25, 2013 at 21:10
• I'm thinking if they are independent of $r_1$, perhaps switch the order of integration, factor out $e^{-\beta r_2}$ outside both integrals and trying to integrate w.r.t $r_1$ you have $\int e^{ar^2 + br}rdr$ and try to substitute $u = ar^2+br$ and reduce this to $\int e^{u^2}udu$ and $\int e^{u^2}du$, where the first is just chain rule, and the second is not analytic but is a well studied special function. Jul 25, 2013 at 21:13
• But then the dependence on $\theta$ is very ugly. I strongly suspect there is no expression in terms of common special functions. See wolframalpha.com/input/… Jul 25, 2013 at 21:16

$\int_0^{r_0}\int_a^br_1e^{-\beta(r_1^2+r_2^2-2r_1r_2\cos\theta)}~d\theta~dr_1$

$=\int_0^{r_0}r_1e^{-\beta(r_1^2+r_2^2)}\int_a^be^{2\beta r_1r_2\cos\theta}~d\theta~dr_1$

$=\int_0^{r_0}r_1e^{-\beta(r_1^2+r_2^2)}\int_a^b\sum\limits_{n=0}^\infty\dfrac{2^{2n}\beta^{2n}r_1^{2n}r_2^{2n}\cos^{2n}\theta}{(2n)!}d\theta~dr_1+\int_0^{r_0}r_1e^{-\beta(r_1^2+r_2^2)}\int_a^b\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}\beta^{2n+1}r_1^{2n+1}r_2^{2n+1}\cos^{2n+1}\theta}{(2n+1)!}d\theta~dr_1$

For $\int\cos^{2n}\theta~d\theta$ , where $n$ is any non-negative integer,

$\int\cos^{2n}\theta~d\theta=\dfrac{(2n)!\theta}{4^n(n!)^2}+\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin\theta~\cos^{2k-1}\theta}{4^{n-k+1}(n!)^2(2k-1)!}+C$

This result can be done by successive integration by parts, e.g. as shown as http://hk.knowledge.yahoo.com/question/question?qid=7012022000808

For $\int\cos^{2n+1}\theta~d\theta$ , where $n$ is any non-negative integer,

$\int\cos^{2n+1}\theta~d\theta$

$=\int\cos^{2n}\theta~d(\sin\theta)$

$=\int(1-\sin^2\theta)^n~d(\sin\theta)$

$=\int\sum\limits_{k=0}^nC_k^n(-1)^k\sin^{2k}\theta~d(\sin\theta)$

$=\sum\limits_{k=0}^n\dfrac{(-1)^kn!\sin^{2k+1}\theta}{k!(n-k)!(2k+1)}+C$

$\therefore\int_0^{r_0}r_1e^{-\beta(r_1^2+r_2^2)}\int_a^b\sum\limits_{n=0}^\infty\dfrac{2^{2n}\beta^{2n}r_1^{2n}r_2^{2n}\cos^{2n}\theta}{(2n)!}d\theta~dr_1+\int_0^{r_0}r_1e^{-\beta(r_1^2+r_2^2)}\int_a^b\sum\limits_{n=0}^\infty\dfrac{2^{2n+1}\beta^{2n+1}r_1^{2n+1}r_2^{2n+1}\cos^{2n+1}\theta}{(2n+1)!}d\theta~dr_1$

$=\int_0^{r_0}r_1e^{-\beta(r_1^2+r_2^2)}\left[\sum\limits_{n=0}^\infty\dfrac{\beta^{2n}r_1^{2n}r_2^{2n}\theta}{(n!)^2}+\sum\limits_{n=0}^\infty\sum\limits_{k=1}^n\dfrac{4^{k-1}\beta^{2n}r_1^{2n}r_2^{2n}((k-1)!)^2\sin\theta~\cos^{2k-1}\theta}{(n!)^2(2k-1)!}\right]_a^b~dr_1+\int_0^{r_0}r_1e^{-\beta(r_1^2+r_2^2)}\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^k2^{2n+1}\beta^{2n+1}r_1^{2n+1}r_2^{2n+1}n!\sin^{2k+1}\theta}{(2n+1)!k!(n-k)!(2k+1)}\right]_a^b~dr_1$

$=\int_0^{r_0}r_1e^{-\beta(r_1^2+r_2^2)}\left(\sum\limits_{n=0}^\infty\dfrac{\beta^{2n}r_1^{2n}r_2^{2n}(b-a)}{(n!)^2}+\sum\limits_{n=0}^\infty\sum\limits_{k=1}^n\dfrac{4^{k-1}\beta^{2n}r_1^{2n}r_2^{2n}((k-1)!)^2(\sin b~\cos^{2k-1}b-\sin a~\cos^{2k-1}a)}{(n!)^2(2k-1)!}\right)dr_1+\int_0^{r_0}r_1e^{-\beta(r_1^2+r_2^2)}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^k2^{2n+1}\beta^{2n+1}r_1^{2n+1}r_2^{2n+1}n!(\sin^{2k+1}b-\sin^{2k+1}a)}{(2n+1)!k!(n-k)!(2k+1)}dr_1$

$=\int_0^{r_0}\sum\limits_{n=0}^\infty\dfrac{\beta^{2n}r_1^{2n+1}e^{-\beta r_1^2}r_2^{2n}e^{-\beta r_2^2}(b-a)}{(n!)^2}dr_1+\int_0^{r_0}\sum\limits_{n=0}^\infty\sum\limits_{k=1}^n\dfrac{4^{k-1}\beta^{2n}r_1^{2n+1}e^{-\beta r_1^2}r_2^{2n}e^{-\beta r_2^2}((k-1)!)^2(\sin b~\cos^{2k-1}b-\sin a~\cos^{2k-1}a)}{(n!)^2(2k-1)!}dr_1+\int_0^{r_0}\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^k2^{2n+1}\beta^{2n+1}r_1^{2n+2}e^{-\beta r_1^2}r_2^{2n+1}e^{-\beta r_2^2}n!(\sin^{2k+1}b-\sin^{2k+1}a)}{(2n+1)!k!(n-k)!(2k+1)}dr_1$

$=\int_0^{r_0}\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^m\beta^{m+2n}r_1^{2m+2n+1}r_2^{2n}e^{-\beta r_2^2}(b-a)}{m!(n!)^2}dr_1+\int_0^{r_0}\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^m4^{k-1}\beta^{m+2n}r_1^{2m+2n+1}r_2^{2n}e^{-\beta r_2^2}((k-1)!)^2(\sin b~\cos^{2k-1}b-\sin a~\cos^{2k-1}a)}{m!(n!)^2(2k-1)!}dr_1+\int_0^{r_0}\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{m+k}2^{2n+1}\beta^{m+2n+1}r_1^{2m+2n+2}r_2^{2n+1}e^{-\beta r_2^2}n!(\sin^{2k+1}b-\sin^{2k+1}a)}{m!(2n+1)!k!(n-k)!(2k+1)}dr_1$

$=\left[\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^m\beta^{m+2n}r_1^{2m+2n+2}r_2^{2n}e^{-\beta r_2^2}(b-a)}{m!(n!)^2(2m+2n+2)}\right]_0^{r_0}+\left[\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^m4^{k-1}\beta^{m+2n}r_1^{2m+2n+2}r_2^{2n}e^{-\beta r_2^2}((k-1)!)^2(\sin b~\cos^{2k-1}b-\sin a~\cos^{2k-1}a)}{m!(n!)^2(2k-1)!(2m+2n+2)}\right]_0^{r_0}+\left[\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{m+k}2^{2n+1}\beta^{m+2n+1}r_1^{2m+2n+3}r_2^{2n+1}e^{-\beta r_2^2}n!(\sin^{2k+1}b-\sin^{2k+1}a)}{m!(2n+1)!k!(n-k)!(2m+2n+3)(2k+1)}\right]_0^{r_0}$

$=\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^m\beta^{m+2n}r_0^{2m+2n+2}r_2^{2n}e^{-\beta r_2^2}(b-a)}{2m!(n!)^2(m+n+1)}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^m4^{k-1}\beta^{m+2n}r_0^{2m+2n+2}r_2^{2n}e^{-\beta r_2^2}((k-1)!)^2(\sin b~\cos^{2k-1}b-\sin a~\cos^{2k-1}a)}{2m!(n!)^2(2k-1)!(m+n+1)}+\sum\limits_{m=0}^\infty\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^{m+k}2^{2n+1}\beta^{m+2n+1}r_0^{2m+2n+3}r_2^{2n+1}e^{-\beta r_2^2}n!(\sin^{2k+1}b-\sin^{2k+1}a)}{m!(2n+1)!k!(n-k)!(2m+2n+3)(2k+1)}$

• My God thanks a million for this - i've been away from my computer for a few days over the weekend so I'm just working through this - I'll comment in a bit. Alex :)
– apg
Jul 29, 2013 at 19:44
• You've used fubini's theorem near the beginning: $\int_{-r_2\cos\theta}^{r_0-r_2\cos\theta}\int_{\sin a}^{\sin b}r_2~e^{-\beta(r_1^2+r_2^2\theta^2)}~d\theta~dr_1$ this you say becomes $r_2\int_{-r_2\cos\theta}^{r_0-r_2\cos\theta}e^{-\beta r_1^2}~dr_1\int_{\sin a}^{\sin b}e^{-\beta r_2^2\theta^2}~d\theta$ but the limits of the intital expression depend on theta so this step is wrong - is there a way of saving the rest of the expression?
– apg
Jul 30, 2013 at 15:10
• sorry scrap that r2 in front of the second expression
– apg
Aug 1, 2013 at 17:10