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UnkleRhaukus

  • 3 years ago

\[\begin{align*} f(x)&=\frac{\sin(ax)}x\\ \\F(p)&=\mathcal L\left\{\frac{\sin(ax)}x\right\}\\ \\&=\int_p^\infty\mathcal L\left\{\sin(ax)\right\} \cdot\text dp'\\ \\&=\int_p^\infty\frac{a}{p'^2+a^2} \cdot\text dp'\\ \\&=\left.\arctan\left(\frac {p'}a\right) \right|_p^\infty\\ \\&=\frac \pi2-\arctan\left(\frac{p}{a}\right)\\ \end{align*}\]

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  1. mahmit2012
    • 3 years ago
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  2. mahmit2012
    • 3 years ago
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  3. UnkleRhaukus
    • 3 years ago
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    ok~!

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