The integral exists (in the usual sense) iff the denominator has no real roots. From the factorization we see that this happens iff .
The roots are the simple poles of the integrand and thus the residua are simple to compute, e.g. and so on. The curve we are going to use is the upper or lower half circle with diameter . The curve integral along tends to zero like , so, by residue theorem .
Now we have to distinguish several cases, depending on the placement of the roots with respect to the real line. It is sufficient to consider four cases as the other can be recovered from these by relabeling of the roots or by taking the opposite halfcircle.
1) . Then , because each lower half cirlce contains no pole.
2) . Then .
3) . Then .
4) . Then .
Note that the conditions 1)-4) can be written explicitly by writing and and computing the imaginary parts of the roots. Also, the results can be simplified as we know the roots. For example, if and , then we are in case 4) and after some simplification we get .
using
.
and thus the residua are simple to compute, e.g.
and so on. The curve we are going to use is the upper or lower half circle with diameter 
. The curve integral along ![$\{Re^{i\varphi},\;\varphi\in[0,\pi]\;\text{or}\;[\pi,2\pi]\}$](/mathtex/ed/ed5af5b8fe3979346725e5ea0afabcb5.gif "kopírovat do textarea")
tends to zero like 
, so, by residue theorem
.
.
, because each lower half cirlce contains no pole.
.
.
.
.
.
.
and 
and computing the imaginary parts of the roots. Also, the results can be simplified as we know the roots. For example, if 
and 
, then we are in case 4) and after some simplification we get 
.