Indefinite Integral of a Function multiplied by Heaviside I want to do the following integral:
$$\int^\ x*H(x-a) dx $$
In mathematica I get that $$\int^\ x*H(x-a) dx = \frac{1}2*(x-a)*(x+a)*H(x-a)$$ 
where H(x-a) Is the heaviside function.
But by hand I can't find  the same result.What I'm trying to do is that:
$$\int^\ x*H(x-a) dx = \frac{1}2*x^2*H(x-a)+\frac{1}2\int^\ x^2*Dirac(x-a) dx
= \frac{1}2*x^2*H(x-a)+\frac{a^2}2 $$
That is in no way near the answer the software gave me. How can I compute the initial integral by hand to match the software?
I Thank you very much for your help

Actually, your solution is not that far off if you keep in mind that the indefinite integral is only determined up to the addition of a constant. However, here is my way to compute the integral:
First off, we need to clarify: The Heaviside function I'll use is defined by
$$H(x) = \left\lbrace \begin{array}{cl} 1 & , ~ x \geq 0 \\ 0 & , ~ x < 0 \end{array}\right.$$
An indefinite integral is just a function that can be written as a definite integral with a variable as the upper bound:
$$\left( \int x \cdot H(x-a) ~ dx \right)(x_0) = \int\limits_a^{x_0} x \cdot H(x-a) ~ dx$$
Now we just have to divide the upper term into two cases: 
If $x_0  \geq a$ then $H(x-a) = 1$ for all $x \in [a,x_0]$ and therefore the (right) intgral equals
$$\int\limits_a^{x_0} x~dx = \frac{1}{2}(x_0)^2 - \frac{1}{2} a^2 = \frac{1}{2} (x_0 - a)(x_0+a)$$
For $x_0 <a$ we have $H(x-a) = 0$ for all $x \in [x_0, a]$ and therefore the integral equals $0$. 
So if we merge both cases one may write 
$$\int\limits_a^{x_0} x \cdot H(x-a) ~ dx = \frac{1}{2} (x_0-a)(x_0 + a)H(x_0 - a)$$
which is the solution you obtained using Mathematica.
