What is the point of path integral for boson and fermion? I am a beginner to study QFT and confused about path integral for boson or fermion.
I have read about the path integral for single particle, and finished some problems. But I cannot understand the next chapter which is about path integral for boson and fermion. 
What is the difference between the two kinds of path integrals? What is the point of path integral for boson and fermion?
I find that there are big differences in forms between single particle, boson and fermion. I do not understand why the book uses different forms to discuss them. Even the path integral for the spin system has a new form to discuss.
In my opinion, I think the biggest difference between single particle and boson (fermion) is statistics, but how to consider about the statistical properties in path integral? Just using one symbol for differential, $D$, to take the place of the original symbol, $d$, is enough?
 A: 
What is the point of path integral for boson and fermion?

The fields that enter into the bosonic path integral are just commuting $c$-number fields - in other words, the field values are just plain old complex numbers, not quantum operators.  So you don't need to keep track of the ordering of the fields, which makes them much easier to work with.

Could you please tell me the difference between the two kind of path integrals?

The fields in the bosonic path integral are just plain old complex-number functions on spacetime.  The fields in the fermionic path integral are Grassmann-number-valued.  When you reverse the order of multiplication of two Grassmann numbers, the formal value of the product changes by a minus sign.  So in this case the ordering of the fields does matter, but only up to a sign, so you don't need to muck around with complicated commutators or anticommutators.  So while not quite as easy to work with as bosonic fields, they're still a lot easier to work with than quantum-operator-valued fields are.
By the way, you never need to use the path integral - it's mathematically equivalent to the canonically-quantized Hamiltonian formalism with operator-valued fields, which looks a lot more like what you do in nonrelativistic QM.  The path integral formalism's just much more mathematically convenient to work with, although admittedly more abstract and less physically intuitive (especially in the fermionic case).  (Also, the path integral formalism's manifestly Lorentz invariant, while the Lorentz invariance of the canonical Hamiltonian formalism is much harder to see, because the Hamiltonian transforms under Lorentz transformations as the $T_{00}$ component of the rank-two stress-energy tensor, rather than as a Lorentz scalar like the Lagrangian density.)
A: I think as you have said, the biggest difference is the statistics. The number of boson on one quantum state is infinite while the number of fermion on one quantum state is limited to one. This lead to different path integral in order to take into account this difference. For example the grassman algebra for fermion could satisfy the requirement that each state can only have one fermion and it become zero when two fermion occupy the same quantum state. 
