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Since $\alpha^2-\beta^2$ is prime, then $\alpha, \beta$ have to be consecutive. Since $\alpha>\beta$, we obtain that $\alpha-\beta=1$ and thus, $\alpha^2-\beta^2=(\alpha-\beta)(\alpha+\beta)=(\alpha+\beta).$ Hence, $\displaystyle \int_{\alpha^2-\beta^2}^{\alpha+\beta}\frac{f^2(t)+f^4(t)}{f^{10}(t)+1...