Transforming an integral equation using taylor series

Question

Can I transform the equation using a taylor series, as the integration seems to impossible to be done by itself?

Answer 1

This is not a complete answer. Using a identity cos(t/2)^2=(1+cos(t))/2 and substituting: t=2m,m=ArcCos(n) I have a integral:

$\int_{0}^{1}\!{\frac {\int_{0}^{1}\! \sqrt{4\,{n}^{2}{s}^{2}+8\,Rns+4 \,{R}^{2}+{s}^{2}}\,{\rm d}s}{ \sqrt{-{n}^{2}+1}}}\,{\rm d}n $

Using Mathematica Solving integral for s:

$\int _0^1\left(-\frac{4 n R^2}{\sqrt{1-n^2} \left(4 n^2+1\right)}-\frac{2 R^2 \log \left(2 \left(\sqrt{4 n^2+1}+2 n\right) R\right)}{\sqrt{1-n^2} \left(4 n^2+1\right)^{3/2}}+\frac{2 R^2 \log \left(\sqrt{\left(4 n^2+1\right) \left(4 (n+R)^2+1\right)}+4 n (n+R)+1\right)}{\sqrt{1-n^2} \left(4 n^2+1\right)^{3/2}}+\frac{2 n^2 \sqrt{\left(4 n^2+1\right) \left(4 (n+R)^2+1\right)}}{\sqrt{1-n^2} \left(4 n^2+1\right)^{3/2}}+\frac{2 n R \sqrt{\left(4 n^2+1\right) \left(4 (n+R)^2+1\right)}}{\sqrt{1-n^2} \left(4 n^2+1\right)^{3/2}}+\frac{\sqrt{\left(4 n^2+1\right) \left(4 (n+R)^2+1\right)}}{2 \sqrt{1-n^2} \left(4 n^2+1\right)^{3/2}}\right)dn$

First component of the integral:

$\int _0^1-\frac{4 n R^2}{\sqrt{1-n^2} \left(4 n^2+1\right)}dn=-\frac{2 R^2 \sinh ^{-1}(2)}{\sqrt{5}}$

The Second and Third component of the integral can't solve (Mathematica and MAPLE).

Fourth component of the integral:

$\int _0^1\frac{2 n^2 \sqrt{\left(4 n^2+1\right) \left(4 (n+R)^2+1\right)}}{\sqrt{1-n^2} \left(4 n^2+1\right)^{3/2}}dn=4\,{\frac {1}{\sqrt {4\,{R}^{2}+1} \left( {R}^{2}-3/4+i \right) \left( 2\,R+2+i \right) ^{2}} \left( -2/5\,\sqrt {-2\,R+i} \left( -1/ 2+3/8\,i+1/2\,{R}^{3}+ \left( 1+i \right) {R}^{2}- \left( 1/8-3/2\,i \right) R \right) \sqrt {2\,R-2+i}R\sqrt {i+2\,R}\sqrt {-2\,R-2+i}{ \it EllipticPi} \left( {\frac {i\sqrt {2\,R-2+i}}{\sqrt {2\,R+2+i}}},{ \frac {-2-11\,i- \left( 6+8\,i \right) R}{10\,R-10+5\,i}},{\frac {i \sqrt {3+4\,i-4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}}}} \right) -1/5\, \sqrt {-2\,R+i}\sqrt {2\,R-2+i}R\sqrt {i+2\,R} \left( -1+3/4\,i-{R}^{3 }-2\,{R}^{2}- \left( 7/4-i \right) R \right) \sqrt {-2\,R-2+i}{\it EllipticPi} \left( {\frac {i\sqrt {2\,R-2+i}}{\sqrt {2\,R+2+i}}},{ \frac {-10+5\,i- \left( 6-8\,i \right) R}{10\,R-10+5\,i}},{\frac {i \sqrt {3+4\,i-4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}}}} \right) + \sqrt {-2\,R+i} \left( -1+3/4\,i+i{R}^{2}- \left( 1-2\,i \right) R \right) \sqrt {2\,R-2+i}R\sqrt {i+2\,R}\sqrt {-2\,R-2+i}{\it EllipticPi} \left( {\frac {i\sqrt {2\,R-2+i}}{\sqrt {2\,R+2+i}}},{ \frac {2\,R+2+i}{2\,R-2+i}},{\frac {i\sqrt {3+4\,i-4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}}}} \right) +2/5\,\sqrt {-2\,R+i}\sqrt {2\,R-2+i} \sqrt {i+2\,R} \left( -{\frac{15}{32}}-5/8\,i+i{R}^{4}- \left( 1-5/4\, i \right) {R}^{3}- \left( {\frac{7}{8}}+3/4\,i \right) {R}^{2}- \left( 1/2+{\frac {19\,i}{16}} \right) R \right) \sqrt {-2\,R-2+i}{ \it EllipticF} \left( {\frac {i\sqrt {2\,R-2+i}}{\sqrt {2\,R+2+i}}},{ \frac {i\sqrt {3+4\,i-4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}}}} \right) + \left( -1/4\,\sqrt {-2\,R+i} \left( i{R}^{2}-5/4\,i-R \right) \sqrt {2\,R-2+i}\sqrt {i+2\,R}\sqrt {-2\,R-2+i}{\it EllipticE } \left( {\frac {i\sqrt {2\,R-2+i}}{\sqrt {2\,R+2+i}}},{\frac {i\sqrt {3+4\,i-4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}}}} \right) + \left( iR+ {R}^{2}-5/4 \right) \left( {R}^{2}+1/4 \right) \right) \left( {R}^{ 2}+2\,R+5/4 \right) \right) } $

Fifth component of the integral:

$\int _0^1\frac{2 n R \sqrt{\left(4 n^2+1\right) \left(4 (n+R)^2+1\right)}}{\sqrt{1-n^2} \left(4 n^2+1\right)^{3/2}}dn={\frac {8\,i\sqrt {-2\,R+i}R\sqrt {i+2\,R}}{\sqrt {4\,{R}^{2}+1}\sqrt {2\,R-2+i}\sqrt {-2\,R-2+i} \left( 10\,R+10+5\,i \right) } \left( \left( -2\,{R}^{3}- \left( 2+3\,i \right) {R}^{2}+ \left( 1-2\,i \right) R \right) {\it EllipticPi} \left( {\frac {i\sqrt {2\,R-2+i}}{ \sqrt {2\,R+2+i}}},{\frac {-2-11\,i- \left( 6+8\,i \right) R}{10\,R-10 +5\,i}},{\frac {i\sqrt {3+4\,i-4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}} }} \right) + \left( -1+2\,i-2\,{R}^{2}- \left( 2-i \right) R \right) R {\it EllipticPi} \left( {\frac {i\sqrt {2\,R-2+i}}{\sqrt {2\,R+2+i}}}, {\frac {-10+5\,i- \left( 6-8\,i \right) R}{10\,R-10+5\,i}},{\frac {i \sqrt {3+4\,i-4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}}}} \right) + \left( -5/2\,{\it EllipticPi} \left( {\frac {i\sqrt {2\,R-2+i}}{ \sqrt {2\,R+2+i}}},{\frac {2\,R+2+i}{2\,R-2+i}},{\frac {i\sqrt {3+4\,i -4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}}}} \right) +{\it EllipticF} \left( {\frac {i\sqrt {2\,R-2+i}}{\sqrt {2\,R+2+i}}},{\frac {i\sqrt { 3+4\,i-4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}}}} \right) \left( {R}^{ 2}-2\,R+5/4 \right) \right) \left( 2\,R+2+i \right) \right) } $

Sixth component of the integral:

$\int _0^1\frac{\sqrt{\left(4 n^2+1\right) \left(4 (n+R)^2+1\right)}}{2 \sqrt{1-n^2} \left(4 n^2+1\right)^{3/2}}dn={\frac {16\,i\sqrt {-2\,R+i}\sqrt {i+2\,R}}{\sqrt {4\,{R}^{2}+1}\sqrt {2\,R-2+i}\sqrt {-2\,R-2+i} \left( 10\,R+10+5\,i \right) } \left( \left( -1-i/2+i{R}^{2}- \left( 3/2-i \right) R \right) R{\it EllipticPi} \left( {\frac {i\sqrt {2\,R-2+i}}{\sqrt {2\,R+2+i}}},{ \frac {-2-11\,i- \left( 6+8\,i \right) R}{10\,R-10+5\,i}},{\frac {i \sqrt {3+4\,i-4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}}}} \right) - \left( 1+i/2+i{R}^{2}+ \left( 1/2+i \right) R \right) R{\it EllipticPi} \left( {\frac {i\sqrt {2\,R-2+i}}{\sqrt {2\,R+2+i}}},{ \frac {-10+5\,i- \left( 6-8\,i \right) R}{10\,R-10+5\,i}},{\frac {i \sqrt {3+4\,i-4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}}}} \right) -1/4\, {\it EllipticF} \left( {\frac {i\sqrt {2\,R-2+i}}{\sqrt {2\,R+2+i}}},{ \frac {i\sqrt {3+4\,i-4\,{R}^{2}}}{\sqrt {-3+4\,i+4\,{R}^{2}}}} \right) \left( {R}^{2}-2\,R+5/4 \right) \left( 2\,R+2+i \right) \right) }$

where: $i$ is represents the imaginary unit.

$EllipticPi$ gives the complete elliptic integral of the third kind.

$EllipticF$ gives the elliptic integral of the first kind.

Answer 2

I think it is better to go one step further to write the integrand as:

$$ \int \int \sqrt{\left(\frac{s}{2}\right)^2+\left( R+ s \cos \frac{t}{2} \right)^2} ds dt $$

Integral on $s$ can be taken (using Mathematica/Wolfram Alpha for example). Then see if you can simplify it more.

Another way that I can think of is changing coordinates to polar coordinates which is of course messy!

Answer 3

If $R$ is sufficiently large then we can rewrite the integrand to do a Talyor expansion. We start by pulling a factor of $R^2$ out of the square root.

$$ \sqrt{\frac{s^2}{4}+R^2+2Rs\cos(t/2)+s^2\cos^2(t/2)}$$ $$= R \sqrt{1+\color{red}{\frac{2s\cos(t/2)}{R}+\frac{s^2/4+s^2\cos^2(t/2)}{R^2}}}$$

$$= R \sqrt{1+\color{red}{\epsilon}}=R(1+\frac12\epsilon-\frac32\epsilon^2+\cdots)$$

As indicated abouve, we can expand $\sqrt{1+\epsilon}$ in a Mclauren series. Each of the terms of this series can be integrated, though it would be a pain in the neck to do so. The series will only converge if $|\epsilon| < 1$; this requirement leads us to a lower bound for $R$. We will use the fact that $0<s<1$, $|\cos(t)|<1$, and $R>1$.

$$|\epsilon| = \frac{2s|\cos(t/2)|}{R}+\frac{s^2/4+s^2\cos^2(t/2)}{R^2} < \frac{2}{R} + \frac{3/4}{R^2} < \frac{7/4}{R}$$

So we have at least that the series will converge if $R>7/4$, though we could find tighter lower bound than this.