Let $D'(I)$ be the space of distributions on an open interval $I$, and let $D(I)$ be the space of test functions on $I$.
I got the following homework assignment:
"Define $u\in D'(\mathbb{R})$ be defined through $\left\langle u,\phi\right\rangle=-\int_{0}^{\infty}\ln(t)\phi'(t)dt$ for $\phi\in D(\mathbb{R})$. Determine a function $f:\mathbb{R}\rightarrow\mathbb{C}$ such that $u=f$ on $(-\infty,0)$ and on $(0,\infty)$. Is it meaningful to ask if $u=f$ on $\mathbb{R}?$"
When trying to do this I used
$\left\langle u,\phi\right\rangle=-\int_{0}^{\infty}\ln(t)\phi'(t)dt=/\text{Partial integration}/=-\left[\ln(t)\phi(t)\right]_0^{\infty}+\int_0^{\infty}\frac{1}{t}\phi(t)dt$,
but here I got a bit stuck, and I don't know how to move on. The answers sheet says:
"For example $f(t)=\frac{1}{t}$ for $t>0$, and $f(t)=0$ for $t\leq 0$; No, it's not useful, since $f$ is not locally integrable on $\mathbb{R}$".
But then I get when I calculate:
$\left\langle u,\phi\right\rangle=\int_{-\infty}^{\infty}f(t)\phi(t)dt=\int_{0}^{\infty}\frac{1}{t}\phi(t)dt=\left[\ln(t)\phi(t)\right]_0^{\infty}-\int_{0}^{\infty}\ln(t)\phi'(t)dt$.
Are they somehow saying that $\left[\ln(t)\phi(t)\right]_0^{\infty}=0$? Why is that? I get what they mean by $f$ not being locally integrable on $\mathbb{R}$.
Roll back to the definition of a distribution. If we say that $u=f$ on $(-\infty,0)$ and on $(0,+\infty)$, it means for any test function $\phi$ with support in $(-\infty,0)\cup(0,+\infty)$ you have $$\langle u,\phi\rangle=\int_{-\infty}^0 f(t)\phi(t)dt + \int_0^{\infty} f(t)\phi(t)dt.$$ Now, by taking into account the support of $\phi$, you can conclude that, indeed, $\lim_{x\to +0}\ln (t) \phi(t) = \lim_{x\to +\infty}\ln (t) \phi(t)=0$ and hence the choice of $f$ from your textbook fits.