The convolution of two functions $\displaystyle f_1$ and $\displaystyle f_2$ is defined by
$\displaystyle (f_1 \ast f_2)(t) = \int_{-\infty}^\infty f_1(\tau)f_2(t-\tau) \;dx$
The way you have defined $\displaystyle f_1$ means that it is zero outside the interval $\displaystyle [0,1]$. So, the product of $\displaystyle f_1$ and $\displaystyle f_2$ 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 $\displaystyle g(x)$, and your notation is very inconsistent. Some places you use the variable $\displaystyle t$, other places $\displaystyle x$. So, I'm not sure whether this answers your question.
I suggest you start out with drawing a picture of the two signals. If you mirror and timeshift $\displaystyle f_1$ with x, you end up with these limits ($\displaystyle f_1$ has the width 1 which means the right end point is x after you mirror and shift it, while the left one is x-1):
$\displaystyle x < -\pi/2 : (f_1 * f_2)(x) = 0$
$\displaystyle -\pi/2 \leq x < 1 - \pi/2: (f_1 * f_2)(x) = \int_{-\pi/2}^{x}f_1(x-\tau)f_2(\tau)d\tau$
$\displaystyle 1 - \pi/2 \leq x < \pi/2: (f_1 * f_2)(x) = \int_{x-1}^{x}f_1(x-\tau)f_2(\tau)d\tau$
$\displaystyle \pi/2 \leq x < 1 + \pi/2: (f_1 * f_2)(x) = \int_{x-1}^{\pi/2}f_1(x-\tau)f_2(\tau)d\tau$
$\displaystyle x \geq 1+\pi/2: (f_1 * f_2)(x) = 0$
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:
$\displaystyle \int f_1(t-y)f_2(y) dy$ 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.
$\displaystyle x\geq 2: (f*f)(x)=\int_1^{x-1}d\tau$
Again, I suggest you draw the two signals. The definition of f here means that it's constantly 1 for $\displaystyle [1, \infty)$. 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 $\displaystyle f_1$ is the function defined by $\displaystyle f_1(x) = \begin{cases}1&\text{if }0\leqslant x\leqslant 1\\0&\text{otherwise}\end{cases}$, and g is any other function, then $\displaystyle (f_1*g)(t)$ is the mean value of g on the interval [t–1,t].
For the function $\displaystyle f_2(x) = \begin{cases}\tan x&\text{if }-\pi/2\leqslant x\leqslant \pi/2\\0&\text{otherwise}\end{cases}$, its integral over any interval containing $\displaystyle \pm\pi/2$ diverges. Thus $\displaystyle f_1*f_2$ is infinite (or if you prefer, undefined) in those intervals specified in your answer.