Stuck on integration by parts question

Question

Here's what I have so far, but I'm stuck with what to do after this

Answer 1

For any compactly supported function $\phi$ you get $$ \int_{-∞}^∞ f(s)\phi'(x-s)\,ds=\left[f(s)·(-ϕ(x−s))\right]_{-∞}^∞+\int_{-∞}^∞ f'(s)\phi(x-s)\,ds \\=\int_{-∞}^∞ f'(s)\phi(x-s)\,ds $$ so that $$ \int_{-∞}^∞ f(s)\frac{d^n}{ds^n}\phi(x-s)\,ds =(-1)^n\int_{-∞}^∞ f(s)\phi^{(n)}(x-s)\,ds \\=(-1)^n\int_{-∞}^∞ f^{(n)}(s)\phi(x-s)\,ds $$ and for $ϕ\to\delta$ as distributions the last term converges to $(-1)^nf^{(n)}(x)$.

Answer 2

Proceeding formally, we have

$$\begin{align} \int_{-\infty}^\infty f(\tau)\,\frac{d^2\,\delta(t-\tau)}{d\tau^2}\,d\tau&=\int_{-\infty}^\infty \left(\frac{d}{d\tau}\left(f(\tau)\frac{d\,\delta(t-\tau)}{d\tau}\right)-\frac{df(\tau)}{d\tau}\frac{d\,\delta(t-\tau)}{d\tau}\right)\,d\tau\\\\ &=-\int_{-\infty}^\infty \frac{df(\tau)}{d\tau}\frac{d\,\delta(t-\tau)}{d\tau}\,d\tau\\\\ &=-\int_{-\infty}^\infty \left(\frac{d}{d\tau}\left(\frac{df(\tau)}{d\tau}\delta(t-\tau)\right)-\frac{d^2f(\tau)}{d\tau^2}\delta(t-\tau)\right)\,d\tau\\\\ &=\int_{-\infty}^\infty \frac{d^2f(\tau)}{d\tau^2}\delta(t-\tau)\\\\ &=\frac{d^2f(t)}{dt^2} \end{align}$$