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===Average distance to an arbitrary external point=== [[File:Cjcdiscex.svg|thumb|The average distance from a disk to an external point]]Turning to an external location, we can set up the integral in a similar way, this time obtaining <math display="block">b(q) = \frac{2}{3\pi} \int_0^{\textrm{sin}^{-1}\tfrac{1}{q}} \biggl\{ s_{+}(\theta)^3-s_{-}(\theta)^3\biggr\} \textrm{d}\theta</math> where the law of cosines tells us that {{math|''s''<sub>+</sub>(θ)}} and {{math|''s''<sub>–</sub>(θ)}} are the roots for {{math|''s''}} of the equation <math display="block">s^2-2qs\,\textrm{cos}\theta+q^2\!-\!1=0.</math> Hence <math display="block">b(q) = \frac{4}{3\pi} \int_0^{\textrm{sin}^{-1}\tfrac{1}{q}} \biggl\{ 3q^2\textrm{cos}^2\theta \sqrt{1-q^2 \textrm{sin}^2\theta} + \Bigl( 1-q^2 \textrm{sin}^2\theta\Bigr)^{\tfrac{3}{2}} \biggl\} \textrm{d}\theta. </math> We may substitute {{math|''u'' {{=}} ''q'' sinθ }} to get <math display="block">\begin{align}b(q) &= \frac{4}{3\pi} \int_0^1 \biggl\{ 3\sqrt{q^2-u^2} \sqrt{1-u^2} + \frac{(1-u^2)^{\tfrac{3}{2}}}{\sqrt{q^2-u^2}} \biggr\} \textrm{d}u \\[0.6ex] &= \frac{4}{3\pi} \int_0^1 \biggl\{ 4\sqrt{q^2-u^2} \sqrt{1-u^2} - \frac{q^2-1}{q} \frac{\sqrt{1-u^2}}{\sqrt{q^2-u^2}} \biggr\} \textrm{d}u \\[0.6ex] &= \frac{4}{3\pi} \biggl\{ \frac{4q}{3} \biggl( (q^2+1)E(\tfrac{1}{q^2})-(q^2-1)K(\tfrac{1}{q^2}) \biggr) - (q^2-1) \biggl(qE(\tfrac{1}{q^2})-\frac{q^2-1}{q}K(\tfrac{1}{q^2}) \biggr) \biggr\} \\[0.6ex] &= \frac{4}{9\pi} \biggl\{ q(q^2+7)E(\tfrac{1}{q^2}) - \frac{q^2-1}{q}(q^2+3)K(\tfrac{1}{q^2}) \biggr\} \end{align}</math> using standard integrals.<ref>[[Gradshteyn and Ryzhik]] 3.155.7 and 3.169.9, taking due account of the difference in notation from Abramowitz and Stegun. (Compare A&S 17.3.11 with G&R 8.113.) This article follows A&S's notation.</ref> Hence again {{math|''b''(1) {{=}} {{sfrac|32|9π}}}}, while also<ref>Abramowitz and Stegun, 17.3.11 et seq.</ref> <math display="block">\lim_{q \to \infty} b(q) = q + \tfrac{1}{8q}.</math>
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