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8 votes
2 answers
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integral transform of Fibonacci polynomials is integral

The Fibonacci polynomials are defined recursively by $F_0(x)=0, F_1(x)=1$ and $F_n(x)=xF_{n-1}(x)+F_{n-2}(x)$, for $n\geq2$. While computing certain integrals, I observe the following (numerically) ...
T. Amdeberhan's user avatar
27 votes
2 answers
2k views

Are these two new ways of representing odd zeta values as integrals known?

This is inspired by the same beautiful integral expression for $\zeta(3)$ as this question, but goes in a slightly different direction. Writing the original integral in the form $$\int_0^1\frac{x(1-x)}...
Wolfgang's user avatar
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