As usual in such examples there is no need to integrate against a test function. It is a simple consequence of the fact that if a sequence (or net) of distributions converges in the distributional sense, then the sequence obtained by differentation also converges. In particular, this applies if the original sequence consists of functions which converge in pretty well any sensible classical sense, e.g., locally $L^1$ as is the case here.

In this example we consider the functions which are defined as $x^{\epsilon}$ on the positive real axis, and $0$ elsewhere. They converge to the Heaviside function in the above sense and so we can differentiate to obtain the required result. Most of the representations of $\delta$-sequences can be obtained in this way: consider the primitives and show they form a sequence which converges to the Heaviside function in a suitable sense. The result then follows as above.

The first example (Sokhotzky) can be proved in the same way in one line, if you can integrate $\dfrac{\epsilon}{x^2+\epsilon^2}.$

As usual in such examples, there is no need to integrate against a test function. One can simply use the fact that if a sequence (or net) of distributions converges in the distributional sense, then so does the one obtained by differentiating term by term. In particular, this applies when the sequence consists of functions which converge in pretty well any sensible classical sense, e.g., locally $L^1$ as in the case in point, that of the functions which are defined as $x^{\epsilon}$ on the positive real axis, and $0$ elsewhere. They converge to the Heaviside function and we can differentiate to obtain the required result.

Most of the examples of $\delta$-sequences in the literature can be verified in this way: consider the terms’ primitives and show that they converge to the Heaviside function. The result then follows as above.

The first example (Sokhotsky) in the question can be proved in one line, after integrating $\dfrac{\epsilon}{x^2+\epsilon^2}.$

As usual in such examples there is no need to integrate against a test function. It is a simple consequence of the fact that if a sequence (or net) of distributions converges in the distributional sense, then the sequence obtained by differentation also converges. In particular, this applies if the original sequence consists of functions which converge in pretty well any sensible classical sense, e.g., locally $L^1$ as is the case here.

In this example we consider the functions which are defined as $x^{\epsilon}$ on the positive real axis, and $0$ elsewhere. They converge to the Heaviside function in the above sense and so we can differentiate to obtain the required result. Most of the representations of $\delta$-sequences can be obtained in this way: consider the primitives and show they form a sequence which converges to the Heaviside function in a suitable sense. The result then follows as above.

As usual in such examples there is no need to integrate against a test function. It is a simple consequence of the fact that if a sequence (or net) of distributions converges in the distributional sense, then the sequence obtained by differentation also converges. In particular, this applies if the original sequence consists of functions which converge in pretty well any sensible classical sense, e.g., locally $L^1$ as is the case here.

In this example we consider the functions which are defined as $x^{\epsilon}$ on the positive real axis, and $0$ elsewhere. They converge to the Heaviside function in the above sense and so we can differentiate to obtain the required result. Most of the representations of $\delta$-sequences can be obtained in this way: consider the primitives and show they form a sequence which converges to the Heaviside function in a suitable sense. The result then follows as above.

The first example (Sokhotzky) can be proved in the same way in one line, if you can integrate $\dfrac{\epsilon}{x^2+\epsilon^2}.$