Electromagnetics Homework: Asymptotic Evaluation of Integrals Solution

This document provides detailed solutions to an electromagnetics homework assignment involving the asymptotic evaluation of integrals. The first question utilizes integration by parts to evaluate an integral. The second question employs the stationary-phase method to approximate the Bessel function Jn(x) for large x. The third question applies Laplace's method to asymptotically evaluate the modified Bessel function of the first kind, I0(Ω), for large Ω. Finally, the fourth question uses Watson's Lemma to determine the first two leading terms of the asymptotic approximation to a given integral as Ω becomes large. Each solution includes detailed steps and justifications for the methods used.

Running head: ELECTROMAGNETICS 1
Electromagnetics
Name
Institution

ELECTROMAGNETICS 2
Question 1
I ( Ω ) =∫
0
1
e−x cos ( Ωx ) dx
Let e− x=u so that du=−e− x dx and
cos ( Ωx ) =dv so that v=∫ cos ( Ωx ) dx= 1
Ω sin ( Ωx )
Using integration by parts,
∫udv=uv−¿∫ vdu ¿
∫udv= e− x
Ω sin ( Ωx ) −¿∫ −e−x
Ω sin ( Ωx ) dx ¿
¿ e−x
Ω sin ( Ωx ) + 1
Ω ∫e− x sin ( Ωx ) dx … equation 1
Apply integration by parts to solve ∫ e−x sin ( Ωx ) dx by letting
e− x=u so that du=−e− x dx
sin ( Ωx )=dv so that v=∫ sin ( Ωx ) dx=−1
Ω cos ( Ωx )
∫ e−x sin ( Ωx ) dx=−e− x
Ω sin (Ωx )−∫ e−x
Ω cos ( Ωx ) dx
∫ e−x sin ( Ωx ) dx=−e− x
Ω sin (Ωx )− 1
Ω ∫e− x cos ( Ωx ) dx
But we know that, ∫ e−x cos ( Ωx ) dx =I (Ω)

ELECTROMAGNETICS 3
∫ e−x sin ( Ωx ) dx=−e− x
Ω sin ( Ωx ) − I ( Ω )
Ω … equation 2
Substituting equation 2 into equation 1 we obtain,
I ( Ω ) = e−x
Ω sin ( Ωx ) + 1
Ω ( −e−x
Ω sin ( Ωx ) − I ( Ω )
Ω )
I ( Ω )= e−x
Ω sin ( Ωx ) +−e−x
Ω2 sin ( Ωx ) − I (Ω )
Ω2
I ( Ω )+ I ( Ω )
Ω2 = e− x
Ω sin ( Ωx )− e−x
Ω2 sin ( Ωx )
I ( Ω ) (1+ 1
Ω2 )= e−x
Ω sin ( Ωx )− e− x
Ω2 sin (Ωx )
(1+ 1
Ω2 )= ( Ω2+1
Ω2 )
I ( Ω )=
( e− x
Ω sin ( Ωx )− e−x
Ω2 sin ( Ωx ) )÷ ( Ω2+1
Ω2 )
¿ [ Ωe−x sin ( Ωx )−e− x cos ( Ωx )
Ω2+ 1 ]x=0
x=1
¿ {Ωe−x sin ( Ωx ) −e−x cos ( Ωx ) }− {Ωe−x sin ( Ωx )−e− x cos (Ωx ) }
¿ {Ωe−1 sin ( Ω )−e−1 cos ( Ω ) }− {Ωe−0 sin ( 0 )−e−0 cos ( 0 ) } /(Ω¿¿ 2+1)¿
¿ Ωe−1 sin ( Ω ) −e−1 cos (Ω )+1
Ω2+1
¿ Ω sin ( Ω )−cos ( Ω )+ e1
e1(Ω2+1)

ELECTROMAGNETICS 4
I ( Ω )=Ω sin ( Ω )−cos ( Ω )+e1
e1 (Ω2+1)
Question 2
Jn (x )= 1
π ∫
0
π
cos ( nθ−xsin(θ) ) dθ
Jn (x )= 1
π ℜ {∫
0
π
e j ( nθ−xsin(θ) ) dθ }
Jn (n)= 1
π ℜ {∫
0
π
e j ( nθ−nsin (θ) ) dθ }
¿ 1
π ℜ {∫
0
π
e jn ( θ−sin(θ )) dθ }
f ( θ ) =1
g ( θ )=θ−sin ( θ )
g ' ( θ )=1−cos ( θ )
g ' ' ( θ )=sin (θ )=0
g ' ' ( θ0 ) =sin ( θ0 ) =0 , θ0=0
g' '' (θ0 ) =cos ( θ0 )=cos ( 0 )=1>0
I ( x ) f (θ ¿¿ 0)e
jx f (θ¿¿0 )+ j(π / 2 p) { p !
x |g p
( θ0 )| }
1
p × {Γ ( 1
p )
p }…1 ¿
¿
f (θ¿¿ 0)=1 ¿

ELECTROMAGNETICS 5
g ( θ0 ) =0−sin ( 0 )=0
p=3 , π
2 p = π
6 , p !=3 ( 2 ) ( 1 )=6
gp ( θ0 ) =g' '' ( θ0 ) =cos ( 0 ) =1> 0
Substituting these values into equation 1 we obtain,
Jn (n) 1 e
jx f (θ¿¿0 )+ j(π / 2 p) { p !
x |g p
(θ0 )| }1
p × {Γ ( 1
p )
p }¿
Jn (n) 1
π ℜ {e jπ / 6 ¿¿
{ 6
n|1| }1
3 × { Γ ( 1
3 )
3 }} as n → ∞
Jn (n) 1
π 2
2
3
{6
n }1
3 Γ ( 1
3 ) as n → ∞
Jn ( n )= 1
π (2¿¿ 2
3 )(6¿¿ 1
3 )(n¿ ¿ −1
3 ) Γ ( 1
3 )¿ ¿ ¿ as n → ∞
Question 3
I ( Ω )= 1
2 π ∫
0
2 π
eΩsinθ dθ
The equation is the form
I ( Ω )= 1
2 π ∫
a
b
f (x) eΩg(x) dx
g' ( x0 ) =0 , x0 ∈ ( a , b )

ELECTROMAGNETICS 6
g ( x0 ) > g ( x )=0 , x0 ∈ ( a , b )
I ( Ω ) f ( θ0 ) eΩg ( θ0 )
√ −2 π
Ω g' ' ( θ0 )
f ( θ ) =1 , f ( θ0 ) =1
g ( θ )=sin (θ )
g ' ( θ )=cos ( θ )
g ' ( θ0 ) =cos ( θ0 )=0 , θ0 = π
2
g' '
( θ0 )=−sin ( θ0 )=−sin ( π
2 )=−1
I ( Ω ) = 1
2 π (1)eΩ(0 )
√ −2 π
Ω ( −1 )
I ( Ω )= 1
2 π √ 2 π
Ω = √ 2 π
Ω (2 π )2 = √ 2 π
4 Ωπ2 = √ 1
2Ωπ as Ω → ∞
Question 4
I ( Ω )=∫
−1
1
e−Ω s2
ds
From the equation we have,
f ( s ) =cos (s ) and g ( s )=−s2
g ' ( s )=−2 s and s0 =0
h ( x )=f ( s ( x ) ) ds
dx ( x )

ELECTROMAGNETICS 7
x2=g ( s0 ) −g ( s ) =−s2 ¿s=0 + s2 =0+ s2=s2
x2=s2
Hence, x=s
f ( s ( x ) ) =cos ( s ( x ) ) =cos ( x)
ds
dx = dx
dx =1
h ( x ) =cos ( x ) ( 1 ) =cos (x )
But we know that h ( x ) ∑
n=−0,2,4
an xn
as x → 0
h ( x ) =cos ( x ) =1− x2
2! + x4
4 ! …
which implies that,
a0=1 , a2=−1
2 , a4= 1
4
Recall:
I ( Ω ) e−Ω g ( s0 ) ∑
n=0,2,4
an
n+1
2
Γ ( n+1
2 )
Hence, we have:
I ( Ω ) eΩ ( 0 )
( a0
√Ω Γ ( 1
2 )+ a2
Ω
3
2
Γ ( 3
2 )+…
)

ELECTROMAGNETICS 8
I ( Ω ) ( 1
√ Ω Γ ( 1
2 )−
1
2
Ω
3
2
Γ ( 3
2 )+ … )
Γ ( 1
2 )= √π , Γ ( 3
2 )=1 √π
2
Hence,
I ( Ω ) = √ π
√ Ω − ( 1
2
Ω
3
2 ) 1 √ π
2 = √ π
Ω − √ π
4 Ω
3
2
Therefore, the first two leading terms are I ( Ω )= √ π
Ω and I ( Ω ) =− √ π
4 Ω
3
2