I have this function $h(t)=\frac{t}{1+t^2}\theta(t)$, and I have to find its unit step response. By the use of this post I took the convolution of the function, $h(t)$, which I consider composed of two convolving functions, $\frac{t}{1+t^2}$ and the unit step function, $\theta(t)$. So the convolution theorem gives
\begin{equation} \int_0^t\theta(t-\tau)\frac{\tau}{1+\tau^2}d\tau \end{equation}
However, there are two issues I am not sure about, first, how does one remove $\theta(t-\tau)$ and include it in the boundaries instead? Then, once this is done, assuming I am right, I get:
\begin{equation} \int_0^t \frac{\tau}{1+\tau^2}d\tau=t \arctan t-\int_0^t\arctan\tau d\tau=\arctan\tau-\big[ t\arctan t-\frac{1}{2}\ln |1+t^2| \big]_0^t=\frac{1}{2}\big(\ln|1+t^2|-\ln1\big) \end{equation}
When I plot this I get
But this does not seem right, when the plot of the function $h(t)$ is:
The part before zero should be different in the first plot, not gradually decreasing?
Where is the error in this procedure?
Thanks
UPDATE:
Correction: have to multiply in $\theta(t)$
\begin{equation} \int_0^t \frac{\tau}{1+\tau^2}d\tau=t \arctan t-\int_0^t\arctan\tau d\tau=\arctan\tau-\big[ t\arctan t-\frac{1}{2}\ln |1+t^2| \big]_0^t=\frac{1}{2}\big(\ln|1+t^2|\big)\theta(t) \end{equation}
Gives the plot
Which by Mathematica codes is still different:
The step response is given by the convolution
$$y(t) = \int_{-\infty}^\infty\theta(t-s)\dfrac{s}{1+s^2}\theta(s)\mathrm{d}s=\dfrac{t-s}{1+(t-s)^2}\theta(t-s)\theta(s)\mathrm{d}s.$$
The first expression is much easier to consider and reduces to
$$y(t)=\int_0^t\dfrac{s}{1+s^2}\mathrm{d}s=\dfrac{1}{2}[\log(1+s^2)]^t_0\theta(t)=\dfrac{1}{2}\log(1+t^2)\theta(t).$$