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Math 412 Group Work #12 Spring 2009 SOLUTIONS
Problem 1. You may recall from elementary calculus that the integrals ∫ (^) b
a
et^ cos t dt and
∫ (^) b
a
et^ sin t dt
require a tricky (double) application of Integration by Parts to evaluate. Now, using complex analysis, you’ll evaluate them both easily!
(a): Express (^) ∫ b
a
e(1+i)t^ dt
as a sum of two integrals.
SOLUTION: We have ∫ (^) b
a
e(1+i)t^ dt =
∫ (^) b
a
et(cos t + i sin t)dt =
∫ (^) b
a
et^ cos t dt + i
∫ (^) b
a
et^ sin t dt.
2
(b): Evaluate (^) ∫ b
a
e(1+i)t^ dt
directly, and use the answer to evaluate both of the two integrals you wrote down in part (a).
SOLUTION: We have ∫ (^) b
a
e(1+i)t^ dt =
1 + i
e(1+i)t
b
a
1 + i
[
e(1+i)b^ − e(1+i)a
]
1 − i 2
[
eb(cos b + i sin b) − ea(cos a + i sin a)
]
− i
[
(eb^ cos b − ea^ cos a) + i(eb^ sin b − ea^ sin a)
]
[
(eb^ cos b − ea^ cos a) +
(eb^ sin b − ea^ sin a)
]
[
(eb^ sin b − ea^ sin a) −
(eb^ cos b − ea^ cos a)
]
Matching the real and imaginary parts with part (a), we conclude that ∫ (^) b
a
et^ cos t dt =
(eb^ cos b − ea^ cos a) +
(eb^ sin b − ea^ sin a)
and (^) ∫ (^) b
a
et^ sin t dt =
(eb^ sin b − ea^ sin a) −
(eb^ cos b − ea^ cos a).
2
(c): Evaluate (^) ∫ π
0
et^ cos t dt and
∫ (^) π
0
et^ sin t dt.
SOLUTION: We simply plug a = 0 and b = π into each integral formula obtained in part (b) obtain. The results are ∫ (^) π
0
et^ cos t dt = −
(eπ^ + 1) and
∫ (^) π
0
et^ sin t dt =
(eπ^ + 1).
Problem 2. Recall the Mean Value Theorem for Definite Integrals from elementary calculus:
“If f is continuous real-valued function on [a, b], then there exists a number c ∈ [a, b] such that
f (c) =
b − a
∫ (^) b
a
f (t) dt.”
Now you will see that this theorem cannot be extended to complex-valued functions. Consider the interval [0, 2 π] and let f (t) = eit. Does the theorem hold in this situation?
SOLUTION: NO. We find that ∫ (^2) π
0
eit^ dt =
i
eit
2 π
0
i
Thus, 1 2 π − 0
∫ (^2) π
0
f (t)dt = 0,
but eic^6 = 0 for any value of c ∈ C. Thus, the Mean Value Theorem for Definite Integrals fails to hold for complex-valued functions. 2
