A problem in understanding distributions exercise

[Goldbeetle]

I'm reading the first chapters of "A Guide to Distribution Theory and Fourier Transforms".
On page 10, Exercises 3,6,7 the distribution is defined in terms of integrals. The first one is always without integrand (there's only the integral sign). What does that mean? Am I missing something? The book can be consulted online in Google Books.

Thanks for any help!

Goldbeetle

 

[BruceG]

I think it is just you intergrate over two disconnected ranges (-inf,-a) + (+a,+inf).

 

[arkajad]

Exactly

$$\int_{-\infty}^{-a}+\int_{a}^{\infty}\;f(x)dx=\int_{-\infty}^{-a}\;f(x)\;dx+\int_{a}^{\infty}\;f(x)\;dx$$

 

[Goldbeetle]

OK, thanks.

 

[blue_raver22]

Hi Goldbeetle,

It is common in distribution theory to define distributions in terms of integrals. This is because distributions are generalized functions that do not have a well-defined pointwise value, so they cannot be evaluated at a single point like regular functions can. Instead, distributions are defined by how they act on test functions, which are smooth functions with compact support. This is similar to how we define integrals in terms of limits of Riemann sums, rather than evaluating them at a single point.

In the exercises you mentioned, the integral without an integrand is likely meant to be interpreted as an integral over all of space, or an integral over the entire domain of the distribution. This is because distributions are defined over all of space, and the integral is used to represent how the distribution acts on test functions over its entire domain.

I hope this helps clarify the concept of distributions and their definition in terms of integrals. Keep up the good work studying distribution theory and Fourier transforms!