You can use the above method of residues to evaluate obnoxious integrals of the form
∫ ∫
∞ p(x) R p(x)
−∞ q(x)dx ≡ Rl→im∞ −R q(x)dx
provided the degree of p
(x)
is two less than the degree of q
(x )
. This can be done by using the contour γ_{R}
which goes from
(− R,0)
to
(R,0)
along the real line and then on the semicircle of radius R from
(R,0)
to
(− R,0)
.
PICT
Letting C_{R} be the circular part of this contour, for large R,
|∫ |
|| p(z) || CRk--
|CR q(z)dz| ≤ πR Rk+2
which converges to 0 as R →∞. Therefore, it is only a matter of taking large enough R to enclose all the
roots of q
(z)
which are in the upper half plane, finding the residues at these points and then computing the
contour integral. Then you would let R →∞ and the part of the contour on the semicircle will disappear
leaving the Cauchy principal value integral which is desired. There are other situations which will work just
as well. You simply need to have the case where the integral over the curved part of the contour converges
to 0 as R →∞.
Here is an easy example.
Example 16.8.1Find∫_{−∞}^{∞}
1
x2+1
dx
You know from calculus that the answer is π. Lets use the method of residues to find this. The function
-21-
z +1
has poles at i and −i. We don’t need to consider −i. It seems clear that the pole at i is of order 1 and
so all we have to do is take
You might observe that this is a lot easier than doing the usual partial fractions and trig substitutions etc.
Now here is another tedious example.
Example 16.8.3Find∫_{−∞}^{∞}
x+2
(x2+1)(x2+4)2
dx
The poles of interest are located at i,2i. The pole at 2i is of order 2 and the one at i is of order 1. In
this case, the partial fractions expansion is
Sometimes you don’t blow up the curves and take limits. Sometimes the problem of interest reduces
directly to a complex integral over a closed curve. Here is an example of this.
Example 16.8.4The integral is
∫ π
--cosθ--dθ
0 2 + cosθ
This integrand is even and so it equals
∫ π
1 --cosθ--dθ.
2 −π 2+ cosθ
For z on the unit circle, z = e^{iθ}, z =
1z
and therefore, cosθ =
12
( )
z + 1z
. Thus dz = ie^{iθ}dθ
and so dθ =
dz
iz
. Note that this is done in order to get a complex integral which reduces to
the one of interest. It follows that a complex integral which reduces to the integral of interest
is
Other rational functions of the trig functions will work out by this method also.
Sometimes we have to be clever about which version of an analytic function that reduces to a real
function we should use. The following is such an example.
Example 16.8.5The integral here is
∫ ∞
-lnx--dx.
0 1 +x4
It is natural to try and use the contour in the following picture in which the small circle has radius r
and the large one has radius R.
PICT
However, this will create problems with the log since the usual version of the log is not defined on the
negative real axis. This difficulty may be eliminated by simply using another branch of the logarithm as in
Example 16.1.3. Leave out the ray from 0 along the negative y axis and use this example to define L
(z)
on
this set. Thus L
(z)
= ln
|z|
+ iarg _{1}
(z)
where arg _{1}
(z)
will be the angle θ, between −
π
2
and
3π
2
such that
z =
|z|
e^{iθ}. Then the function used is f
(z)
≡
-L(z)
1+z4
. Now the only singularities contained in this contour
are
1√ - 1√ - 1 √- 1 √-
2 2+ 2i 2,−2 2 + 2i 2
and the integrand f has simple poles at these points. Thus res
which is probably not the first thing you would thing of. You might try to imagine how this could be
obtained using elementary techniques.
Example 16.8.6The Fresnel integrals are
∫ ∞ ∫ ∞
cosx2dx, sinx2dx.
0 0
To evaluate these integrals we will consider f
(z)
= e^{iz2
} on the curve which goes from the origin to the
point r on the x axis and from this point to the point r
( )
1√+2i
along a circle of radius r, and from there
back to the origin as illustrated in the following picture.
PICT
Thus the curve is shaped like a slice of pie. The angle is 45^{∘}. Denote by γ_{r} the curved part. Since f is
analytic,
∫ ∫ ∫ ( ( )) ( )
iz2 r ix2 r it 1√+2i 2 1+-i
0 = γr e dz + 0 e dx − 0 e √2 dt
∫ 2 ∫ r 2 ∫ r 2 (1 + i)
= eiz dz + eixdx − e−t -√-- dt
∫γr ∫0 √0-( ) 2
= eiz2dz + reix2dx − --π 1+√-i + e(r)
γr 0 2 2
where e
(r)
→ 0 as r →∞. This used ∫_{0}^{∞}e^{−t2
}dt =
√-
-π-
2
. Now examine the first of these integrals.
| | | |
||∫ iz2 || ||∫ π4 i(reit)2 it ||
| γ e dz| = || 0 e rie dt||
r ∫ π
≤ r 4 e−r2sin2tdt
0
r ∫ 1 e−r2u
= 2 √-----2du
0 1− u
r ∫ r−(3∕2) 1 r( ∫ 1 1 ) 1∕2
= - √-----2du + - √-----2- e−(r )
2 0 1− u 2 0 1− u
which converges to zero as r →∞. Therefore, taking the limit as r →∞,
√ π( 1+ i) ∫ ∞ 2
-2- -√-- = eix dx
2 0
and so the Fresnel integrals are given by
∫ ∞ √ π- ∫ ∞
sinx2dx = -√--= cosx2dx.
0 2 2 0
The following example is one of the most interesting. By an auspicious choice of the contour it is
possible to obtain a very interesting formula for cotπz known as the Mittag Leffler expansion of
cotπz.
Example 16.8.7Let γ_{N}be the contour which goes from −N −
12
−Ni horizontally to N +
12
−Niand from there, vertically to N +
12
+Ni and then horizontally to −N−
12
+Ni and finally verticallyto −N −
12
− Ni. Thus the contour is a large rectangle and the direction of integration is in thecounter clockwise direction.