The question is about Calculus of Residues. (Integration of functions using complex functions). z is a complex number. One of the photos shows the solution. Use only the given solution. Solve parts a, b and c.
The question is about Calculus of Residues. (Integration of functions using complex functions). z is a complex number. One of the photos shows the solution. Use only the given solution. Solve parts a, b and c.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The question is about Calculus of Residues. (
z is a
One of the photos shows the solution. Use only the given solution.
Solve parts a, b and c.
![Example 21.3.1. Let us evaluate [* cos ax dx/(x² + 1)² where a + 0. This
integral is the real part of the integral I2 = [° _ eia¤
close in the UHP. Then we proceed as for integrals of rational functions. Thus, we
dx/(x² + 1)2. When a > 0, we
have
eiaz
Ia = f.
(2²
+ 1)2
dz =
2πί Res[f(i)]
for
a > 0,
because there is only one singularity in the UHP at z = i which is a pole of order 2.
We next calculate the residue:
eiaz
Res[f(i)] = lim
z→i dz
d
(z – i)².
(z – i)?(z + i)².
(z + i)iae?az
(z + i)3
eiaz
2eiaz
d
= lim
2-i dz | (z + i)²
-a
e
= lim
(1+a).
4i
zi
Substituting this in the expression for I2, we obtain I2 = (T/2)e»(1+a) for a > 0.
When a < 0, we have to close the contour in the LHP, where the pole of order
2 is at z = -i and the contour is taken clockwise. Thus, we get
iaz
dz = -2ni Res[f(-i)]
1)2
I2 =
for a < 0.
(z2 +
C
For the residue we obtain
eiaz
eª
(1
4i
d
Res[f(-i)] = lim
z--i dz
(z + i)²
(2 – i)? (z + i)² ]
а)
and the expression for I2 becomes I2 = (T/2)eª(1 – a) for a < 0. We can combine
the two results and write
COS ax
de
(x² + 1)²
Re(I2) = I2 = (1+ |a|)e-lal.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4cb9c102-5e1b-4a9f-b872-4df2c89baccb%2F43ce4175-2ce1-440b-b37c-c39a2f69fe6e%2Fo8hjwbl_processed.png&w=3840&q=75)
Transcribed Image Text:Example 21.3.1. Let us evaluate [* cos ax dx/(x² + 1)² where a + 0. This
integral is the real part of the integral I2 = [° _ eia¤
close in the UHP. Then we proceed as for integrals of rational functions. Thus, we
dx/(x² + 1)2. When a > 0, we
have
eiaz
Ia = f.
(2²
+ 1)2
dz =
2πί Res[f(i)]
for
a > 0,
because there is only one singularity in the UHP at z = i which is a pole of order 2.
We next calculate the residue:
eiaz
Res[f(i)] = lim
z→i dz
d
(z – i)².
(z – i)?(z + i)².
(z + i)iae?az
(z + i)3
eiaz
2eiaz
d
= lim
2-i dz | (z + i)²
-a
e
= lim
(1+a).
4i
zi
Substituting this in the expression for I2, we obtain I2 = (T/2)e»(1+a) for a > 0.
When a < 0, we have to close the contour in the LHP, where the pole of order
2 is at z = -i and the contour is taken clockwise. Thus, we get
iaz
dz = -2ni Res[f(-i)]
1)2
I2 =
for a < 0.
(z2 +
C
For the residue we obtain
eiaz
eª
(1
4i
d
Res[f(-i)] = lim
z--i dz
(z + i)²
(2 – i)? (z + i)² ]
а)
and the expression for I2 becomes I2 = (T/2)eª(1 – a) for a < 0. We can combine
the two results and write
COS ax
de
(x² + 1)²
Re(I2) = I2 = (1+ |a|)e-lal.
Transcribed Image Text:(3)
(1) /
(h) /
(k)
21.4. Evaluate the following integrals in which a and b are nonzero real con-
stants:
2x2 + 1
dx
dx
(a) /
(c) /
dx.
+ 5x2 + 6
(b)
6.x4 + 5x2 + 1'
x4
x4
+
1'
COS x dr
COs ax
dx
(d)
(e)
(x² + a²)²(x² + b²)'
(f)
dx.
(x² + b²)²
po0 2x²
(x² + 1)² °
x²dx
dx
1
dx.
x6 +1
(x² + 1)²(x² + 2)
(x2 + a²)² ¨
x dx
(J) L Ta² + 4x + 13)2 "
poo p³ sin ax
dx.
x6 +1
x2 +1
dx.
x2 + 4
(1)
x COs x dx
x sin x dx
dx
(n)
(0)
x2
2x + 10
x2 – 2x + 10
x² + 1
x2 dx
dx
(4) /
COs ax
dx.
x2 + b²
(x2 + 4)² (x² + 25)'
(x² + 4)2 "
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