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Borwein integral
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<p>In <a href="/facts/Mathematics/pxTouaz4">mathematics</a>, a Borwein integral is an <a href="/facts/Integral/V7lyV997">integral</a> whose unusual properties were first presented by mathematicians <a href="/facts/David_Borwein/IwlJbIID">David Borwein</a> and <a href="/facts/Jonathan_Borwein/j5xCll2M">Jonathan Borwein</a> in 2001. Borwein integrals involve products of sinc ( a x ) {\displaystyle \operatorname {sinc} (ax)} , where the <a href="/facts/Sinc_function/lSvDtHy0">sinc function</a> is given by sinc ( x ) = sin ( x ) / x {\displaystyle \operatorname {sinc} (x)=\sin(x)/x} for x {\displaystyle x} not equal to 0, and sinc ( 0 ) = 1 {\displaystyle \operatorname {sinc} (0)=1} . </p><p>These integrals are remarkable for exhibiting apparent patterns that eventually break down. The following is an example. </p> ∫ 0 ∞ sin ( x ) x d x = π 2 ∫ 0 ∞ sin ( x ) x sin ( x / 3 ) x / 3 d x = π 2 ∫ 0 ∞ sin ( x ) x sin ( x / 3 ) x / 3 sin ( x / 5 ) x / 5 d x = π 2 {\displaystyle {\begin{aligned}&\int _{0}^{\infty }{\frac {\sin(x)}{x}}\,dx={\frac {\pi }{2}}\\[10pt]&\int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\,dx={\frac {\pi }{2}}\\[10pt]&\int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}{\frac {\sin(x/5)}{x/5}}\,dx={\frac {\pi }{2}}\end{aligned}}} <p>This pattern continues up to </p> ∫ 0 ∞ sin ( x ) x sin ( x / 3 ) x / 3 ⋯ sin ( x / 13 ) x / 13 d x = π 2 . {\displaystyle \int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/13)}{x/13}}\,dx={\frac {\pi }{2}}.} <p>At the next step the pattern fails, </p> ∫ 0 ∞ sin ( x ) x sin ( x / 3 ) x / 3 ⋯ sin ( x / 15 ) x / 15 d x = 467807924713440738696537864469 935615849440640907310521750000 π = π 2 − 6879714958723010531 935615849440640907310521750000 π ≈ π 2 − 2.31 × 10 − 11 . {\displaystyle {\begin{aligned}\int _{0}^{\infty }{\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/15)}{x/15}}\,dx&={\frac {467807924713440738696537864469}{935615849440640907310521750000}}~\pi \\[5pt]&={\frac {\pi }{2}}-{\frac {6879714958723010531}{935615849440640907310521750000}}~\pi \\[5pt]&\approx {\frac {\pi }{2}}-2.31\times 10^{-11}.\end{aligned}}} <p>In general, similar integrals have value π/2 whenever the numbers 3, 5, 7… are replaced by positive real numbers such that the sum of their reciprocals is less than 1. </p><p>In the example above, 1/3 + 1/5 + … + 1/13 < 1, but 1/3 + 1/5 + … + 1/15 > 1. </p><p>With the inclusion of the additional factor 2 cos ( x ) {\displaystyle 2\cos(x)} , the pattern holds up over a longer series, </p> ∫ 0 ∞ 2 cos ( x ) sin ( x ) x sin ( x / 3 ) x / 3 ⋯ sin ( x / 111 ) x / 111 d x = π 2 , {\displaystyle \int _{0}^{\infty }2\cos(x){\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/111)}{x/111}}\,dx={\frac {\pi }{2}},} <p>but </p> ∫ 0 ∞ 2 cos ( x ) sin ( x ) x sin ( x / 3 ) x / 3 ⋯ sin ( x / 111 ) x / 111 sin ( x / 113 ) x / 113 d x ≈ π 2 − 2.3324 × 10 − 138 . {\displaystyle \int _{0}^{\infty }2\cos(x){\frac {\sin(x)}{x}}{\frac {\sin(x/3)}{x/3}}\cdots {\frac {\sin(x/111)}{x/111}}{\frac {\sin(x/113)}{x/113}}\,dx\approx {\frac {\pi }{2}}-2.3324\times 10^{-138}.} <p>In this case, 1/3 + 1/5 + … + 1/111 < 2, but 1/3 + 1/5 + … + 1/113 > 2. The exact answer can be calculated using the general formula provided in the next section, and a representation of it is shown below. Fully expanded, this value turns into a fraction that involves two 2736 digit integers. </p> π 2 ( 1 − 3 ⋅ 5 ⋯ 113 ⋅ ( 1 / 3 + 1 / 5 + ⋯ + 1 / 113 − 2 ) 56 2 55 ⋅ 56 ! ) {\displaystyle {\frac {\pi }{2}}\left(1-{\frac {3\cdot 5\cdots 113\cdot (1/3+1/5+\dots +1/113-2)^{56}}{2^{55}\cdot 56!}}\right)} <p>The reason the original and the extended series break down has been demonstrated with an intuitive mathematical explanation. In particular, a <a href="/facts/Random_walk/08v0jmfv">random walk</a> reformulation with a causality argument sheds light on the pattern breaking and opens the way for a number of generalizations. </p>