Does the 1; 470u on C3 denote a 1F cap in parallel with a 470u? The 470u should be enough to get the desired response, but the graph is showing near dc response, which seems incorrect.
Good question! Anything after ";" is treated as a comment by SPICE, so it behaves like a 1F capacitor in the simulation. It's a workaround to get rid of some artifacts that skew the THD. As you've noticed, it falsely boosts low frequency performance.
Here's a version with the VAS mistake that @jxdking pointed out corrected. I'm seeing about 0.07% THD and about 60kHz bandwidth, which is fairly respectable. As you can see, you need to connect the collector of the first transistor in the VAS to ground instead of in a Darlington configuration. This takes care of the weird bandwidth limiting you were seeing. This amplifier would probably sound quite OK, although it's missing some safety mechanisms and other miscellaneous things like Boucherot cells and output coils.
Thank you very much, it did increase the bandwidth very much. However, i am curious that how to caculate the gain of differential amplifier with both current source and current mirror and the gain of VAS which seems like CE configuration, but i dont know how to caculate the Ic to get r_π of hybrid-pi model.
On the distortion analysis, are you using Fourier or the small signal distortion method? As I mentioned before, I think the only valid way to measure distortion in a AB type amplifier is fourier which uses transient analysis. The output devices are switching which invalidates the small signal distortion method.
For the differential amplifier with a current mirror... It's complicated. The equation is at the bottom of this article: https://wiki.analog.com/university/courses/electronics/text/chapter-12 I have no idea how that's derived. There are people a lot smarter than me here who might know.
If we are talking about the overall loop gain of the amplifier, it's a lot easier. A Class A/B/AB amplifier is pretty much just a high powered opamp, so the gain is determined by the resistors R8 and R9 in your diagram as A=1+(R8/R9).