Complex Analysis - Maths KU

–R
y
i
C R
Figure 6.26 First contour used to
evaluate (23)
–R
y
–i
C R
Figure 6.27 Second contour used to
evaluate (23)
R
R
x
x
C
6.7 Applications 383
z = ±i.From here on the procedure used is basically the same as that used
to evaluate trigonometric integrals in the preceding section by the theory of
residues.The contour C shown in Figure 6.26 encloses the simple pole z = i
in the upper plane and consists of the interval [−R, R] on the real axis and
a semicircular contour CR, where R>1.Formally, we have
�
C
1
π(1 + z2 ) e−izx �
1
dz =2πi Res
π(1 + z2 ) e−izx �
,i
= e x . (24)
Obviously the result in (24) is not the function f that we started with in
Example 4.A more detailed analysis in this case would reveal that the contour
integral along CR approaches zero as R →∞only if we assume that x 0, we then have − lim =2πiRes(z = −i) or lim
R→∞
−R
R→∞
−R =
− 2πi Res(z = −i).By combining (23), (24), and (25), we arrive at
F −1 {F (α)} = 1
� ∞
2π −∞
2
1+α 2 e−iαx dα =
⎧
⎨
e
⎩
x , x < 0
e−x ,x>0
which agrees with (21).Note that when x = 0 in (23) conventional integration
gives the value 1, which is f(0) in (21).
Remarks
(i) The two conditions of piecewise continuity and exponential order are
sufficient but not necessary for the existence of F (s) =� {f(t)}.
For example, the function f(t) =t −1/2 is not piecewise continuous
on [0, ∞) (Why not?); nevertheless � � t −1/2� exists.