Some Integrals, A Sum

Calculus Level pending

Assuming x x and y y are both in ( − 1 , 1 ) (-1,1) ,

∫ 0 1 2 ( ∫ 0 y ( ∑ k = 0 ∞ x k ∫ 0 1 cos ⁡ k ( π x ) d x ) d x ) d y = π a b \int_0^{\frac{1}{2}} \left(\int_0^y \left(\sum _{k=0}^{\infty } \frac{x^k}{\int_0^1 \cos ^k(\pi x) \, dx}\right) \, dx\right) \, dy = \frac{{\pi}^a}{b}

where a , b a,b are positive integers. Submit a + b a+b .


The answer is 74.

Christopher Criscitiello by Christopher Criscitiello , 25, USA

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1 solution

Hints:

∫ 0 1 cos ⁡ k ( π x ) d x = ( ( − 1 ) k + 1 ) Γ ( k + 1 2 ) π k Γ ( k 2 ) \int_0^1 \cos ^k(\pi x) \, dx=\frac{\left((-1)^k+1\right) \Gamma \left(\frac{k+1}{2}\right)}{\sqrt{\pi } k \Gamma \left(\frac{k}{2}\right)}

∑ k = 0 ∞ x k ∫ 0 1 cos ⁡ k ( π x ) d x = − 1 − x 2 − x sin ⁡ − 1 ( x ) 1 − x 2 ( x 2 − 1 ) \sum _{k=0}^{\infty } \frac{x^k}{\int_0^1 \cos ^k(\pi x) \, dx}=\frac{-\sqrt{1-x^2}-x \sin ^{-1}(x)}{\sqrt{1-x^2} \left(x^2-1\right)}

∫ 0 y − 1 − x 2 − x sin ⁡ − 1 ( x ) 1 − x 2 ( x 2 − 1 ) d x = − 1 − y 2 sin ⁡ − 1 ( y ) y 2 − 1 \int_0^y \frac{-\sqrt{1-x^2}-x \sin ^{-1}(x)}{\sqrt{1-x^2} \left(x^2-1\right)} \, dx=-\frac{\sqrt{1-y^2} \sin ^{-1}(y)}{y^2-1}

∫ 0 z − 1 − y 2 sin ⁡ − 1 ( y ) y 2 − 1 d y = 1 2 sin ⁡ − 1 ( z ) 2 \int_0^z -\frac{\sqrt{1-y^2} \sin ^{-1}(y)}{y^2-1} \, dy=\frac{1}{2} \sin ^{-1}(z)^2

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