- #1

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## Main Question or Discussion Point

Hello--

I have a function:

[tex]

u(t,\tau)=\frac{1}{\pi}\int_{0}^{\infty}\! G(\omega)\, d\omega

[/tex]

[tex]

G(\omega)=4\sqrt{\pi}\frac{\omega^{2}}{\omega_{0}^{3}}\mbox{exp}\left(-\frac{\omega^{2}}{\omega_{0}^{2}}\right)\mbox{cos\left(\omega t-\left(\frac{\omega}{\omega_{0}}\right)^{-\gamma}\omega\tau\right)\mbox{exp}\left(-\frac{1}{2Q}\left(\frac{\omega}{\omega_{0}}\right)^{-\gamma}\omega t\right)}

[/tex]

Now what I would like to do is to take the first time derivative of [tex]u(t,\tau)[/tex] to obtain the roots of [tex]\partial u(t,\tau) / \partial t = 0[/tex], where [tex]\partial u(t,\tau) / \partial t[/tex] is the derivative with respect to [tex]t[/tex].

How would I get started? I think that I need to somehow get rid of the improper integral so that I can then take the first derivative. I've noticed that for [tex]\omega \rightarrow \infty[/tex], [tex]G(\omega) \rightarrow 0[/tex].

I have a function:

[tex]

u(t,\tau)=\frac{1}{\pi}\int_{0}^{\infty}\! G(\omega)\, d\omega

[/tex]

[tex]

G(\omega)=4\sqrt{\pi}\frac{\omega^{2}}{\omega_{0}^{3}}\mbox{exp}\left(-\frac{\omega^{2}}{\omega_{0}^{2}}\right)\mbox{cos\left(\omega t-\left(\frac{\omega}{\omega_{0}}\right)^{-\gamma}\omega\tau\right)\mbox{exp}\left(-\frac{1}{2Q}\left(\frac{\omega}{\omega_{0}}\right)^{-\gamma}\omega t\right)}

[/tex]

Now what I would like to do is to take the first time derivative of [tex]u(t,\tau)[/tex] to obtain the roots of [tex]\partial u(t,\tau) / \partial t = 0[/tex], where [tex]\partial u(t,\tau) / \partial t[/tex] is the derivative with respect to [tex]t[/tex].

How would I get started? I think that I need to somehow get rid of the improper integral so that I can then take the first derivative. I've noticed that for [tex]\omega \rightarrow \infty[/tex], [tex]G(\omega) \rightarrow 0[/tex].