# The Riemann Zeta Function Using The Pythagorean Theorem

#### by C. D. Chester

begin{aligned} int_0^infty dfrac{mathrm{e}^{-bx}sinleft(axright)}{a} : dx= dfrac{1}{b^2+a^2} :; a ne 0, b>0 end{aligned}

This result can be easily found using Integration by Parts twice. What makes this result so peculiar is the relation to the Pythagorean Theorem in its result. Therefor if:

begin{aligned} a^2+b^2=c^2 leadsto dfrac{1}{b^2+a^2} = dfrac{1}{c^2} end{aligned}

begin{aligned} sqrt{int_0^infty dfrac{mathrm{e}^{-bx}sinleft(axright)}{a}} : dx = dfrac{1}{c} end{aligned}

begin{aligned} leadsto Sigma_{c=1}^infty left(left[{int_0^infty dfrac{mathrm{e}^{-bx}sinleft(axright)}{a}} : dx right]^{frac{k}{2}}right) = Sigma_{c=1}^infty left (dfrac{1}{c^k} right) = zeta (k) end{aligned}

### C. D. Chester

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