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$$
\mbox{In spherical coordinates,}\quad
\delta\pars{\vec{r}}={\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}}\quad
\mbox{such that}
$$
\begin{align}
\color{#66f}{\large\int_{{\mathbb R}^{3}}\delta\pars{\vec{r}}\,\dd^{3}\vec{r}}
&=\int_{0^{-}}^{\infty}\dd r\,r^{2}\int_{0}^{\pi}\dd\theta\,\sin\pars{\theta}
\int_{0}^{2\pi}\dd\phi\,{\delta\pars{r}\delta\pars{\cos\pars{\theta}}\delta\pars{\phi} \over r^{2}}
\\[3mm]&=\underbrace{\bracks{\int_{0^{-}}^{\infty}\delta\pars{r}\,\dd r}}
_{\ds{=\ 1}}\
\underbrace{\bracks{%
\int_{0}^{\pi}\delta\pars{\cos\pars{\theta}}\sin\pars{\theta}\,\dd\theta}}
_{\ds{=\ 1}}\
\underbrace{\bracks{\int_{0}^{2\pi}\delta\pars{\phi}\,\dd\phi}}_{\ds{=\ 1}}\
\\[3mm]&=\ \color{#66f}{\Large 1}
\end{align}
$$\mbox{Note that}\quad
\int_{{\mathbb R}^{3}}\delta\pars{\vec{r} - \vec{r}_{0}}\,\dd^{3}\vec{r}
=\int_{{\mathbb R}^{3}}\delta\pars{\vec{r}}\,\dd^{3}\vec{r}
$$
Yes it is fine to say that $\frac{1}{\lambda}\int_0^D \delta(x)\sin(\lambda x) = \sin(0) = 0$.
You can think of the dirac delta as just being an infinite mass on the point $0$. Any other parts of the integral become irrelevant.
Also, you have to choose which condition to use at $t=0, x=0$ surely.
Best Answer
We start from $$ \delta(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ix\xi} \, d\xi . $$
If $g$ is analytic then $$ g(x) = \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{k!}x^k $$ and we define $$ g(-i\partial_x) = \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{k!}(-i\partial_x)^k . $$
Note now that $$ g(-i\partial_x) e^{ix\xi} = \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{k!}(-i\partial_x)^k e^{ix\xi} = \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{k!}(-i\cdot i\xi)^k e^{ix\xi} = \sum_{k=0}^{\infty} \frac{g^{(k)}(0)}{k!}\xi^k e^{ix\xi} = g(\xi) e^{ix\xi} . $$
Therefore, applying $g(-i\partial_x)$ on $\delta$ gives $$ g(-i\partial_x) \delta(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} g(-i\partial_x) e^{ix\xi} \, d\xi = \frac{1}{2\pi} \int_{-\infty}^{\infty} g(\xi) e^{ix\xi} \, d\xi = \frac{1}{\sqrt{2\pi}} \tilde{g}(x) $$ from which we conclude $$ \delta(x) = \frac{1}{\sqrt{2\pi}} g(-i\partial_x)^{-1} \tilde{g}(x) . $$