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I was asked to find the Bilateral Laplace Transform of the following distribution $f(x) = e^{-x^{2}/2}\delta '(2-3x)$. I simply went on to calculate \begin{equation*} \mathcal{L}(f(x))(z) = \langle e^{-x^{2}/2}\delta ' (2-3x) | e^{-zx} \rangle. \end{equation*} Then, I took \begin{equation*} \langle e^{-x^{2}/2}\delta ' (2-3x) | e^{-zx} \rangle = \langle \delta ' (2-3x) | e^{-x^{2}/2}e^{-zx} \rangle = \langle \delta ' (2-3x) | e^{-zx - x^{2}/2} \rangle. \end{equation*} So as to perform a change of variables and apply the Delta centered at zero. The change I took was \begin{equation*} -s = 2-3x, \quad \textrm{where} \quad ds = 3dx \quad \textrm{and} \quad x = \frac{2+s}{3}. \end{equation*} the dirac delta transforms into $\delta' (-s)$. But since the dirac delta is even, then $\delta '(-s) = \delta '(s)$ and we would get \begin{equation*} \frac{1}{3} \left\langle \delta '(s) | e^{k(z,s)} \right\rangle, \quad k(z,s) = -z\left( \frac{2+s}{3} \right) -\frac{1}{2}\left( \frac{2+s}{3} \right) ^{2}. \end{equation*} After that, it would be a matter of applying the properties of the generalized derivative so as to evaluate directly \begin{align*} -\frac{1}{3} \left\langle \delta (s) | e^{k(z,s)}\frac{\partial k}{\partial s} \right\rangle &= -\frac{1}{3}\left\langle \delta (s) | \left( -\frac{1}{3}z - \frac{1}{9}(2+s) \right) e^{k(z,s)} \right\rangle. \\[1ex] &= -\frac{1}{3} \left( -\frac{1}{3}z -\frac{2}{9} \right) e^{k(z,0)} \\[1ex] &= \left( \frac{1}{9}z + \frac{2}{27} \right)e^{-2z/3}e^{-2/9} \end{align*} This is my result. I thought it'd be a good idea to check with WolframAlpha just to verify, but their result is \begin{equation*} \left( \frac{1}{3}z + \frac{2}{9} \right) e^{-2z/3}e^{-2/9} \end{equation*} And I have no clue how they get their result. I've been banging my head for hours and can't find the mistake. Any help would be appreciated.

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$\delta'$ is odd not even.

Let $$T(x) = \delta(2-3x)= \frac13 \delta(2/3-x)=\frac13 \delta(x-2/3),\qquad T'(x) = -3 \delta'(2-3x)$$ thus $$\langle \delta'(2-3x),\varphi(x) \rangle = -\frac13\langle T'(x),\varphi(x) \rangle=\frac13\langle T(x),\varphi'(x) \rangle\\=\frac19\langle\delta(x-2/3) ,\varphi'(x)\rangle=\frac19 \varphi'(2/3)$$ Finally $$\langle e^{-zx}\delta'(2-3x), e^{-x^2/2} \rangle=\langle \delta'(2-3x), e^{-zx}e^{-x^2/2} \rangle$$

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  • $\begingroup$ Oh boy, I totally forgot about it being odd. This is a step forward, but this yields the WolframAlpha result with the wrong sign. $\endgroup$ – Samuel Alonso Oct 22 '17 at 20:09

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