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Mathematics
General Math
Math Proof Training and Practice
Math Challenge - December 2021
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[QUOTE="julian, post: 6574673, member: 142346"] In post original post, #17, I got the partial solution: \begin{align*} \int_{-\infty}^\infty \dfrac{| \sin (\alpha x) |}{1 + x^2} dx & = 2 - 4 \sum_{k=1}^\infty \frac{e^{-2 k \alpha}}{4k^2-1} \nonumber \\ & \equiv 2 + 2 \sum_{k=1}^\infty \left( \frac{1}{2k+1} - \frac{1}{2k-1} \right) e^{-2 k \alpha} \end{align*} I now write it is closed form. \begin{align*} \int_{-\infty}^\infty \dfrac{| \sin (\alpha x) |}{1 + x^2} dx & = 2 + 2 e^\alpha \sum_{k=1}^\infty \frac{ (e^{-\alpha})^{2k+1} }{2k+1} - 2 e^{-\alpha} \sum_{k=0}^\infty \frac{(e^{-\alpha})^{2 k + 1}}{2k+1} \nonumber \\ &= 2 + 2 ( e^\alpha \tanh^{-1} (e^{-\alpha}) - 1) - 2 e^{-\alpha} \tanh^{-1} (e^{-\alpha}) \nonumber \\ &= 2 e^\alpha \tanh^{-1} (e^{-\alpha}) - 2 e^{-\alpha} \tanh^{-1} (e^{-\alpha}) \nonumber \\ & = 4 \sinh \alpha \tanh^{-1} (e^{-\alpha}) \end{align*} EDIT: The series expansion for ##\tanh^{-1} x## only converges if ##|x| \leq 1## and so the above formula applies when ##e^\alpha \geq 1##, i.e. when ##\alpha \geq \ln (1) = 0##. I really do need to go to bed now! [/QUOTE]
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Forums
Mathematics
General Math
Math Proof Training and Practice
Math Challenge - December 2021
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