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SCHWEITZER 1958 Problem 10

    Prove that the function f(x)=∫−∞∞(sin⁡θθ)2kcos⁡(2xθ)dθf(x)= \int_{-\infty}^{\infty} \left (\frac{\sin\theta}{\theta} \right )^{2k}\cos (2x\theta) d\theta where kk is a positive integer, satisfies the following conditions:(i) f(x)=0f(x)=0 if ∣x∣≥k\mid x \mid \geq k and f(x)≥0f(x) \geq 0 elsewhere;(ii) in interval (l,l+1)(l,l+1) (l=−k,−k+1,…,k−1)(l= -k, -k+1, \dots , k-1) the function f(x)f(x) is a polynomial of degree 2k−12k-1 at most. (R. 7)

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