Why do we use functional integration in QFT?

Recently I learned functional integral's formalism in quantum field theory. I have realized that I don't understand why exactly do we introduce it. We have the expression for $S$-matrix, then we may rewrite it in through functional integral.

But why do we use functional integration? I read that it helps to "conserve" gauge symmetries for massless fields and that it avoids infinite results after integration through internal lines. But these are formal arguments. I don't understand it clearly.

Can you add some information which can help me?

• Some of the benefits of path integral formulation are - i) Perturbation series is easy to derive compared to Hilbert space formalism ii) All objects in path integral formulation are classical; one doesn't need to introduce quantum operatorts iii) In Hilbert space formalism one needs to choose a time slice which breaks the manifest Lorentz invariance; this is not a problem in path integral formalism iv) After Wick rotation to Euclidean space, the path integral formalism is completely equivalent to the partition function formalism of stat mech; hence use of RG methods etc is more intuitive. – user10001 Apr 2 '14 at 19:38
• @user10001, (iii) is a myth FWIW, since Lorentz invariance is handled representation theoretically. Further, when you make a choice of which component is the time component, you make a similar choice in the path integral approach which supposedly "breaks the manifest Lorentz invariance". – Alex Nelson Apr 2 '14 at 21:17
• @AlexNelson You may be right; I am not sure. However its true that in path integral formalism all the spacetime coordinates t,x,y,z are treated on equal footing which is more in the spirit of relativity. In particular, path integral formalism doesn't require the spacetime to be of the form Space X Time (as is necessarily required in Hamiltonian formalism) and hence can be used to define theories on more general manifolds including compact ones. – user10001 Apr 2 '14 at 22:46
• @user10001 ...but when you say "$x^0$ is the $t$-coordinate" when doing computations in the path integral approach, you just performed the same space+time splitting as in the canonical approach. The remark that this destroys Lorentz invariance would be true if it depends on a particular choice of splitting. Since neither the canonical nor path-integral approaches explicitly depends on this choice, both are equally as Lorentz-invariant. This is particular important in quantum gravity, see, e.g., arXiv:0809.0097 for a review. – Alex Nelson Apr 2 '14 at 22:55
• @AlexNelson Its true that we choose a time coordinate but in all the expressions time and spatial coordinates appear on equal footing. There are no expressions such as Exp(itH) in which time 'appears' to be different from spatial coordinates. – user10001 Apr 2 '14 at 23:07