Electromagnetics

This article discusses solutions to problems in Electromagnetics, including integration by parts, Bessel functions, and Laplace transforms. The article provides step-by-step solutions to each problem and includes relevant equations and formulas. The subject matter is relevant to students studying Electromagnetics in college or university courses.

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