I think you have a few issues that you need to address.
The primary issue is that the Q of your tank is very low. The Q of a parallel RLC circuit is:
$$Q=R\sqrt{\frac{C}{L}}$$
Considering that your resistive feedback is effectively in parallel with the tank, we can calculate Q:
$$ 1500\Omega \sqrt{\frac{1nF}{15\mu H}}=Q= 12 $$
The Q of 12 is VERY low, so this is probably your main issue. Increasing the magnitude of R1 and R2 should be sufficient for your circuit to oscillate properly. See my example simulation here. If your Q is too low, you will need more gain to compensate for the losses.
A second potential problem is that the op-amp you selected, the 741, has a bandwidth that is a bit low for the frequency you selected. The datasheet indicates a bandwidth of 1.5 MHz, and your oscillator frequency is 1.3 MHz. This may result in your op-amp not providing enough gain for the oscillator to function properly. There are MANY op-amps that would provide an improvement over a 741.
Another possible issue is that oscillators are a bit tricky to get working in simulators. While it sounds like this is not an issue for you, it is a potential pitfall. Often, the random noise that usually starts oscillators in reality does not occur in a simulator. Often a noise source or impulse is required to kick-start the oscillator.
First a comment.
Please label components in a circuit diagram with unique letters/numbers - not just values. Make them easy to read by rotating labels in the same direction if possible. It gets very confusing when describing a circuit by values alone.
I suspect that the waveform your circuit generates is by accident rather than design. You have two types of oscillator in your circuit.
(1) The relaxation (RC) oscillator.
The first (dominant) circuit is a relaxation RC oscillator controlled by the R4,47k and C2,10uF capacitor. If you remove the LC circuit (and its 10uF capacitor) the circuit will oscillate just fine at a low frequency.
2 A damped LC oscillator.
Looking at the waveform it should be obvious that the higher frequency (LC) oscillation is coming at the end of the low frequency pulse, i.e. its trailing edge. This is due to a small amount of feedback through R3, 100k resistor causing a small change in voltage between R1 and R2. This small step voltage is just enough to cause the LC tank circuit to oscillate for a few cycles and then die out (damped oscillation). So we get this. The first few oscillations are large enough to switch the comparator but because the phase change of the feedback is wrong the oscillation is not maintained and it dies out.
Oscillator circuits must satisfy two conditions known as Barkhausen conditions:
The first condition is that the magnitude of the loop gain (Aβ) must be unity. This means the product of gain of amplifier 'A' and the gain of feedback network 'β' has to be unity.
The second condition is that the phase shift around the loop must be 360° or 0°. This means, the phase shift through the amplifier and feedback network has to be 360° or 0°.
As a suggestion (I haven't built this so no guarantees) you could try this configuration based upon the op amp (dual supply) version of the LC oscillator circuit re-configured for a single supply. R1,R2 divide the supply to give a mid-way voltage. C1 decouples this voltage. At least it should point you in the right direction.
)
Best Answer
The equation you have should be correct. I derive it this way:
Note that the only assumption I'm making here is that the output swing of the first opamp is symmetrical about zero, ±\$V_{O1}\$. It doesn't matter whether this swing is related to the supply voltage or not.
The voltage \$V_{FB}\$ at the + input of the first opamp is:
$$V_{FB} = \frac{R3 \cdot V_{O1} + R2 \cdot V_{O2}}{R2 + R3}$$
Solve this for \$V_{O2}\$:
$$V_{O2} = \frac{(R2 + R3)}{R2} V_{FB} - \frac{R3}{R2} V_{O1}$$
The circuit switches when \$V_{FB}\$ reaches 0, so this condition defines the peak-to-peak output voltage in terms of the peak-to-peak swing of the comparator:
$$V_{O2(PP)} = - \frac{R3}{R2} V_{O1(PP)} = - \frac{R3}{R2} 2 V_{O1}$$
Now consider the integrator, whose output voltage with respect to time is:
$$V_{O2(PP)} = - \frac{V_{O1}}{R C} t$$
To get the half-period of the output waveform, set these two equations equal to each other. What this says is that the step change in \$V_{FB}\$ caused by \$V_{O1}\$ must be equal to the height of the slope change caused by \$V_{O2}\$:
$$- \frac{R3}{R2} 2 V_{O1} = - \frac{V_{O1}}{R C} t$$
Solve for t:
$$t = \frac{R3}{R2} 2 R C$$
The frequency is 1/2t:
$$f = \frac{R2}{R3} \frac{1}{4 R C}$$