3
Integrals and derivatives
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You should have at your disposal the basic theorems of first and second year
calculus, such as the Mean Value Theorem, Taylor’s Theorem, Fundamental
Theorem of Calculus, and so on.
3.1
Easier problems
Problem 1. Prove that if f : [0, 1] → R is continuous, then
Z π
Z π
xf (sin x)dx = π
f (sin x)dx.
0
0
Solution
Split the integral into the ranges [0, π/2] and [π/2, π] and
make the substitution x = π − t.
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Problem 2. Determine an explicit formula in z for
Z 1
x−1
dx.
ln x
0
Solution
Consider instead
Z
1
I(α) =
0
xα − 1
dx.
ln x
So we want I(1). We can differentiate under the integral sign with respect
to α (Leibniz Rule) to get
Z
0
I (α) =
1
xα dx =
0
1
.
α+1
R
Therefore I(α) = I 0 (α) = ln(α + 1) + c for an appropriate constant c. Now
we have I(0) = 0 and so c = 0 and I(α) = ln(α + 1). Letting α converge to
1, we get I(1) = ln 2 which is the integral we wanted.
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Problem 3. Find the exact value of
Z π
I=
ln(sin x)dx.
0
Solution
Write sin x = 2 sin(x/2) cos(x/2) so
ln(sin x) = − ln 2 + ln(sin(x/2)) + ln(cos(x/2)).
1
Integrating this, we get
Z π
Z
I = π ln 2 + 2
ln(sin(x/2)) = π ln 2 + 4
0
π/2
ln(sin y)dy = π ln 2 + 2I
0
since the integrals if sin(x/2) and cos(x/2) are equal because sin x = cos(π/2−
x). So I = −π ln 2.
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Problem 4. Prove that there is at most one continuous real-valued function
f satisfying, for all reals x and y:
Z xZ y
f (x, y) = 1 +
f (u, v)dvdu.
0
0
Solution
Suppose f and g are two such functions. Since f (u, v)−g(u, v)
is continuous on [0, x] × [0, y], we know that |f (u, v) − g(u, v)| ≤ K on
[0, x] × [0, y] for some constant K. Therefore
Z xZ y
Kdvdu = Kxy.
|f (x, y) − g(x, y)| ≤
0
0
Repeat this argument to get
Z
xZ y
|f (x, y) − g(x, y)| ≤
0
0
1
Kuvdvdu = K(xy)2 .
2
Repeat again until we get
|f (x, y) − g(x, y)| ≤
K(xy)n
.
n!
Taking limits as n → ∞, we get f (x, y) = g(x, y). This is valid for any
(x, y) ∈ R2 , so f = g.
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Problem 5. Let f : R → R be differentiable and integrable, and let a ∈ R+ .
Prove that
Z ∞
f (x) − f (ax)
dx = f (0) ln a.
x
0
Solution
This is trivial if a = 1. Suppose a 6= 1; we can assume
0 < a < 1. The integral is exactly
Z ∞Z x 0
f (u)
dudx.
x
0
ax
Now interchange the order of integration.
2
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Problem 6. Determine
n
X
lim
n→∞
Solution
k=1
n
.
n2 + k 2
Write the sum as
n
1X
1
.
n
1 + k 2 /n2
k=1
This is a Riemann sum for the function
f (x) =
1
1 + x2
on the interval [0, 1]. Therefore it converges to
Z
0
1
1
π
dx = arctan 1 = .
2
1+x
4
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3.2
Homework - Three Short Problems
1.Determine
Z
0
1
ln(1 + x2 )
dx.
x
2. Let f (x) = x(1 + x) and let t ∈ (0, 1). Prove that after at most k =
d−2 log te iterations of f we exceed 1, that is,
f ◦ f ◦ f ◦ · · · ◦ f (t) > 1.
|
{z
}
k times
3. Suppose that f : R → R is a differentiable function and that
lim [f (x) + f 0 (x)] = 0.
x→∞
Prove that limx→∞ f (x) = 0.
3