In mathematics, a Borwein integral is an integral involving products of sinc(ax), where the sinc function is given by sinc(x) = sin(x)/x for x not equal to 0, and sinc(0) = 1.
These integrals are remarkable for exhibiting apparent patterns which, however, eventually break down. An example is as follows,
∫ 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 This pattern continues up to
∫ 0 ∞ sin ( x ) x sin ( x / 3 ) x / 3 ⋯ sin ( x / 13 ) x / 13 d x = π / 2 . Nevertheless, at the next step the obvious pattern fails,
∫ 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 . 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.
In the example above, 1/3 + 1/5 + … + 1/13 < 1, but 1/3 + 1/5 + … + 1/15 > 1.
An example for a longer series,
∫ 0 ∞ 2 cos ( x ) sin ( x ) x sin ( x / 3 ) x / 3 ⋯ sin ( x / 111 ) x / 111 d x = π / 2 , but
∫ 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 , is shown in together with an intuitive mathematical explanation of the reason why the original and the extended series break down. In this case, 1/3 + 1/5 + … + 1/111 < 2, but 1/3 + 1/5 + … + 1/113 > 2.
Given a sequence of real numbers, a 0 , a 1 , a 2 , . . . , a general formula for the integral
∫ 0 ∞ ∏ k = 0 n sin ( a k x ) a k x d x
can be given. To state the formula, one will need to consider sums involving the a k . In particular, if γ = ( γ 1 , γ 2 , . . . , γ n ) ∈ { ± 1 } n is an n -tuple where each entry is ± 1 , then we write b γ = a 0 + γ 1 a 1 + γ 2 a 2 + ⋯ + γ n a n , which is a kind of alternating sum of the first few a k , and we set ϵ γ = γ 1 γ 2 ⋯ γ n , which is either ± 1 . With this notation, the value for the above integral is
∫ 0 ∞ ∏ k = 0 n sin ( a k x ) a k x d x = π 2 a 0 C n
where
C n = 1 2 n n ! ∏ k = 1 n a k ∑ γ ∈ { ± 1 } n ϵ γ b γ n sgn ( b γ )
In the case when a 0 > | a 1 | + | a 2 | + ⋯ + | a n | , we have C n = 1 .
Furthermore, if there is an n so that for each k = 0 , . . . , n − 1 we have 0 < a n < 2 a k and a 1 + a 2 + ⋯ + a n − 1 < a 0 < a 1 + a 2 + ⋯ + a n − 1 + a n , which means that n is the first value when the partial sum of the first n elements of the sequence exceed a 0 , then C k = 1 for each k = 0 , . . . , n − 1 but
C n = 1 − ( a 1 + a 2 + ⋯ + a n − a 0 ) n 2 n − 1 n ! ∏ k = 1 n a k
The first example is the case when a k = 1 2 k + 1 . Note that if n = 7 then a 7 = 1 15 and 1 3 + 1 5 + 1 7 + 1 9 + 1 11 + 1 13 ≈ 0.955 but 1 3 + 1 5 + 1 7 + 1 9 + 1 11 + 1 13 + 1 15 ≈ 1.02 , so because a 0 = 1 , we get that
∫ 0 ∞ sin ( x ) x sin ( x / 3 ) x / 3 ⋯ sin ( x / 13 ) x / 13 d x = π 2
which remains true if we remove any of the products, but that
∫ 0 ∞ sin ( x ) x sin ( x / 3 ) x / 3 ⋯ sin ( x / 15 ) x / 15 d x = π 2 ( 1 − ( 3 − 1 + 5 − 1 + 7 − 1 + 9 − 1 + 11 − 1 + 13 − 1 + 15 − 1 − 1 ) 7 2 6 ⋅ 7 ! ⋅ ( 1 / 3 ⋅ 1 / 5 ⋅ 1 / 7 ⋅ 1 / 9 ⋅ 1 / 11 ⋅ 1 / 13 ⋅ 1 / 15 ) )
which is equal to the value given previously.
