Obtain the derivative of the distribution defined by
$\rho [t] = \int_0^\infty \frac{t(x)}{\sqrt{x}}dx$, and express your answer in the form of an integral over $x$ of a formula that involves $t(x)$ but not $t'(x)$.
Progress: So we know for distribution $\rho'[t] = -\rho[t']$. Applying this we have $-\int_0^\infty \frac{t'(x)}{\sqrt{x}}dx$. We (naively) apply integration by parts to obtain $\left.\frac{t(x)}{\sqrt{x}}\right|_0^\infty - \frac{1}{2}\int_0^\infty \frac{t(x)}{x^{3/2}} dx$. However now this integral doesn't converge, since t(x) can be any smooth function, replace it by it's maximum and you see the integral no longer converges. How do we rectify this?
Let's apply integration by parts less naively, then.
$$ \begin{split} -\int_0^\infty \frac{t'(x)}{\sqrt{x}}dx &= \lim_{a\to 0^+}-\int_a^\infty \frac{t'(x)}{\sqrt{x}}dx \\ &= \lim_{a\to 0^+} -\left.\frac{t(x)}{\sqrt{x}}\right|_a^\infty - \frac{1}{2}\int_a^\infty \frac{t(x)}{x^{3/2}} dx \\ &= \lim_{a\to 0^+} \frac{t(a)}{\sqrt{a}}- \frac{1}{2}\int_a^\infty \frac{t(x)}{x^{3/2}} dx \end{split} $$ Since $t$ is smooth, $t(a)-t(0)$ is $O(a)$, which allows us to continue with $$ \begin{split} \dots & =\lim_{a\to 0^+} \frac{t(0)}{\sqrt{a}}- \frac{1}{2}\int_a^\infty \frac{t(x)}{x^{3/2}} dx \\ & =\lim_{a\to 0^+} \frac{1}{2}\int_a^\infty \frac{t(0)-t(x)}{x^{3/2}} dx \\ & =\frac{1}{2}\int_0^\infty \frac{t(0)-t(x)}{x^{3/2}} dx \end{split} $$ The last integral converges. (why?)