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for
8.6 Integrands with Branch Points
We now show how to evaluate certain
improper real integrals involving the integrand
. Since
the complex function
is
multivalued, so we must first specify the branch to be used.
Let
be a real number with
. In
this section we use the branch of
corresponding to the branch of the logarithm
(see Equation (5-20)) as follows:
,
where
and
. Note
that this is not the traditional principal branch of
and that, as defined, the function
is analytic in the domain
.
Theorem 8.7. Let
P(x) and Q(x)
be polynomials of degree m and
n, respectively,
where
. If
, and
Q(x) has a zero of order at most
1 at the origin, and
,
then
,
where
are
the non-zero poles of
.
![]()
Figure 8.7 The contour C that encloses the nonzero poles
of
.
Proof of Theorem 8.7 is in the book.
Complex
Analysis for Mathematics and Engineering
Example
8.23. Evaluate
, where
.
![[Graphics:Images/IntegralsBranchPointsMod_gr_23.gif]](integralsbranchpoints/IntegralsBranchPointsMod/Images/IntegralsBranchPointsMod_gr_23.gif)
Solution. The function
has
a nonzero pole at the point
,
and the denominator has a zero of order at most 1
(in fact, exactly 1) at the
origin. Calculating the residue we get
![[Graphics:Images/IntegralsBranchPointsMod_gr_26.gif]](integralsbranchpoints/IntegralsBranchPointsMod/Images/IntegralsBranchPointsMod_gr_26.gif)
Using Theorem 8.7, we have
We can apply the preceding ideas to other multivalued functions.
Example
8.24. Evaluate
, where
.
![[Graphics:Images/IntegralsBranchPointsMod_gr_45.gif]](integralsbranchpoints/IntegralsBranchPointsMod/Images/IntegralsBranchPointsMod_gr_45.gif)
Solution. We use the function
. Recall
that
,
where
and
. The
path C of integration will consist of
the segments
of the x axis together with the upper
semicircles
and
, for
, as
shown in Figure 8.8.
![]()
Figure 8.8 The contour C for the integrand
.
We chose the branch
because
it is analytic on C and its interior,
hence so is the function f(z) which
has a simple pole in the upper half-plane at the point
.
![[Graphics:Images/IntegralsBranchPointsMod_gr_57.gif]](integralsbranchpoints/IntegralsBranchPointsMod/Images/IntegralsBranchPointsMod_gr_57.gif)
This choice enables us to apply the residue theorem properly (see
the hypotheses of Theorem 8.1), and we get
Keeping in mind the branch of logarithm that we're using, we then
have
(8-33)
If
, then
by the ML inequality (Theorem 6.3)
![[Graphics:Images/IntegralsBranchPointsMod_gr_61.gif]](integralsbranchpoints/IntegralsBranchPointsMod/Images/IntegralsBranchPointsMod_gr_61.gif)
and L'Hôpital's rule yields
. A
similar computation shows that
.
We use these results when we take limits Equations
(8-33) to get
Equating the real parts in this equation gives
![[Graphics:Images/IntegralsBranchPointsMod_gr_65.gif]](integralsbranchpoints/IntegralsBranchPointsMod/Images/IntegralsBranchPointsMod_gr_65.gif)
Exercises for Section 8.6. Integrands with Branch Points
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