Singularities/Residues/ Applications 717by using (9.11-51). Equating imaginary parts in (9.11-52) the result
in (9.11-50) is obtained anew.
Type IX. f 0
00
f(x) dx, where J(x) satisfies the following conditions:- J(x) has an extension f(z) that is meromorphic in C.
- J(z) has infinitely many poles all of which lie on a ray z = bt with
0 < 8 :=:; Arg b :=:; rr - 8.
3. f(-z) = J(z)
4. limR-+oo lf(±R + iy)I = 0 for 0 :=:; y :=:; c, c > 0.
In this case it is not convenient in general to make use of half-circles to
join the endpoints of the real interval [-R, R], unless the resulting series of
residues converges (see Example 2 below). Instead, we use rectangles with
vertices (-R,O), (R,O), (R,c), and (-R,c), with an appropriate choice of
c, so as to have the integral along the upper boundary of the rectangle thesame as that along the lower boundary, except for a constant factor. If f(z)
happens to have a pole at c' + ic (-R < c' < R), then an indentation of
the upper boundary at that point will be necessary. Condition ( 4) above is
imposed in order that the integrals along the vertical sides of the rectangle
tend to zero as R -t oo.
Examples 1. To evaluate J 0
00
( sinh ax/ sinh x) dx, -1 < a < 1. We shall
assume 0 < a < 1 since the case a = 0 is trivial, and for a negative a we have
[00
sinhax dx = - [00
sinh(-ax) dx. } 0 sinh x } 0 sinh x
The integrand has the obvious extension f(z) = sinhaz/ sinhz, which
is even and has poles at the points bn = inrr (n = ±1, ±2, ... ). The point
b 0 is not a pole, but a removable singularity, since
1. sinhaz
1m ---=a
z-+O sinhz
This implies that the given integral is proper at the lower limit.
Now consider the complex integral
J
sinhaz dz
sinhz
c+where c+ is the boundary or the rectangle with vertices at (-R, 0), (R, 0),
( R, irr ), and ( -R, irr) described in the positive sense, with an indentation
,-: z - irr = reilJ, 0 2:: () 2:: -7r, so as to avoid the pole b 1 = irr (Fig. 9.23).
Since there are no poles inside c+ we have
J
R sinhax d 1.,,. sinha(R+iy) ·a
-R sm. h x x +. h(R. ) z y
0 sm + zy.