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∫0∞x5dxe5x−1=1abζ(c)Γ(c) \displaystyle \int_{0}^{\infty} \dfrac{x^{5}\mathrm{d}x}{e^{5x} -1} = \dfrac{1}{a^{b}}\zeta(c)\Gamma(c)\\ ∫0∞e5x−1x5dx=ab1ζ(c)Γ(c)
With a,b,ca,b,ca,b,c are positive integers with prime number aaa, find 56abc2\dfrac{5}{6}abc^265abc2
Details and Assumptions \text{ Details and Assumptions } Details and Assumptions 1.)ζ(x)\zeta(x) ζ(x) is the Riemann Zeta Function 2.)Γ(x)\Gamma(x)Γ(x) is the Gamma Function
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