(1) Evaluate I

13 Pages·2011·0.12 MB·English

Assignment 8
(1) Evaluate I = R2π(a2+sin2θ)−2dθ using contour integrals. Because this is a function of trigonometric
0
functions, we can take a contour, C, around the unit circle and replace sinθ with (z−1/z)/(2i). Thus, I is
exactly this contour integral around the unit circle
I 1 dz I 24z4 dz
I = =
C (a2−(z−1/z)2/4)2 iz C (cid:0)4a2z2−(z2−1)2(cid:1)2 iz
16I z3dz
=
i C (cid:0)(z2−1)−2az(cid:1)2(cid:0)(z2−1)+2az(cid:1)2
16I z3dz
=
i (z−z )2(z−z )2(z−z )2(z−z )2
C 1 2 3 4
√ √
wherez =a± a2+1andz =−a± a2+1areallsecondorderpolesofourfunction. Becausea>1,
1,2 √ 3,4 √
the poles z =a+ a2+1 and z =−a− a2+1 lie outside the unit circle and should not be included in
1 4
the calculation of the residues. Also, note that z =−z and z =−z
4 1 3 2
From the residue theorem, this contour integral becomes
16 ( d (cid:18) z3 (cid:19)(cid:12)(cid:12) d (cid:18) z3 (cid:19)(cid:12)(cid:12) )
I = 2πi (cid:12) + (cid:12)
i dz (z−z )2(z−z )2(z−z )2 (cid:12) dz (z−z )2(z−z )2(z−z )2 (cid:12)
1 3 4 z=z2 1 2 4 z=z3
=32π(cid:8) 3z 2(z −z )−2(z −z )−2(z −z )−2 − 2z 3(z −z )−3(z −z )−2(z −z )−2
2 2 1 2 3 2 4 2 2 1 2 3 2 4
− 2z 3(z −z )−2(z −z )−3(z −z )−2 − 2z 3(z −z )−2(z −z )−2(z −z )−3
2 2 1 2 3 2 4 2 2 1 2 3 2 4
+ 3z 2(z −z )−2(z −z )−2(z −z )−2 − 2z 3(z −z )−3(z −z )−2(z −z )−2
3 3 1 3 2 3 4 3 3 1 3 2 3 4
− 2z 3(z −z )−2(z −z )−3(z −z )−2 − 2z 3(z −z )−2(z −z )−2(z −z )−3 (cid:9)
3 3 1 3 2 3 4 3 3 1 3 2 3 4
=32πz 2(z −z )−2(2z )−2(z +z )−2
2 2 1 2 2 1
(cid:8) 3−2z (z −z )−1−2z (2z )−1−2z (z +z )−1
2 2 1 2 2 2 2 1
+3+2z (−z −z )−1+2z (−2z )−1+2z (−z +z )−1(cid:9)
2 2 1 2 2 2 2 1
(cid:20) (cid:21)
32π z (z +z )+z (z −z )
= 1− 2 2 1 2 2 1
(z 2−z 2)2 (z 2−z 2)
2 1 2 1
= 32π (cid:2)−z 2−z 2(cid:3)
(z 2−z 2)3 2 1
2 1
√
Using z 2+z 2 =4a2+2 and z 2−z 2 =4a a2+1, this becomes
1 2 1 2
Z 2π dθ π 2a2+1
I = =
(a2+sin2θ)2 a3 (a2+1)3/2
0
(2) Evaluate I =R∞x/(1+x3)dx. We want a contour that includes the positive half of the real axis. The
0
denominator has zeros at −1=eiπ,eiπ/3, and e−iπ/3. Note that including the negative real axis will give us
acontributiontothecontourintegralquitedifferentfromI,soitwouldbewisetoavoidusingthataspartof
the contour. We might also imagine that part of our contour will be at least some part of a circle with fixed
radius, R, that will eventually go to ∞. The best choice is hinted at by the x3 in the denominator. Take as
our contour the following three parts: (1) the positive real axis, (2) the large sector from θ =0 to θ =2π/3
and (3) the line or ray with constant θ = 2π/3. This contour encloses a third of the complex plane (in the
limitthattheradius,R,inpart(2)goesto∞)aswellasthepoleatz =eiπ/3. Thereasonwetakepart(3)is
thatonthereturntripalongtheray, thedenominatorinthecomplexplane, 1+z3 =1+(rei2π/3)3 =1+r3
1
and we will recover a real integral similar to I. We now have the contour integral
( )
I z Z R r Z 2π/3 Reiθ Z 0 rei2π/3
dz = lim dr+ Rieiθdθ+ ei2π/3dr
C 1+z3 R→∞ 0 1+r3 0 1+R3e3iθ R 1+(rei2π/3)3
Z ∞ r Z ∞ r
= dr−ei4π/3 dr
1+r3 1+r3
0 0
=−2iei2π/3sin(2π/3)I
where in the second line, the second term goes to zero.
Now using the residue theorem, we evaluate the contour integral
I z (cid:18) z (cid:19) (cid:12)(cid:12)
dz =2πiRes (cid:12)
1+z3 1+z3 (cid:12)
C eiπ/3
(cid:12)
(cid:16) z (cid:17) (cid:12)
=2πi (cid:12)
3z2 (cid:12)
eiπ/3
2πi
=
3eiπ/3
Combining these results, we get
√
Z ∞ x π 2 3
I = dx= = π
(1+x3) 3sin(2π/3) 9
0
R∞
(3) This integral, I = sinh(ax)/sinh(πx)dx, can be written with limits from −∞ to ∞ provided we
0
multiplytheresultby1/2. Wecandothissincetheintegrandisanevenfunction. Thepolesofthefunction
sit along the imaginary axis at all integer multiples of i with the exception of z = 0. One possibility for
a good contour is a rectangle that extends from (−R,0) to (R,0) along the real axis, from (R,0) to (R,1)
along the line x = R, from (R,1) to (−R,1) along the line y = 1, and then from (−R,1) to (−R,0) along
the line x=−R. Taking this as our contour, we get
I sinh(az) (Z R sinh(ax) Z 1 sinh(cid:0)a(R+iy)(cid:1)
dz = lim dx+ dy
(cid:0) (cid:1)
C sinh(πz) R→∞ −R sinh(πx) 0 sinh π(R+iy)
Z −R sinh(cid:0)a(x+i)(cid:1) Z 0 sinh(cid:0)a(−R+iy)(cid:1) )
+ dx+ dy
(cid:0) (cid:1) (cid:0) (cid:1)
sinh π(x+i) sinh π(−R+iy)
R 1
The integrals over y which are the vertical end pieces of the contour, C, with R fixed go to 0 as R → ∞
provided a<π. This is seen by considering only the integrands
(cid:0) (cid:1)
sinh a(±R+iy) ea(±R+iy)−e−a(±R+iy)
lim = lim
(cid:0) (cid:1)
R→∞sinh π(±R+iy) R→∞eπ(±R+iy)−e−π(±R+iy)
= lim eR(a−π)
R→∞
=0 if a<π
Our contour integral now is
I sinh(az) Z ∞ sinh(ax) Z −∞ sinh(cid:0)a(x+i)(cid:1)
dz = dx+ dx
(cid:0) (cid:1)
sinh(πz) sinh(πx) sinh π(x+i)
C −∞ ∞
Z ∞ sinh(ax) Z −∞ sinh(ax)cosh(ia)+cosh(ax)sinh(ia)
= dx+ dx
sinh(πx) sinh(πx)cosh(iπ)+cosh(πx)sinh(iπ)
−∞ ∞
Z ∞ sinh(ax) Z ∞ sinh(ax) Z ∞ cosh(ax)
= dx+cos(a) dx+isin(a) dx
sinh(πx) sinh(πx) sinh(πx)
−∞ −∞ −∞
2
wherewehaveusedcosh(iπ)=−1andsinh(iπ)=0. Nownotethatthethirdintegraliszerobecauseweare
integrating an odd function over a symmetric interval. The second integral is just some constant times I.
Now,usingtheresiduetheorem,wecanevaluatethecontourintegral. Theonlypolethatweneedconsideris
that at z =i. In fact, it sits on the contour; and as a simple pole, we need take only half of its contribution.
The point z =0, it should be mentioned, is not a pole. This can be seen by taking the limit of the function
as z →0. The result is a finite value, a/π. The contour integral is then
I sinh(az) (cid:18)sinh(az)(cid:19) (cid:12)(cid:12)
dz =πiRes (cid:12)
sinh(πz) sinh(πz) (cid:12)
C z=i
sinh(ia)
=iπ
πcosh(iπ)
=sin(a)
Putting it all together, we have
Z ∞ sinh(ax) 1 sina
I = dx=
sinh(πx) 2 1+cosa
0
(4)Theintegral, I =R∞ eax/cosh(x)dx,with0<a<1isverysimilartothepreviousproblemsinceboth
−∞
deal with exponentials. Since the poles of the function lie at half integer multiples of π on the imaginary
axis: z = i(2n+1)π/2, we can use a contour similar to that in problem 3. The only difference will be to
take the top of the contour to run from (R,π) to (−R,π) along the line y = π instead of using y = 1. In
this case only a single pole is entirely enclosed in the contour. Taking this as our contour, we get
(
I eaz Z R eax Z π/2 ea(R+iy)
dz = lim dx+ dy
(cid:0) (cid:1)
C coshz R→∞ −R coshx 0 cosh R+iy
)
Z −R ea(x+iπ) Z 0 ea(−R+iy)
+ dx+ dy
(cid:0) (cid:1) (cid:0) (cid:1)
cosh x+iπ cosh −R+iy
R 1
As in the previous problem, the integrals over y which are the vertical end pieces of the contour, C, with R
fixed go to 0 as R→∞ provided a<1. The proof of this is exactly as before.
The contour integral is now
I eaz Z ∞ eax Z −∞ ea(x+iπ)
dz = dx+ dx
(cid:0) (cid:1)
coshz coshx cosh x+iπ
C −∞ ∞
Z ∞ eax Z −∞ eax
= dx+eiaπ dx
coshx coshxcosh(iπ)−sinhxsinh(iπ)
−∞ ∞
Z ∞ eax Z ∞ eax
= dx+eiaπ dx
coshx coshx
−∞ −∞
where we have used cosh(iπ) = −1 and sinh(iπ) = 0. The second integral is just some constant times I.
Now, using the residue theorem, we can evaluate the contour integral. The only pole that we need consider
is that at z =iπ/2. The contour integral is then
I eaz (cid:18) eaz (cid:19) (cid:12)(cid:12)
dz =2πiRes (cid:12)
coshz coshz (cid:12)
C z=i
eiaπ/2
=2iπ
sinh(iπ/2)
=2πeiaπ/2
Putting it all together, we have
Z ∞ eax π
I = dx=
(cid:0) (cid:1)
coshx cos πa/2
−∞
3
(5) To find the sum, S = P∞ (n2 −a2)−1, use a contour, C, as from class, that avoids all the poles on
n=1
the positive and negative real axis. The contour integral with this contour can be considered in four parts:
(1) a semicircle in the upper half plane, (2) a contour (closed) around the poles at negative integers, (3) a
semicircle in the lower half plane, and (4) a contour (closed) around the poles at the positive integers. Parts
(1) and (3) will go to zero in the limit the radii of the semicircles go to ∞. Parts (2) and (4) give sums over
all the integers because the function πcot(πz) is chosen to give poles with residue 1 at all integer values (in
addition to the contribution from the poles at z =±a. So we get
I 1 πcos(πz) X−1 1 (cid:18) 1 (cid:19) (cid:12)(cid:12)
dz =−2πi −2πiRes πcot(πz) (cid:12)
z2−a2 sin(πz) n2−a2 z2−a2 (cid:12)
C n=−∞ z=−a
∞ (cid:18) (cid:19) (cid:12)
X 1 1 (cid:12)
−2πi −2πiRes πcot(πz) (cid:12)
n2−a2 z2−a2 (cid:12)
n=1 z=a
∞
X 1 2πcot(πa)
=−4πi −2πi
n2−a2 2a
n=1
Evaluating the original contour integral with the residue theorem, the only pole we need consider is that at
z =0. No other poles are contained within the full contour, C. Thus we have
I 1 πcos(πz) (cid:18) 1 πcos(πz)(cid:19) (cid:12)(cid:12)
dz =2πiRes (cid:12)
z2−a2 sin(πz) z2−a2 sin(πz) (cid:12)
C z=0
2πi
=−
a2
Equating these two results we find
∞
X 1 1 π
= − cot(πa)
n2−a2 2a2 2a
n=1
(6)Tofindthesum,S =P∞ (−1)n(n2−a2)−1,usethesamecontourfrom(5)butdropthecos(πz)inthe
n=1
contour integral since we want an alternating series. Otherwise, this is very similar to the previous problem.
Applying it here, we get
I 1 π X−1 (−1)n (cid:18) 1 (cid:19) (cid:12)(cid:12)
dz =−2πi −2πiRes πcsc(πz) (cid:12)
z2−a2sin(πz) n2−a2 z2−a2 (cid:12)
C n=−∞ z=−a
X∞ (−1)n (cid:18) 1 (cid:19) (cid:12)(cid:12)
−2πi −2πiRes πcsc(πz) (cid:12)
n2−a2 z2−a2 (cid:12)
n=1 z=a
X∞ (−1)n 2πcsc(πa)
=−4πi −2πi
n2−a2 2a
n=1
Evaluating the original contour integral with the residue theorem, the only pole we need consider is that at
z =0. No other poles are contained within the full contour, C. Thus we have
I 1 π (cid:18) 1 π (cid:19) (cid:12)(cid:12)
dz =2πiRes (cid:12)
z2−a2sin(πz) z2−a2 sin(πz) (cid:12)
C z=0
2πi
=−
a2
Equating these two results we find
X∞ (−1)n 1 π
= −
n2−a2 2a2 2asin(πa)
n=1
4
(7)ThesumS =P∞ (−1)n(n2−a2)−p will,ofcourse,besimilartothepreviousproblem. Thetwisthere
n=1
is that the poles at z =±a will be poles of order p. But, otherwise, we can take over our approach from the
previous problem. We have
I 1 π X−1 (−1)n (cid:18) 1 π (cid:19) (cid:12)(cid:12)
dz =−2πi −2πiRes (cid:12)
(z2−a2)psin(πz) (n2−a2)p (z2−a2)psin(πz) (cid:12)
C n=−∞ z=−a
X∞ (−1)n (cid:18) 1 π (cid:19) (cid:12)(cid:12)
−2πi −2πiRes (cid:12)
(n2−a2)p (z2−a2)psin(πz) (cid:12)
n=1 z=a
X∞ (−1)n π
=−4πi −2πi
(n2−a2)p 2asin(πa)
n=1
X∞ (−1)n
=−4πi
(n2−a2)p
n=1
2π2i " dp−1 (cid:18)csc(πz)(cid:19) (cid:12)(cid:12) dp−1 (cid:18)csc(πz)(cid:19) (cid:12)(cid:12) #
− (cid:12) + (cid:12)
(p−1)! dzp−1 (z−a)p (cid:12) dzp−1 (z+a)p (cid:12)
z=−a z=a
Evaluating the original contour integral with the residue theorem, the only pole we need consider is that at
z =0. No other poles are contained within the full contour, C. Thus we have
I 1 π (cid:18) 1 π (cid:19) (cid:12)(cid:12)
dz =2πiRes (cid:12)
(z2−a2)p sin(πz) (z2−a2)p sin(πz) (cid:12)
C z=0
2πi
=
(−a2)p
Equating these two results we find
X∞ (−1)n (−1)p+1 π " dp−1 (cid:18)csc(πz)(cid:19) (cid:12)(cid:12) dp−1 (cid:18)csc(πz)(cid:19) (cid:12)(cid:12) #
= − (cid:12) + (cid:12)
(n2−a2)p 2a2p 2(p−1)! dzp−1 (z−a)p (cid:12) dzp−1 (z+a)p (cid:12)
n=1 z=−a z=a
This is a general formula which we could work out for general p. But to save on the headache, let’s take
p=2. Working this out, we find
X∞ (−1)n 1 π 1 (cid:0) (cid:1)
=− + 1+πacot(πa)
(n2−a2)2 2a4 4a3 sin(πa)
n=1
(8) For the sum, S =P∞ (n2−a2)−1(n2−b2)−1, use the usual contour for the integral
n=1
I πcot(πz)dz X−1 −2πi " 2π2icot(πz) (cid:12)(cid:12) 2π2icot(πz) (cid:12)(cid:12) #
= −Res (cid:12) + (cid:12)
(z2−a2)(z2−b2) (n2−a2)(n2−b2) (z2−a2)(z2−b2)(cid:12) (z2−a2)(z2−b2)(cid:12)
C n=−∞ −a −b
X∞ −2πi (cid:20) 2π2icot(πz) (cid:12)(cid:12) 2π2icot(πz) (cid:12)(cid:12) (cid:21)
+ −Res (cid:12) + (cid:12)
(n2−a2)(n2−b2) (z2−a2)(z2−b2)(cid:12) (z2−a2)(z2−b2)(cid:12)
n=1 a b
∞
X 1 2πcot(πa) 2πcot(πb)
=−4πi −2πi −2πi
(n2−a2)(n2−b2) 2a(a2−b2) (b2−a2)2b
n=1
Evaluating the original contour integral with the residue theorem, the only pole we need consider is that at
z =0. No other poles are contained within the full contour, C. Thus we have
I πcot(πz) (cid:18) πcot(πz) (cid:19) (cid:12)(cid:12)
dz =2πiRes (cid:12)
(z2−a2)(z2−b2) (z2−a2)(z2−b2) (cid:12)
C z=0
2πi
=
a2b2
5
Equating, we find
∞ (cid:20) (cid:21)
X 1 1 π 1 1
=− + cot(πb)− cot(πa)
(n2−a2)(n2−b2) 2a2b2 2(a2−b2) b a
n=1
Arfken 7.1.1 Find the residues of the following functions.
(a) 1/(z2+a2) has simple poles at z =±ia. The residues are
(cid:12) (cid:12)
1 (cid:12) 1 (cid:12)
(z∓ia)(cid:12) = (cid:12)
z2+a2 (cid:12) z±ia(cid:12)
z=±ia z=±ia
1
=±
2ia
(b) 1/(z2+a2)2 has second order poles at z =±ia. The residues are
(cid:18) (cid:19)(cid:12) (cid:12)
d 1 (z∓ia)2 (cid:12)(cid:12) = −2 (cid:12)(cid:12)
dz (z2+a2)2 (cid:12) (z±ia)3(cid:12)
z=±ia z=±ia
1
=±
4ia3
(c) z2/(z2+a2)2 has second order poles at z =±ia. The residues are
d (cid:18) z2 (z∓ia)2(cid:19)(cid:12)(cid:12)(cid:12) =(cid:18) 2z +z2 −2 (cid:19)(cid:12)(cid:12)(cid:12)
dz (z2+a2)2 (cid:12) (z±ia)2 (z±ia)3 (cid:12)
z=±ia z=±ia
i
=∓
4a
(d) sin(1/z)/(z2+a2) has simple poles at z =±ia and an essential singularity at z =0. The residues are
(cid:12) (cid:12)
sin(1/z) (cid:12) sin(1/z)(cid:12)
(z∓ia)(cid:12) = (cid:12)
z2+a2 (cid:12) z±ia (cid:12)
z=±ia z=±ia
sinh(1/a)
= .
2a
For z =0 there are no convenient tricks and we must use the full Laurent expansion given by
∞
sin(cid:0)1(cid:1) 1 = X a zn
z z2+a2 n
n=−∞
where
1 I (cid:0)1(cid:1) 1 dz
a = sin
n 2πi z z2+a2 zn+1
C
1 I X∞ (−1)k (cid:0)1(cid:1)( X∞ (−1)pz2p dz
= 2k+1)
2πi (2k+1)! z a2p+2 zn+1
k=0 p=0
X∞ X∞ (−1)k+p 1 I
= z2p−2k−n−2dz
(2k+1)!a2p+22πi
k=0p=0
X∞ X∞ (−1)k+p
= δ
(2k+1)!a2p+2 2p−2k−n−2,−1
k=0p=0
X∞ (−1)2k+(n+1)/2
=
(2k+1)!a2k+n+3
k=0
∞
(cid:0)i(cid:1)n+1 1 X 1 1
=
a a (2k+1)! a2k+1
k=0
(cid:0)i(cid:1)n+1 1 1
= sinh
a a a
6
Reading off the residue, a , we get a−1 sinh(a−1) for the essential singularity at z =0.
−1
(e) zeiz/(z2+a2) has simple poles at z =±ia The residues are
zeiz (cid:12)(cid:12) zeiz (cid:12)(cid:12)
(z∓ia)(cid:12) = (cid:12)
z2+a2 (cid:12) z±ia(cid:12)
z=±ia z=±ia
1
= e∓a
2
The point at z = ∞ (let w = 1/z) is an essential singularity. To get its residue, we must be a bit creative.
Transforming to w, our function is
1 wei/w
f(z)=f( )=g(w)=
w 1+a2w2
If we ask for the Laurent series of g(w) around w =0, the a term of that series, will be the a term of the
1 −1
Laurent series for f(z) around z =∞. So let’s find a of g(w):
1
1 I wei/w dw
a (w =0)=
1 2πi 1+a2w2 w2
1 X∞ X∞ ik I
= (−a2)l w−k+2l−1dw
2πi k!
k=0l=0
X∞ ik(−1)k/2(a2)k/2
=
k!
k=0
X∞ (−a)k
=
k!
k=0
=e−a
This, then, is the residue of our original function at the essential singularity at z =∞.
(f) zeiz/(z2−a2) has simple poles at z =±a The residues are
zeiz (cid:12)(cid:12) zeiz (cid:12)(cid:12)
(z∓a)(cid:12) = (cid:12)
z2−a2 (cid:12) z±a(cid:12)
z=±a z=±a
1
= e±ia
2
As in (e), there is an essential singularity as well at z =∞. The only difference here from the previous
problem is the sign of the denominator. Applying that here, we have (skipping some steps)
X∞ ik(a2)k/2
a (w =0)=
1 k!
k=0
X∞ (ia)k
=
k!
k=0
=eia
(g) eiz/(z2−a2) has simple poles at z =±a The residues are
eiz (cid:12)(cid:12) eiz (cid:12)(cid:12)
(z∓a)(cid:12) = (cid:12)
z2−a2 (cid:12) z±a(cid:12)
z=±a z=±a
1
=± e±ia
2a
7
Again, we also have an essential singularity at z =∞. Following the two previous examples, we write
1 w2ei/w
f(z)=f( )=g(w)=
w 1−a2w2
and we find the n=1 term of the Laurent series of g(w) expanded about w =0 since it corresponds to the
a term of the Laurent series of f(z) expanded about z =∞:
−1
1 I w2ei/w dw
a (w =0)=
1 2πi 1−a2w2 w2
1 X∞ X∞ ik I
= (a2)l w−k+2ldw
2πi k!
k=0l=0
X∞ X∞ ik
= a2lδ
k! 2l,k−1
k=0l=0
1 X∞ (ia)k
=
a k!
k=1
1"X∞ (ia)k #
= −1
a k!
k=0
= 1(cid:2)eia−1(cid:3)
a
This, then, is the residue of our original function at the essential singularity at z =∞.
(h) z−k/(z+1) (with 0<k <1) has a simple pole at z =−1 and a branch point at z =0. The residue at
z =−1 is (taking the positive real axis as our cut line – thus 0<argz <2π)
z−k (z+1)(cid:12)(cid:12)(cid:12) =(−1)−k
z+1 (cid:12)
z=−1
=(eiπ)−k
=e−ikπ
At the branch point at z =0, no residue is defined. The point at z =∞ is also a branch point.
8
Arfken 7.1.6FindthegeneratingfunctionfortheBesselandHermitefunctionsgiveng(t,x)=P f (x)tn
n n
where f (x) is the integral representation of the function.
n
(a) Bessel function, J (x)
n
g(t,x)= X∞ tn 1 I e(x/2)(t0−1/t0)t0−n−1dt0
2πi
n=0
= X∞ tn 1 I X∞ 1 (cid:16)− x (cid:0)1−t02(cid:1)(cid:17)kt0−n−1dt0
2πi k! 2t0
n=0 k=0
= 1 X∞ tnX∞ 1 (cid:16)−x(cid:17)kI Xk k! (−t02)mt0−k−n−1dt0
2πi k! 2 m!(k−m)!
n=0 k=0 m=0
= X∞ tnX∞ 1 (cid:16)−x(cid:17)k Xk k! (−1)m 1 I t02m−k−n−1dt0
k! 2 m!(k−m)! 2πi
n=0 k=0 m=0
∞ ∞ k
X X 1 (cid:16) x(cid:17)k X k!
= tn − (−1)mδ
k! 2 m!(k−m)! n,2m−k
n=0 k=0 m=0
∞ k
X 1 (cid:16) x(cid:17)k X k!
= − (−1)mt2m−k
k! 2 m!(k−m)!
k=0 m=0
∞ k
X 1 (cid:16) x(cid:17)k X k!
= − (−t2)m
k! 2t m!(k−m)!
k=0 m=0
∞
X 1 (cid:16) x(cid:17)k
= − (1−t2)k
k! 2t
k=0
∞
X 1 (cid:16)x (cid:17)k
= (t2−1)
k! 2t
k=0
=e(x/2)(t−1/t)
Thereisanotherwaytodothisproblem(aswellaspartd). Onecanviewf (x)=J (x)asthex-dependent
n n
coefficients of a Laurent expansion for the complex (in t) function g(t,x):
∞
X
g(t,x)= a (x)(t−t )n
n 0
−∞
where
1 I g(t0,x)dt0
a (x)=
n 2πi (t0−t )n+1
C 0
Because we are given
1 I
a (x)=J (x)= e(x/2)(t0−1/t0)t0−n−1dt0
n n 2πi
if we set t =0, on comparing this with the general form for a (x), we can just read off the interior function
0 n
as g(t,x)
g(t,x)=e(x/2)(t−1/t)
9
(d) Hermite function, H (x) (NOTE: There is a mistake in the problem: there should be no n!.)
n
g(t,x)= X∞ tn 1 I e−t02+2t0xt0−n−1dt0
2πi
n=0
= X∞ tnex2 I e−(t0−x)2t0−n−1dt0
2πi
n=0
= X∞ tnex2 I X∞ 1(cid:0)−(t0−x)2(cid:1)kt0−n−1dt0
2πi k!
n=0 k=0
= X∞ tnex2 X∞ (−1)k I X2k (2k)! t02k−m(−x)mt0−n−1dt0
2πi k! m!(2k−m)!
n=0 k=0 m=0
= X∞ tnex2 X∞ (−1)k X2k (2k)! (−x)m I t02k−m−n−1dt0
2πi k! m!(2k−m)!
n=0 k=0 m=0
= X∞ tnex2X∞ (−1)k X2k (2k)! (−x)mδ
k! m!(2k−m)! n,2k−m
n=0 k=0 m=0
=ex2X∞ (−1)k X2k (2k)! t2k−m(−x)m
k! m!(2k−m)!
k=0 m=0
=ex2X∞ (−1)k(t−x)2k =e−t2+2tx
k!
k=0
Arfken 7.1.14
(a) Evaluate I = R∞ cosx/(x2 +a2)dx for a > 0. First, note that cosz in the corresponding contour
a −∞
integral will diverge for a contour with semi-circle that closes in either the upper or lower half planes. The
trick is to do R∞ eix/(x2+a2)dx and take only the real part at the end. If our contour is the real axis and
−∞
a semi-circle in the upper half plane, the integral along the semi-circle will go to zero by Jordan’s Lemma
and we need only evaluate the residue at z =ia. thus
I eiz Z ∞ eix
dz = dx
z2+a2 x2+a2
−∞
ei(ia)
=2πi
2ia
π
= e−a
a
Thisisreal, soI equalsthis. Ifwechangethingstoincludecoskx, wehavetobemorecarefulhowweclose
a
the contour. Doing the same trick, and considering eikz in the contour integral, note that the exponential
can be written eikrcosθ−krsinθ and will decay (and the corresponding integral in θ) in the upper half plane
if k >0. If k <0, then decay will occur only if we close the contour in the lower half plane. The latter case
means that we would encircle the pole at z =−ia instead of z =ia. This would give
I eikz Z ∞ eikx
dz = dx
z2+a2 x2+a2
−∞
(
2πiei(ika) k >0
= 2ia
−2πiei(−ika) k <0
−2ia
π
= e−|k|a
a
10