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Let $f_1(s,\tau)=\delta(e^{(\tau+s)} \sinh \tau )$.

This should be equal to $0$ everywhere unless $\tau=0$, but I think there should be some constant multiplying the delta, i.e. it should be equivalent to $f_2(s,\tau)=c(\tau,s)\delta(\tau)$ where $c(\tau,s)$ is a constant.

How can I manipulate $f_1$ to resemble $f_2$?

Thanks.

Jyrki Lahtonen
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yoki
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1 Answers1

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Consider, for fixed $s$, the function $$ g_s(\tau) = \exp(\tau + s)\sinh \tau $$ we have $$ g_s'(\tau) = \exp(\tau +s)(\sinh\tau + \cosh\tau) = \exp(2\tau + s)$$ Hence, $g_s$ is monotone, therefore we have by substitution for $\phi \in C_c^\infty(\def\R{\mathbf R}\R)$: \begin{align*} \int_\R f_1(s,\tau)\phi(\tau)\, d\tau &= \int_\R \delta\bigl(g_s(\tau)\bigr)\phi(\tau)\, d\tau\\ &= \int_{-\exp(s)/2}^\infty \delta(x)\phi\bigl(g_s^{-1}(x)\bigr)\, g_s^{-1}{}'(x)\, dx\\ &= \phi\bigl(g_s^{-1}(0)\bigr)g_s^{-1}{}'(0) \end{align*} Now, as $g_s(0) = 0$, we have $g_s^{-1}(0) = 0$ and by the rule for the derivative of the inverse, we have $$ g_s^{-1}{}'(0) = \frac 1{g_s'(0)} = \frac 1{\exp(s)} $$ Hence, we may continue \begin{align*} \int_\R f_1(s,\tau)\phi(\tau)\, d\tau &= \phi\bigl(g_s^{-1}(0)\bigr)g_s^{-1}{}'(0) \\ &= \phi(0) \exp(-s)\\ &= \int_\R \delta(\tau)\exp(-s)\phi(\tau)\, d\tau \end{align*} That is, $$ f_1(s,\tau) = \delta(\tau)\exp(-s). $$

martini
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