Help me to survive from a big dilemma(Unit step function)

[sinaphysics]

Consider:

$$u(t)=\begin{cases} 1\quad \quad \quad \quad t>0 \\ 0\quad \quad \quad \quad t<0 \end{cases}$$
Now I want to calculate this:

$$\int _{ 0 }^{ a }{ \frac { u(t)-u(t-a) }{ { t }^{ 2 } } } dt$$
whereas: a>0
What is confusing me is this point that should our answer for the integral include the step function again?

 

[SteamKing]

Consider:

$$u(t)=\begin{cases} 1\quad \quad \quad \quad t>0 \\ 0\quad \quad \quad \quad t<0 \end{cases}$$
Now I want to calculate this:

$$\int _{ 0 }^{ a }{ \frac { u(t)-u(t-a) }{ { t }^{ 2 } } } dt$$
whereas: a>0
What is confusing me is this point that should our answer for the integral include the step function again?

It's not clear what you mean when you say, "should our answer for the integral include the step function again?"

What I see is a step function which is active when t > 0 but which is turned off for t > a, so that there is a non-zero integrand in the interval 0 < t < a.

 

[HallsofIvy]

In this integral t is always between 0 and a so t is positive while t- a is negative. u(t)= 1 and u(t- a)= 0.

Your integral is just $\int_0^a \frac{1}{t^2} dt$