Solving integral of sinusoid involving unit step and dirac delta function I am trying to integrate the following function of $\tau$ from $-\infty$ to $t$. The expected result is $cos(\Omega_0*t)*u(t)$, where $u(t)$ is the unit step function. I'm getting close to the expected result, but not quite. Where am I going wrong?

 A: We have 
$$\begin{align}
\int_{-\infty}^t-\Omega_0\sin(\Omega_0\tau)u(\tau)\,d\tau&=\int_{-\infty}^t\frac{d\cos(\Omega_0\tau)}{d\tau}u(\tau)\,d\tau \\\\
&=\int_{-\infty}^{\infty}\frac{d\cos(\Omega_0\tau)}{d\tau}u(t-\tau)u(\tau)\,d\tau \tag 1\\\\
&=-\int_{-\infty}^{\infty}\frac{du(\tau)u(t-\tau)}{d\tau}\cos(\Omega_0\tau)\,d\tau \tag 2\\\\
&=-\int_{-\infty}^{\infty}\delta(\tau)u(t-\tau)\cos(\Omega_0\tau)\,d\tau+\int_{-\infty}^{\infty}\delta(t-\tau)u(\tau)\cos(\Omega_0\tau)\,d\tau\\\\
&=-u(t)+\cos(\Omega_0t)u(t)
\end{align}$$
as expected!  Note that the development is in the context of Generalized Functions or a Distribution Theoretic Approach.  In that setting, the notation using the integral sign is interpreted as a linear functional and therefore Equations $(1)$ and $(2)$ are more appropriately written as
$$\begin{align}
\left\langle \frac{d\cos(\Omega\tau)}{d\tau},u(\tau)u(t-\tau)\right \rangle&=-\left \langle \cos(\Omega_0\tau),\frac{du(\tau)u(t-\tau)}{d\tau}\right \rangle\\\\
&=-u(t)+\cos(\Omega t)u(t)
\end{align}$$
