I need to solve this integral:
$$-\frac{1}{C \cdot R_1} \int_{0}^{t} {v_{in}(\tau)\cdot e^{-(t-\tau)/(C \cdot R_1)}\cdot d\tau}$$
It came from an electronic circuit, so $v_{in}(\tau)$ comes from $v_{in}(t)$, which is the input voltage. The input voltage doesn't have a form at all, so I left it as a function not defined, but I would like to know how to solve that equation with such a handicap, if it is possible.
I tried to do integration by parts, but I got an integral inside another integral:
$$ \left[ e^{-(t-\tau)/(C\cdot R_1)} \right]\_0^t \cdot \int_0^t{v_{in}(\tau)d\tau} -\frac{1}{C\cdot R_1}\cdot e^{-t/(C\cdot R_1)} \int_0^t{\left( e^{\tau/(C\cdot R_1)}\left(\int_0^t{v_{in}(\tau)d\tau} \right)\right) d\tau} $$
I could continue considering the antiderivative of $v_{in}(\tau)$ as a constant, getting $V_{in}(t)$ and $V_{in}(0)$, but I am not pretty sure if this is right. I got the previous result doing integration by parts considering $dv=v_{in}(\tau)d\tau$. I also tried considering $u=v_{in}(\tau)$, but that yielded a result in which I had to do integration by parts again, and I didn't know if I had to consider $d(v_{in})$ as $u$ or as $dv$.