# The Riemann Zeta Function Using The Pythagorean Theorem

#### by C. D. Chester

\begin{aligned} \int_0^\infty \dfrac{\mathrm{e}^{-bx}\sin\left(ax\right)}{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}\sin\left(ax\right)}{a}} \: dx = \dfrac{1}{c} \end{aligned}

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

### C. D. Chester

35F Army Intelligence Analyst
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