Your integral is just the laplace transformation of evaluated at s=0.
Okay, I'm trying to compute the following integral:
But first, I will try to prove that converge.
The integral (Dirichlet's test)
But what about ?
I know that is our problematic point and also I know that .
Can I say that because that limit exists the function is integrable?
Now to the the second part-computation.
It's from a book by G. M. Fikhtengoltz, The Fundamentals of Mathematical Analysis.
(I'm translating this from Russian into Hebrew and then into English.)
We will represent I in form of sum:
Putting or and the he says the following:
we substituting or , and we get:
Why we need the both substitutions, or ?
To be continued...
A more general form of the indefinite integral is...
This integral has great importance in several applcations, so that almost nobody remembers the 'theoretical bag' hidden in it: the definite integral (1) converges if it is processed as Riemann integral and doesn't converge if it is processed as Lebesgue integral, as for example in...
Lebesgue integration - Wikipedia, the free encyclopedia
All that is well known of course but, proper considering this example, I wonder how to justify motivations regarding the 'superiority' of the Lebesgue integral like that ...
... the Riemann integral is not defined for a wide range of functions and situations of importance in applications (and of interest in theory). For example, the Riemann integral can easily integrate density to find the mass of a steel beam, but cannot accommodate a steel ball resting on it. This motivates other definitions, under which a broader assortment of functions are integrable. The Lebesgue integral, in particular, achieves great flexibility by directing attention to the weights in the weighted sum...