An integral
An integral
Let $\alpha, \beta$ be arbitrary positive integer numbers such that $\alpha>\beta$ and $\alpha^2 - \beta^2$ is prime. If a function $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous then evaluate the integral:
$$\mathcal{J} = \int_{\alpha^2-\beta^2}^{\alpha + \beta} \frac{f^2(t) + f^4(t)}{1+f^{10}(t)} \, {\rm d}t$$
$$\mathcal{J} = \int_{\alpha^2-\beta^2}^{\alpha + \beta} \frac{f^2(t) + f^4(t)}{1+f^{10}(t)} \, {\rm d}t$$
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Re: An integral
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, $\displaystyle \int_{\alpha^2-\beta^2}^{\alpha+\beta}\frac{f^2(t)+f^4(t)}{f^{10}(t)+1}=0.$
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