- #1
Amok
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I had to show that:
[tex] \int_{-\infty}^{\infty} e^{-a \left | t \right |} e^{\omega t} \mathrm{d} t [/tex]
with a positive, is equal to a Lorentzian function.
I simply did this:
[tex]
\int_{-\infty}^{\infty} e^{-a \left | t \right |} e^{i\omega t} \mathrm{d} t = \int_{0}^{\infty} e^{(-a + i\omega) t} \mathrm{d} t + \int_{-\infty}^{0} e^{(a + i\omega)t } \mathrm{d} t
= \frac{-1}{i\omega\ -a} + \frac{1}{i\omega + a} + \lim_{x \to -\infty} e^{(-a + i\omega )x} - \lim_{y \to +\infty} = \frac{2a}{\omega^2 + a^2} + \lim_{x \to +\infty} e^{-ax} lim_{x \to +\infty} e^{i\omega} - \lim_{y \to -\infty} e^{ax} \lim_{y \to -\infty} e^{i\omega} = \frac{2a}{\omega^2 + a^2} [/tex]
Is this correct? I'm asking because I've seen people do this by using Euler's formula and then calculating an integral with a cosine in it, but I don't really see the point of that.
EDIT: can you guys see the whole thing? I can't, what should I do?
[tex] \int_{-\infty}^{\infty} e^{-a \left | t \right |} e^{\omega t} \mathrm{d} t [/tex]
with a positive, is equal to a Lorentzian function.
I simply did this:
[tex]
\int_{-\infty}^{\infty} e^{-a \left | t \right |} e^{i\omega t} \mathrm{d} t = \int_{0}^{\infty} e^{(-a + i\omega) t} \mathrm{d} t + \int_{-\infty}^{0} e^{(a + i\omega)t } \mathrm{d} t
= \frac{-1}{i\omega\ -a} + \frac{1}{i\omega + a} + \lim_{x \to -\infty} e^{(-a + i\omega )x} - \lim_{y \to +\infty} = \frac{2a}{\omega^2 + a^2} + \lim_{x \to +\infty} e^{-ax} lim_{x \to +\infty} e^{i\omega} - \lim_{y \to -\infty} e^{ax} \lim_{y \to -\infty} e^{i\omega} = \frac{2a}{\omega^2 + a^2} [/tex]
Is this correct? I'm asking because I've seen people do this by using Euler's formula and then calculating an integral with a cosine in it, but I don't really see the point of that.
EDIT: can you guys see the whole thing? I can't, what should I do?
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