25.10 Hints and answers
t=−iand then approaches the origin in the fourth quadrant in a curve that
is ultimately antiparallel to the positive real axis. The other contour,C 1 ,isthe
mirror image of this in the real axis; it is confined to the upper half-plane, passes
throught=iand is antiparallel to the realt-axis at both of its extremities. The
contribution toJν(z) from the curveCkis^12 H(νk), the functionHν(k)being known
as a Hankel function.
Using the method of steepest descents, establish the leading term in an asymp-
totic expansion forHν(1)forzreal, large and positive. Deduce, without detailed
calculation, the corresponding result forH(2)ν. Hence establish the asymptotic
form ofJν(z) for the same range ofz.
25.23 Use the method of steepest descents to find an asymptotic approximation, valid
forzlarge, real and positive, to the function defined by
Fν(z)=∫
Cexp(−izsint+iνt)dt,whereνis real and non-negative andCis a contour that starts att=−π+i∞
and ends att=−i∞.25.10 Hints and answers25.1 Apply Kirchhoff’s laws to three independent loops, sayADBA,ADEAand
DBED. Eliminate other currents from the equations to obtainIR=ω 0 CV 0 [(ω 02 −
ω^2 − 2 iωω 0 )/(ω^20 +ω^2 )], whereω^20 =(LC)−^1 ;|IR|=ω 0 CV 0 ; the phase ofIRis
tan−^1 [(− 2 ωω 0 )/(ω^20 −ω^2 )].
25.3 Setccothu 1 =−d,ccothu 2 =+d,|ccosechu|=aand note that the capacitance
is proportional to (u 2 −u 1 )−^1.
25.5 ξ= constant, ellipsesx^2 (a+1)−^2 +y^2 (a−1)−^2 =c^2 /(4a^2 );η= constant, hyperbolae
x^2 (cosα)−^2 −y^2 (sinα)−^2 =c^2. The curves are the cuts−c≤x≤c,y=0and
|x|≥c,y=0.Thecurvesforη=2πare the same as those forη=0.
25.7 (a) For a quarter-circular contour enclosing the first quadrant, the change in the
argument of the function is 0 + 8(π/2) + 0 (sincey^8 + 5 = 0 has no real roots);
(b) one negative real zero; a conjugate pair in the second and third quadrants,
−^32 ,− 1 ±i.
25.9 Evaluate ∫
πcotπz
( 1
2 +z
)( 1
4 +z)dzaround a large circle centred on the origin; residue atz=− 1 /2 is 0; residue at
z=− 1 /4is4πcot(−π/4).
25.11 The behaviour of the integrand for large|z|is|z|−^2 exp [ (2α−π)|z|]. The residue
atz=±m,foreachintegerm,issin^2 (mα)(−1)m/(mα)^2. The contour contributes
nothing.
Required summation = [ total sum−(m= 0 term) ]/2.
25.13 Note that ̄f(s) has no pole ats=0.Fort<0 close the Bromwich contour in the
right half-plane, and fort>1 in the left half-plane. For 0<t<1 the integrand
has to be split into separate terms containinge−sands−1 and the completions
made in the right and left half-planes, respectively. The last of these completed
contours now contains a second-order pole ats=0.f(t)=1−tfor 0<t<1,
but is 0 otherwise.
25.15
∫
Γand∫
γtend to 0 asR→∞andρ→0. Puts=rexpiπands=rexp(−iπ)on
the two sides of the cut and use∫∞
0 exp(−t(^2) x)dt= 1
2 (π/x)
1 / (^2). There are no poles
inside the contour.