Frequency band and signal rise time

When choosing an oscilloscope, the first thing to consider is the relationship between the oscilloscope's bandwidth and signal rise time.
Traditionally, the rise time tr has been said to be tr=0.35/fc for the band fc.
Let's examine this point a little.

Figure 1 shows the most basic CR integrator circuit that determines the band.

The frequency band of this circuit, that is, the fc (cut off frequency) of the primary LPF (Low Pass Filter) is expressed by fc=1/2πCR.
Its transfer function F(s) is expressed as shown in Equation (1) in the same figure.
In the same formula, setting s=jω gives the frequency characteristics.

Here, find the time response when a step signal is applied.
Multiply the transfer function F(s) by the step signal 1/s as shown in equation (2) in the same figure.

To solve equation (2), we apply partial fraction expansion as shown in equation (3) in Figure 2.

Eliminate the denominator of equation (3) as in equation (4) and compare the coefficients of s on both sides to get equation (5).

Substituting equation (5) into equation (3) and inverse Laplace transform yields the time function on the right side of equation (6).

Figure 3 shows the result of calculating equation (6) on the time axis.

Find the times of the 10% and 90% points of the waveform in Figure 3.

Equations (7) and (8) in Figure 4 represent 10% and 90%.

Rearranging these, we obtain equations (9) and (10).
Divide both equations side by side to get equation (11).
Calculating t2-t1 from equation (11) yields equation (12), which yields tr=0.35/fc at the beginning.

I think the actual oscilloscope filter uses a higher-order filter instead of the CR integration circuit.
Assuming this to be a 7th-order Bessel filter, please refer to the following for the results of obtaining the relationship between tr and fc.

Bessel filter response using Laplace transform