29.03.

2025
1 Integrals
We begin by recalling the basic facts about indefinite integrals. Integration is the inverse operation to
differentiation. The fundamental methods for computing integrals are the backward application of the chain
rule, which takes the form
Z Z
′
f (g(x))g (x)dx = f (u)du

and shows up in the guise of the first and second substitutions, and integration by parts
Z Z
u dv = uv − v du,

which comes from the product rule for derivatives. Otherwise, there is Jacobi’s partial fraction decompo-
sition method for computing integrals of rational functions, as well as standard substitutions such as the
trigonometric and Euler’s substitutions.
Now let us turn to our nonstandard examples.
Problem 1. Compute the integrals:
Z Z
sin x cos x
dx and dx.
sin x + cos x sin x + cos x
Problem 2. For a > 0, compute the integral
Z
1
√ dx, x > 0.
x x + xa + 1
2a

Next, definite integrals. Here the limits of integration also play a role.
Problem 3. Let f : [0, 1] → R be a continuous function. Prove that
x
Z x Z
2
xf (sin x) dx = x f (sin x) dx.
0 0

Problem 4. There are special types of integrals that are computed recursively. We illustrate this with a
proof of The Leibniz formula.
π 1 1 1
= 1 − + − + ··· ,
4 3 5 7

2 Riemann Sums
The definite integral of a function is the area under the graph of the function. In approximating the area
under the graph by a family of rectangles, the sum of the areas of the rectangles, called a Riemann sum,
approximates the integral. When these rectangles have equal width, the approximation of the integral by
Riemann sums reads
n Z b
1X
lim f (x̂i ) = f (x)dx,
n→∞ n a
i=1

where each x̂i is a number in the interval a + i−1 i

 
n
(b − a), a + n
(b − a) .
Since the Riemann sum depends on the positive integer n, it can be thought of as the term of a sequence.
Sometimes the terms of a sequence can be recognized as the Riemann sums of a function, and this can prove
helpful for finding the limit of the sequences. As Hilbert advised: “Always start with an easy example.”
 Problem 5. Compute the limit
 
1 1 1
lim + + ··· + .
n→∞ n+1 n+2 2n

Problem 6. (Pólya-Szegő) Denote by Gn the geometric mean of the binomial coefficients

     
n n n
, ,..., .
0 1 n
Prove that p √
n
lim Gn = e.
n→∞

Source: G. Pólya, G. Szegő, Aufgaben und Lehrsätze aus der Analysis (Springer-Verlag, 1964)

3 Inequalities for Integrals

A very simple inequality states that if f : [a, b] → R is a nonnegative continuous function, then
Z b
f (t)dt ≥ 0,
a

with equality if and only if f is identically equal to zero.

Problem 7. Find all continuous functions f : [0, 1] → R satisfying
Z 1 Z 1
1
f (x)dx = + f 2 (x2 )dx.
0 3 0

Fundamental Integral Inequalities

We now list some fundamental inequalities. We will be imprecise as to the classes of functions to which they
apply, because we want to avoid the subtleties of Lebesgue’s theory of integration. The natural consideration
should have the piecewise continuous real-valued functions on some domain D that is an interval of the real
axis or some region in Rn .
The Cauchy-Schwarz Inequality. Let f and g be square integrable functions. Then
Z 2 Z  Z 
2 2
f (x)g(x) dx ≤ f (x) dx g (x) dx .
D D D

Minkowski’s Inequality. If p > 1, then

Z 1/p Z 1/p Z 1/p
p p p
|f (x) + g(x)| dx ≤ |f (x)| dx + |g(x)| dx .
D D D

1 1
Hölder’s Inequality. If p, q > 1 such that p
+ q
= 1, then
Z Z 1/p Z 1/q
p q
|f (x)g(x)| dx ≤ |f (x)| dx |g(x)| dx .
D D D

Problem 8. Prove the Chebyshev’s Inequality. Let f and g be two increasing functions on [a, b].
Then Z b Z b  Z b 
(b − a) f (x)g(x) dx ≥ f (x) dx g(x) dx .
a a a
4 Taylor and Fourier Series
Some functions, called analytic, can be expanded around each point of their domain in a Taylor series:

f ′ (a) f ′′ (a) f (n) (a)

f (x) = f (a) + (x − a) + (x − a)2 + · · · + (x − a)n + · · · .
1! 2! n!
If a = 0, the expansion is also known as the Maclaurin series. Rational functions, trigonometric functions, the
exponential and the natural logarithm are examples of analytic functions. A particular example of a Taylor
series expansion is Newton’s binomial formula:
∞   ∞
a
X a n X a(a − 1) · · · (a − n + 1) n
(1 + x) = x = x ,
n=0
n n=0
n!

which holds true for all real numbers a and for |x| < 1.
Problem 9. Compute the integral Z 1
ln x ln(1 − x) dx.
0

Problem 10. (Rădulescu) Prove that for |x| < 1,

∞
2
X 22k−1 2k
(arcsin x) =
x .
k 2 2k
k=1 k

Source: S. Rădulescu, M. Rădulescu, Theorems and Problems in Mathematical Analysis

(Editura Didactică s, i Pedagogică, Bucharest, 1982)

Problem 11. (Stirling’s Formula) Prove that

√  n n θn
n! = 2πn · e 12n , for some 0 < θn < 1.
e

Fourier Series Expansions

In a different perspective, we have the Fourier series expansions. The Fourier series allows us to write an
arbitrary oscillation as a superposition of sinusoidal oscillations. Mathematically, a function f : R → R that
is continuous and periodic of period T admits a Fourier series expansion
∞   X ∞  
X 2nπ 2nπ
f (x) = a0 + an cos x + bn sin x .
n=1
T n=1
T

This expansion is unique, with coefficients given by:

1 T
Z
a0 = f (x) dx,
T 0

2 T
Z  
2nπ
an = f (x) cos x dx,
T 0 T
2 T
Z  
2nπ
bn = f (x) sin x dx.
T 0 T
Of course, we can require f to be defined only on an interval of length T , and then extend it periodically.
However, if the values of f at the endpoints of the interval differ, then the convergence of the series is
guaranteed only in the interior of the interval.
 Let us discuss a problem from the Soviet Union University Student Contest.
Problem 12. Compute the sum
∞
X cos n
.
n=1
1 + n2
We find even more exciting a fundamental result of ergodic theory that proves that for an irrational number
α, the fractional parts of nα (n ≥ 1) are uniformly distributed in [0, 1]. For example, when α = log10 2, we
obtain as a corollary that on average, the first digit of a power of 2 happens to be 7 as often as it happens to
be 1.
Problem 13. Do you know a power of 2 whose first digit is 7?
Theorem. Let f : R → R be a continuous function of period 1 and let α be an irrational number. Then
Z 1
1
lim (f (α) + f (2α) + · · · + f (nα)) = f (x)dx.
n→∞ n 0

Problem 14. Prove that for every 0 < x < 2π the following formula is valid:
∞
π − x X sin nx
= .
2 n=1
n

Derive from it the formula: ∞

π X sin(2k − 1)x
= , x ∈ (0, π).
4 k=1
2k − 1
1
Problem 15. Use the Fourier series of the function of period 1 defined by f (x) = 2
− x for 0 ≤ x < 1 to
prove Euler’s formula:
∞
π2 X 1
= 2
.
6 n=1
n
Problem 16. Prove that: ∞
π2 X 1
= 2
.
8 k=0
(2k + 1)