Find the solution to the given derivative of the product: $\big(\cos{x}\delta{x}\big)^{(k)}$

Question

Find the solution to the given derivative of the product $$\big(\cos{x}\delta{x}\big)^{(k)}$$

We need to use the differentiation formula for generalized functions:

$$\big(D^\alpha f, \phi)=(-1)^{|\alpha|}(f,D^\alpha\phi)$$ with $a(x)=cos{x}$, $f=\delta(x)$, and $\alpha=k $

$$\big(D^k(\cos{x},\delta(x))=-1^{k}\int\cos{x} D^{(k)}\delta(x)\text{d}x=D^{(k)}\delta(x)$$

Although the answer is right, I can't make sense of that $\cos{x}$ gives $1$ in order to obtain the right answer.

Any ideas on how to get that last integral right?

Thanks

Answer 1

Lemma: $\newcommand{\ph}{\varphi}\cos(x)\cdot \delta(x)=\delta(x)$.

Proof: In order to prove prove that two distributions $T_1$ and $T_2$ are equal, it is necessary and sufficient to show that $(T_1,\ph)=(T_2,\ph)$ for all test functions $\ph\in C_c^\infty(\mathbb R)$. Letting $\ph\in C_c^\infty(\mathbb R)$ be an arbitrary test function, we compute $$ \begin{align} (\cos x\cdot\,\delta(x), \ph(x)) &= (\delta(x), \cos(x) \cdot \ph(x)) \\&= \cos(0)\cdot \ph(0) \\&= \ph(0) \\&= (\delta(x), \ph(x)). \end{align} $$ The first equality is the only one that needs a little justification. It follows from the definition of the product of a distribution with a smooth function (Wikipedia). From the link, we see that if $T$ is a distribution, and $m$ is any smooth function, then the product $m\cdot T$ is defined by $(m(x)\cdot T,\ph(x))=(T,m(x)\cdot \ph(x))$. Conclude by letting $T=\delta$ and $m=\cos$.

This completes the proof that $\cos(x)\cdot \delta(x)=\delta(x)$.$\tag*{$\square$}$

Since $\cos(x)\cdot \,\delta(x) = \delta(x)$, it immediately follows that $(\cos(x)\cdot \, \delta(x))^{(k)} = \delta(x)^{(k)}$. To be complete, we should describe how $\delta^{(k)}(x)$ acts on test functions, but this is easy: $$ (\delta^{(k)}(x),\ph(x))=(-1)^k (\delta(x),\ph^{(k)}(x))=(-1)^k \ph^{(k)}(0). $$

Answer 2

Find the solution to the given derivative of the product $$\big(\cos{x}\delta(x)\big)^{(k)}$$

$$\big(D^\alpha f,\phi\big)=(-1)^{|\alpha|}(f,D^{\alpha}\phi),$$ with $f(x)=\cos{x}$, $a(x)=\delta(x)$ and $\phi(x)$, and $\alpha=k $

$m=1$

Then, by the rule $\big(f(x)a(x),\phi(x)\big)=\big((f(x),a(x)\phi(x)\big)$

$$\big(\cos{x}\delta(x)\big)^{\prime}=\big(\cos{x}'\delta(x),\phi(x)+\cos{x}\delta'(x),\phi(x)\big)$$

$$\big(\cos{x}\delta(x)\big)^{\prime}=\big((\delta'(x),\cos{x}\phi(x))+(\delta(x),\cos{x}'\phi(x)\big)$$

Let us split, consider the first part on the LHS above:

$$\big(\delta'(x),\cos{x}\phi(x)\big)\implies$$ $$- (\delta (x), [\cos x \phi(x)]')= - \big(\delta (x), \cos x \phi'(x)) + (\delta (x), \sin x \phi'(x)\big)= $$ $$-(\delta (x), \cos x \phi'(x))= - \phi'(0)= \big(\delta'(x), \phi(x)\big)=\delta'(x)$$

and consider the second part on the LHS above $$\big(\delta(x),\cos{x}'\phi(x)\big)=\big(\delta'(x),\cos{x}\phi(x)\big)=\delta'(x)$$

We sum up, and obtain:

$$\big(\cos{x}\delta(x)\big)^{\prime}=2\delta'(x)$$