P ( x ) = ∫ − x 2 0 ( n = 1 ∏ ∞ n 2 n 2 + t ) d t
When x ∈ Z and P ( x ) = 0 , the value of P ( x ) can be expressed in the form π β α , where α is a perfect square and β is a prime number. Find α β .
This problem is original. The picture of the graph was produced by Wolfram .
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Hello, Mark. I thought saying P ( x ) besides zero would suffice, but I could see why that would be ambiguous. I just fixed it!
What I have written now is what I really meant to say.
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Yeah, it's fixed now. Interesting problem.
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The question needs re-expressing. The integral is P ( x ) = ∫ − x 2 0 n = 1 ∏ ∞ ( 1 + n 2 t ) d t = 2 ∫ 0 x u n = 1 ∏ ∞ ( 1 − n 2 u 2 ) d u and hence P ( x ) = 2 ∫ 0 x u × π u sin π u d u = π 2 ∫ 0 x sin π u d u = π 2 2 ( 1 − cos π x ) . When x is an integer, this integral is either π 2 4 or zero.
The way the question is written, it wants P ( x ) to take the nonzero value for all integers x , which is not true. The question should either say "When x ∈ Z and P ( x ) = 0 ...", or else "When x ∈ Z is odd ... " to be accurate, and obtain the intended answer of 4 2 = 1 6 .