The convolution of two functions and is defined by
The way you have defined means that it is zero outside the interval . So, the product of and is also zero outside the same interval. Therefore it should be sufficient to integrate from 0 to 1.
However, you have not defined the function , and your notation is very inconsistent. Some places you use the variable , other places . So, I'm not sure whether this answers your question.
Thing with 't' is actually only my attempt to solve it, thus:
for t <-pi/2, -pi/2 + 1> X=<-pi/2, t>
for t <-pi/2 + 1, pi/2> X=<t-1, t>
for t <pi/2, pi/2 + 1> X=<t-1, pi/2>
it shoud be the same as you wrote, but with 't'. Then I tried to solve integral:
with appropriate limits. And it gave me really strange conclusion (for quick calculation i used wolfram)
EDIT: tg(x) is tangens function
I don't know where you got the problem from, but it's poorly written to say the least. I can help you with the last convolution though.
Again, I suggest you draw the two signals. The definition of f here means that it's constantly 1 for . Mirroring it and translating it with x gives the right endpoint the coordinate x-1.
Alright then i understand this second convolution, but i have problem with first.
By the way, function tg(x) is simply tangens function (i forgot that somebody write it as tan(x))
I got
f1*f2(t) = {-infinity for t in <-pi/2, 1 - pi/2>
ln|cos(t-1)| - ln|cos(t)| for t in <1 - pi/2, pi/2>
infinity for t in <pi/2, pi/2 + 1>
0 otherwise
but i am not sure about infinity
Hm but second integral is like: ln|cos(x-1)| - ln|cos(x)| does it mean that f1*f2(x) is not convergent in every possible case?
edit: i calculate those integrals over and over again but still do not know what is result
edit2: could you please write result funcion, becasuse i am getting quite desperate now
is result only that the function f1*f2(x) is divergent ? (withou any calculation?)
That solution is correct. If is the function defined by , and g is any other function, then is the mean value of g on the interval [t–1,t].
For the function , its integral over any interval containing diverges. Thus is infinite (or if you prefer, undefined) in those intervals specified in your answer.