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I want to output the following expression: $tδ^{''}\left(t\right)=-2δ^{'}\left(t\right)$

My attempt, which did not lead to success: $\int_{ }^{ }δ^{'}\left(t-s\right)f^{''}\left(t\right)dt=f^{''}\left(s\right)$

$-\int_{ }^{ }δ^{''}\left(t-s\right)f^{'}\left(t\right)dt=f^{''}\left(s\right)$

$f^{'}\left(t\right)=g^{'}\left(t\right)\left(t-s\right)$

$-\int_{ }^{ }δ^{''}\left(t\right)\cdot t\cdot g^{'}\left(t+s\right)dt=g^{'}\left(s\right)$

$g^{'}\left(s\right)=\int_{ }^{ }g^{'}\left(t\right)δ\left(t-s\right)$

$δ^{''}\left(t\right)t=-δ\left(t\right)$

I probably made a mistake somewhere

Maxim Storm
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    What is the definition of $\int_{-\infty}^\infty t \delta''(t) f(t)dt$ and $\int_{-\infty}^\infty \delta'(t) f(t)dt$ – reuns Mar 01 '22 at 19:21
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    In your very first line, don't you mean to use the idea that $\delta$, not its derivative, is the "identity operator" under convolution? That's not a good start. In any case, I think you are doing more work than is necessary... – paul garrett Mar 01 '22 at 19:49

2 Answers2

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This should be a standard sort of computation, I think simpler than what you wrote. Short of doing the induction that proves, analogously, that $t\delta^{(n)}=-n\cdot \delta^{(n-1)}$, the explicit computations for $n=1,2$ are adequate to suggest the full case. The notation may be slightly different than what is familiar to you, but I claim it is very economical, and accurate. :)

Namely, $$(t\delta')(f)\;=\;\delta'(tf)\;=\;-\delta({d\over dt}(tf))\;=\;-\delta(f+tf')\;=\;-\delta(f)+(0\cdot f'(0)=-\delta(f)$$ That is, $t\delta'=-\delta$. An example of the induction step is $$ (t\delta'')(f) \;=\; \delta''(tf) \;=\; (\delta')'(tf)\;=\; -\delta'({d\over dt}(tf)) \;=\; -\delta'(f+tf') \;=\; -\delta'(f)-\delta'(tf') $$ $$ \;=\; -\delta'(f)-(t\delta')(f') \;=\; -\delta'(f)+\delta(f') \;=\; -\delta'(f)-\delta'(f) \;=\; -2\cdot \delta'(f) $$

paul garrett
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  1. Start with this identity of distributions $$ t\delta(t)~=~0.$$
  2. Differentiate $$ \delta(t)+t\delta^{\prime}(t)~=~0.$$
  3. Differentiate again $$ 2\delta^{\prime}(t)+t\delta^{\prime\prime}(t)~=~0.$$
Qmechanic
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