q is not equal to
\$
\int_{t0}^{t} I(\tau) d\tau + V(t_0)
\$
it has no sense to sum apples (charge) and pears (potential) :)
if you have
\$
V(t)=\frac{1}{C}\int_{t0}^{t} I(\tau) d\tau + V(t_0)
\$
\$ \frac{1}{C} \$ is just multiplying \$\int_{t0}^{t} I(\tau) d\tau \$ and not \$ V(t_0)\$
\$
\int_{t0}^{t} I(\tau) d\tau
\$
is a \$ \Delta Q\$
so \$
V(t)=\frac{1}{C}\int_{t0}^{t} I(\tau) d\tau + V(t_0)=\frac{\Delta Q}{C} +V(t_0)
\$
I hope this is what you were missing :)
Let's check the numbers. You have a R-C circuit starting at 1.28 V decaying to 5.0 V, and want to know how long it will take to get to 3.12 V.
That means the interval in question is decaying 1.979 times towards the final value, which happens in .683 time constants. A time constant is (10 nF)(1 kΩ) = 10 µs, so this decay should take 6.8 µs. You are seeing it take 4.3 µs.
What you are seeing is 37% faster than expected. First look at tolerances. Capacitors can easily be ±20%, sometimes up to 50% off for certain types. Resistors are generally 5% unless stated otherwise. That immediately gives you 25% slop. These things simply aren't that accurate unless you paid big bucks for high accuracy components, particularly the capacitor. Still, the error you are seeing is a bit more than that.
Assuming the component values are exactly as stated, how much extra current would have to be dumped onto the cap to get to the threshold early? In 4.3 µs, the R-C would decay by a factor of 1.54 on its own, which means to 2.58 V. That means something else is dumping enough current onto the cap to raise it 540 mV in 4.3 µs. (540 mV)(10 nF)/(4.3 µs) = 1.26 mA. Now look up what the input current of a 74LS14 is. These kinds of inputs float high, so 1.26 mA could be plausible. I haven't looked it up, that's your job.
Add to that the slop in part values, and there doesn't seem to be much of a mystery here.
Also, that's a long way to go to make a one-shot. Where did you get this circuit from? R3 and R5 are pointless, and nowadays you'd use a HC logic family instead of the archaic LS. Unless you are doing this for learning, just go get a one-shot chip and be done with it.
Best Answer
You could use a FET bus switch, like the SN74CBT3125, however this will add about 5 ohms resistance in series with each capacitor. If that is a problem, you could always use reed relays like this one, each driven by a FET or BJT transistor, which would exactly emulate your current arrangement with the pushbuttons.