21.4. Evaluate the following integrals in which a and b are nonzero real con- stants: 2x2 +1 dx d.x dx. x4 + 5x2 + 6 (c) (b) 6л4 + 5а? + 1 x4 +1
21.4. Evaluate the following integrals in which a and b are nonzero real con- stants: 2x2 +1 dx d.x dx. x4 + 5x2 + 6 (c) (b) 6л4 + 5а? + 1 x4 +1
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 complex analysis and Calculus of Residues.
One of the 2 pictures is an example of a solution.
Please solve parts a, b and c.
![Example 21.3.2. As another example, let us evaluate
x sin ax
dx
xª + 4
where a + 0.
This is the imaginary part of the integral I, = , xeias dæ/(x* +4) which, in terms
of z and for the closed contour in the UHP (when a > 0), becomes
zeiaz
I2 = .
d 2πί ) Resf (2)]
z4 + 4
for a > 0,
(21.6)
j=1
where C is the large semicircle in the UHP. The singularities are determined by the
zeros of the denominator: 24 +4 = 0 or z = 1+i,-1±i. Of these four simple poles
only two, 1+i and -1+i, are in the UHP. We now calculate the residues:
zeiaz
Res[f(1+ i)] = lim (z – 1– i).
z-1+i
(z – 1– i)(z – 1 + i)(z+1 – i)(z+1+i)
(1+i)eia(1+i)
(2i) (2)(2+ 2i)
eia e-c
8i
žečaz
Res[f(-1+i)] =
lim (z +1– i).
z--1+i
(z +1– i)(z+1+ i)(z – 1 – i)(z – 1+i)
(-1+i)e²a(-1+i)
(2i)(-2)(-2+2i)
-ia.-a
8i
Substituting in Equation (21.6), we obtain
I2 = 2ni-
8i
- e-i«) = ie-" sin a.
Thus,
x sin ax
dx
x4 + 4
Im(I2) =
2
-a
sin a
for a > 0.
(21.7)
For a < 0, we could close the contour in the LHP. But there is an easier way of
getting to the answer. We note that -a > 0, and Equation (21.7) yields
a sin ax
dx =
x sin[(-a)x]
dx
sin(-a)
re“ sin a.
x4 + 4
x4 + 4
We can collect the two cases in
x sin ax
dx
x4 + 4
-|a|
2
sin a.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F94dc01b6-0881-4fad-b8f8-3d9bb8d43f15%2F0b9d1b07-1b4c-4cf6-963e-cb635e27c742%2Fyih1pmc_processed.png&w=3840&q=75)
Transcribed Image Text:Example 21.3.2. As another example, let us evaluate
x sin ax
dx
xª + 4
where a + 0.
This is the imaginary part of the integral I, = , xeias dæ/(x* +4) which, in terms
of z and for the closed contour in the UHP (when a > 0), becomes
zeiaz
I2 = .
d 2πί ) Resf (2)]
z4 + 4
for a > 0,
(21.6)
j=1
where C is the large semicircle in the UHP. The singularities are determined by the
zeros of the denominator: 24 +4 = 0 or z = 1+i,-1±i. Of these four simple poles
only two, 1+i and -1+i, are in the UHP. We now calculate the residues:
zeiaz
Res[f(1+ i)] = lim (z – 1– i).
z-1+i
(z – 1– i)(z – 1 + i)(z+1 – i)(z+1+i)
(1+i)eia(1+i)
(2i) (2)(2+ 2i)
eia e-c
8i
žečaz
Res[f(-1+i)] =
lim (z +1– i).
z--1+i
(z +1– i)(z+1+ i)(z – 1 – i)(z – 1+i)
(-1+i)e²a(-1+i)
(2i)(-2)(-2+2i)
-ia.-a
8i
Substituting in Equation (21.6), we obtain
I2 = 2ni-
8i
- e-i«) = ie-" sin a.
Thus,
x sin ax
dx
x4 + 4
Im(I2) =
2
-a
sin a
for a > 0.
(21.7)
For a < 0, we could close the contour in the LHP. But there is an easier way of
getting to the answer. We note that -a > 0, and Equation (21.7) yields
a sin ax
dx =
x sin[(-a)x]
dx
sin(-a)
re“ sin a.
x4 + 4
x4 + 4
We can collect the two cases in
x sin ax
dx
x4 + 4
-|a|
2
sin a.
Transcribed Image Text:21.4. Evaluate the following integrals in which a and b are nonzero real con-
stants:
(e)
2x2 + 1
dx
dx
(a) /
dx.
x4 + 5x² + 6
(b)
6x4 + 5x² + 1
x4 +1
COs x dx
COS аx
dx
(d) . T2 + a²)² (x² + b²)'
(fĐ
(e)
d.x.
(x² + 1)²´
1)2"
(x2 + 62)2
* 2x2
dx
x² dx
(1)/
(k) /
(g) (a2 + 1)²(x² + 2)
1
dx.
x6 +1
(1) /
(x² + a²)² *
0* p3 sin ax
dx.
x6 +1
x2 +1
dx.
x2 + 4
x dx
(6) /
(1)
(x2 + 4x + 13)² *
x sin x dx
(0) /
x COs x dx
dx
(m) /
(n) /
22 — 2л + 10
x² – 2x + 10
x2 + 1
x? dx
dx
(4)
(x2 + 4)² (x² + 25)
COs ax
dx.
x2 + b2
(p)
(*)
(x²
+ 4) *
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