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$$F(t)=\displaystyle\int_{-\infty}^{\infty}\left[e^{-2t^2}\left\{\dot\delta(t-2)\right\}+\delta(t^2-16)\right]\,dt$$ How to get rid of the derivative ? and for the second function i wrote it as $\delta\left\{(t-4)(t+4)\right\}$ and then substitute $t-4=u$ but after that i'm again getting a squared function inside, i don't know how to get rid of it either.

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    $\begingroup$ The Dirac Delta is NOT a function. And one does not SOLVE an integral, one evaluates an integral. $\endgroup$ – Mark Viola Jun 22 '17 at 16:13
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i believe These two will get the job done

$$\delta(t^{2}-a^{2})= \dfrac{1}{2|a|} \left \{\delta(t-a)+\delta(t+a ) \right \}\tag{1}$$

$$\int_{-\infty}^{\infty}x(t).\delta^{n}(t-t_0)dt=(-1)^{n} \left \{ \dfrac{d^{n}}{dt^{n}}x(t)\Big{|}_{t=t_0} \right \}\tag{2}$$

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  • $\begingroup$ Thanks Mark it means a lot to me , everyday i learn a lot going through your answers @MarkViola $\endgroup$ – Zeno San Jun 22 '17 at 16:19
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    $\begingroup$ You're welcome! My pleasure. $\endgroup$ – Mark Viola Jun 22 '17 at 16:25
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Hints: for any test function $f(t)$,

$$ \int_{-\infty}^\infty f(t)\dot{\delta}(t-\tau)dt = -\dot{f}(\tau) $$ This follows from "integration by parts". Then,

$$ \delta(g(t)) = \sum_{\tau:g(\tau) = 0}\frac{\delta(t-\tau)}{\vert \dot{g}(\tau)\vert} $$

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    $\begingroup$ A tempered distribution it is $\endgroup$ – Zeno San Jun 22 '17 at 16:20
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    $\begingroup$ @Lelouch.D.Light (+1) for elucidating my comment. $\endgroup$ – Mark Viola Jun 22 '17 at 16:24
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    $\begingroup$ @MarkViola $\int_{-\infty}^\infty f(t)\dot{\delta}(t-\tau)dt$ is 100% standard abuse of notation, particularly in the physics and engineering literature. I have no issue using it - for me it's just a notation to indicate applying a continuous linear functional. $\endgroup$ – icurays1 Jun 22 '17 at 16:27
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    $\begingroup$ Using the notation is fine. But calling it by something that it is not is objectionable. The quotations around "integration by parts" mollify this, but I just wanted to point alert the OP. $\endgroup$ – Mark Viola Jun 22 '17 at 16:40
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    $\begingroup$ I didn't call it anything, and I specifically put quotes around integration by parts because it isn't. This is clearly a homework problem from an engineering course and I didn't want to launch into a discussion. I gave correct hints at the level and in the language of the question. $\endgroup$ – icurays1 Jun 22 '17 at 16:44

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