1550251515-Classical_Complex_Analysis__Gonzalez_

(jair2018) #1
Singularities/Residues/Applications

On the other hand, from

we get

sinh a( x + i7l") = sinh ax cos a7!" + i cosh ax sin a7!"
sinh(x + i7r) = - sinhx

719

l


r sinha(x. +. i7r) d x = cos a7!" iR -. sinhax --d x + ism.. a7!" iR coshax d x
R smh(x + i7r) r smhx r sinhx

1


-R sinha(x + i7r) d 1R sinhax d .. 1R coshax d


. h(. ) x - cos a7!". h x i sin a7!" .nh x
-r Sill X + Z7!" r Sill X r SI X
and


l


r sinha(x + i7r) d 1-R sinha(x + i7r) d _ JR sinhax d


R sm. h( x + i7!". ) x + -r sm. h( x + i7!". ) x - 2cosa7!" r s1 .nh x x


Hence, (9.11-53) can be written as

1


R sinh ax iR sinh ax
2. h dx + 2 cos a7!" -. -h-dx
o Sill X r Sill X

1


,,. sinha(R+iy) ·a j sinhaz d
+. h(R. ) i y + -.-h- z
0 sm + iy sm z
'Y-

1


0 sinh a( -R + i y). d -0
+ ,,. sm .h(R - + iy .)iy- (9.11-57)

Now, letting r -t 0, then R --+ oo in (9.11-57), and taking (9.11-54),
(9.11-55), and (9.11-56) into account, we obtain

1


(^00) sinhax.
2(1 + cosa7r) -. -h-dx - 7!"Sllla7!" = 0
0 sm x
or
--- x = -7!" == -7!"tan -a7!"
1
(^00) sinhax d 1 sina7r 1 1
0 sinhx^2 l+cosa7!"^2 2
the convergence of the integral at the upper limit being a consequence
of (9.11-57).


2. If f(s) = .C {F(t)}, then the inverse Laplace transform F(t) =


.c-1 {f(s)} is given by

F(t) = 2 ~i 1::: e
8

tf(s)ds

fort > O, and F(t) = 0 fort < 0. This is the so-called complex inversion
integral or Bromwich's integral formula. The integration is supposed to be

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