CÁLCULO GRACELI SINTÉTICO [40]

$\int _{-\infty }^{\infty }f(\tau )e^{\frac {-i\tau \omega }{a}}\,{\frac {d\tau }{a}}$ / 1 / $\mathrm {\omega _{n}\ {\overset {\underset {\mathrm {def} }{}}{=}}\ n{\frac {2\pi }{T}}}$ $sin$ $\mathrm {e^{i2\pi \phi }}$ $T^{\nu }{}_{\alpha }$ $\lambda$ ω ${\mathcal {T}}$ , $\sigma$ , $\psi$

$\mathrm {{\hat {f}}(\omega )={\frac {1}{|a|}}{\hat {g}}{\Bigl (}{\frac {\omega }{a}}{\Bigr )}}$ . /1 / $\mathrm {\omega _{n}\ {\overset {\underset {\mathrm {def} }{}}{=}}\ n{\frac {2\pi }{T}}}$ $sin$ $\mathrm {e^{i2\pi \phi }}$ $T^{\nu }{}_{\alpha }$ $\lambda$ ω ${\mathcal {T}}$ , $\sigma$ , $\psi$

${\mathcal {F}}\{f(at)\}$ $={\frac {1}{|a|}}{\hat {g}}{\Bigl (}{\frac {\omega }{a}}{\Bigr )}$

f(t)=\left({\frac {1}{2\pi }}\right){\int _{-\infty }^{\infty }F(\omega )e^{i\omega t}dw}

/

\mathrm {\omega _{n}\ {\overset {\underset {\mathrm {def} }{}}{=}}\ n{\frac {2\pi }{T}}}

sin

\mathrm {e^{i2\pi \phi }}

T^{\nu }{}_{\alpha }

\lambda

ω

{\mathcal {T}}

,

\sigma

,

\psi

{\mathcal {F}}(\omega )={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }f(t)e^{-i\omega t}\,dt,

f(t)={\frac {1}{\sqrt {2\pi }}}\int _{-\infty }^{\infty }{\mathcal {F}}(\omega )e^{it\omega }\,d\omega .

\phi (\lambda )=\int _{-\infty }^{\infty }f(x)e^{i\lambda x}\,dx.

f(x)=\int _{-\infty }^{\infty }{\hat {f}}(\xi )e^{2i\pi x\xi }\,d\xi

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