17 More topics in integration

Normal distribution of stock returns. Functions defined by integrals. Error func-

tion. Improper integrals (with infinite limits of integration). Convergence test.
Cumulative distribution function.

17.1 Normal distribution of stock returns

Consider stock returns over time. There are certainly good reasons why some
grow, some falls, and trends change. A careful investor can then try to analyse
these reasons to predict the performance of certain stocks and decide whether or
not to invest. As an alternative, one can study their behaviour as a whole. For
example, considering a large set over time and representing it by plotting on the
x-axis the return and on the y-axis the percentage of stocks with that return.
Surprisingly, or perhaps not, if the data refers to a period with no catastrophic
events, one finds (roughly) a normal distribution 1 . A symmetrical bell-shaped line

x
μ-3σ μ-2σ μ -σ μ μ+σ μ+2σ μ+3σ

centred on a certain value µ, called mean, and with a certain average semi-width ,
called standard deviation. 2 is called the variance of the distribution. About 68%
of all values are within one standard deviation from the mean, about 95% within
1
The normal distribution is central in probability theory and statistics. The Central Limit
Theorem states that, under appropriate conditions, the normalized distribution of a sample
mean converges to a normal distribution when the size of the sample grows.

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17 More topics in integration

two standard deviations and about 99% within three standard deviations. The
mean µ and standard deviation are a characteristic of the set of data analyzed
and, in a sense, completely characterize it. This fact can be (and is) used for asset
investments, with the mean quantifying the portfolio return and the standard
deviation the risk , so that knowing these values for a selected asset will make an
investor aware of the expected returns and risks.
The analytic form of the normal distribution is given by
1 (x µ)2
f (x) = p e 2 2 ,
2⇡
1
with the constant p2⇡ introduced in such a way that the area under the bell is
exactly equal to one. In this way, f (x) corresponds to the probability that a stock
randomly drawn from the asset has return x, and the integral
Z b
1 (x µ)2
p e 2 2 dx,
a 2⇡
to the probability that the return is in the interval [a, b]. However, as we already
2
mention in section 15.3, the indefinite integral of functions like e x can not be
evaluated explicitly and therefore neither can the definite integral. How can we
deal with these kinds of integrals?

17.2 Functions defined by integrals

As already seen in section 16.4, the definite integral of a function f in an interval
[c, x], where c 2 Df is a given constant and x 2 Df is an arbitrary point of the
domain of f ,
Z x
F (x) = f (t)dt,
c

provides a rule that associates to each x 2 Df a real number, thus defining a

real function. If the integral can be evaluated explicitly, we simply obtain a new
representation of a known function. If the integral cannot be evaluated explicitly,
we obtain a completely new function that cannot be expressed in algebraic terms.
Example 17.1. For example, if we had not extended algebra to the point of in-
troducing the natural exponential (see sections 5.4, 5.5 and 5.6), the natural log-
arithm, ubiquitous in science, could have been defined as
Z x
1
ln x = dt with x 2 (0, 1),
1 t

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 17.2 Functions defined by integrals

instead of as the inverse function of the natural exponential, and the natural
exponential could have been introduced as the inverse function of the natural
logarithm. The basic properties of the natural logarithm can be obtained from its
integral representation. For example,
Z xy Z x Z xy Z x Z xy
1 1 1 1 1
ln(xy) = dt = dt + dt = dt + dt =
1 t 1 t x t 1 t x t
Z x Z y
1 1
= dt + ds = ln x + ln y,
1 t 1 s

where, after splitting the integral, we have performed the substitution of variable
t
x
= s with dt = xds, s = 1 for t = x, s = y for t = xy, in the second integral.
Similarly we can show that ln xy = ln x ln y and ln xy = y ln x.

As further illustrated in the previous section, functions defined by integrals can

be important in applications. Therefore, we must understand their properties and
develop tools to study them, first and foremost derivatives.

Continuity Let us start by observing that, even if the integrand f is not contin-
uous, the function F is continuous at every point x̄ 2 Df , because F is defined at
x̄ and

lim F (x) = F (x̄),

x!x̄

since the signed area between the graph of f and the x-axis varies with continuity
even if the function has a discontinuity.

Example 17.2. For example, if c = 0 and f is the Heaviside function H(x) (0 for
x < 0 and 1 for x 0), which is discontinuous at 0, we obtain
Z x ⇢
0 for x < 0
h(x) = H(x)dx = ,
0 x for x 0

which is continuous at 0 because limx!0+ h(x) = limx!0 h(x) = h(0) = 0.

Di↵erentiability On the other hand, from this example we see that if f is not
continuous at a point, F is not di↵erentiable at that point.

Example 17.3. Resuming example 17.2 we see that h(x) is not di↵erentiable at 0
because
h(0 + x) h(0) h( x) 0
lim = lim = lim = lim 0 = 0,
x!0 x x!0 x x!0 x x!0

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17 More topics in integration

while
h(0 + x) h(0) h( x) x
lim + = lim + = lim + = lim + 1 = 1.
x!0 x x!0 x x!0 x x!0

The left and right limits being di↵erent, the derivative at 0 does not exist.
It is possible to prove that if f is continuous at a point x, then F is di↵erentiable
at that point. As we have seen already, the derivative of F is then equal to the
integrand,
F 0 (x) = f (x).
Correspondingly, the second derivative of F is then equal to the derivative of f ,
F 00 (x) = f 0 (x),
and so on. If the function is sufficiently regular, this allows us to Taylor ap-
proximate the value of the function at a point with arbitrary precision. Nothing
di↵erent than what we already do for complicated functions such as exponentials,
roots, and logarithms.

Error function Calculations with the normal distribution require the introduc-
tion of the error function
Z x
2 2
erf x = p e t dt,
⇡ 0
which owes its name to its importance in the theory of errors, which are also
normally distributed. The probability that a stock has return in the interval [a, b]
considered in section 17.1, can then be expressed in terms of the error function by
performing the substitution
x µ p a µ b µ
p = t, dx = 2 dt, t = p for x = a and t = p for x = b,
2 2 2
so that
Z b Z b µ
p
p Z b µ
p
1 (x µ)2 1 2
t2 1 2 2
t2
p e 2 2 dx = p e 2 dt = p e dt =
a 2⇡ 2⇡ a
p µ 2 ⇡ a
p µ
2 2

Z Z b µ
!
0 p
1 2 t2 2 2
t2
= p e dt + p e dt =
2 ⇡ a
p µ ⇡ 0
2

 ✓ ◆ ✓ ◆
1 b µ a µ
= erf p erf p .
2 2 2

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 17.2 Functions defined by integrals

The values of the error function can be found on some scientific calculators, which
use its Taylor expansion or more advanced approximation techniques to estimate
them. To Taylor approximate the function it is sufficient to observe that
X1 X1
t2 1 2 n ( 1)n 2n
e = ( t) = t
n=0
n! n=0
n!

so that
Z 1
xX 1 Z
2 ( 1)n 2n 2 X ( 1)n x 2n
erf x = p t dt = p t dt
⇡ 0 n=0
n! ⇡ n=0 n! 0

1 ✓ ◆
2 X ( 1)n 2n+1
2 1 3
1 5
1 7
=p x =p x x + x x + ... ,
⇡ n=0 (2n + 1)n! ⇡ 3 10 42

a series that converges very fast. To graph the function, we observe that
x2
,! erf 0 x = e , which is positive for all x 2 R, so that erf x is monotone
increasing;
2
,! erf 00 x = 2xe x , which is positive for x < 0 and negative for x > 0, so that
erf x is convex for x < 0 and concave for x > 0;

,! erf ( x) = erf x, because its Taylor expansion contains only odd powers,
so that its graph is invariant under double reflection with respect to the x
and y axes;

,! lim erf x = 1, because the area under the bell is equal to one2 , so that erf x
x!1
approaches the horizontal line x = 1 for x ! 1.

Error function
y
1.0

0.5

-3 -2 -1 1 2 3
x

-0.5

-1.0

Z 1
2 t2
p
It can be shown that e dt = ⇡, but the calculation is not yet within our grasp.
1

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17 More topics in integration

Example 17.4. Assume that a stock has returns of 4%, 8%, 3% and 15% over
the last four years. Assuming that stock returns are normally distributed, what is
the probability that the next return will be between 10% and 20%? To answer the
question, it is first necessary to identify the parameters of the distribution. The
way to extract the mean µ and standard deviation from a data set r1 , r2 , ... rn ,
is provided by the statistical formulas
v
n u n
1 X u 1 X
µ= rk and = t (rk µ)2 ,
n k=1 n 1 k=1

which are the more accurate the greater the number of data. For our small data
sample
1
µ= (4 + 8 3 + 15) % = 6%,
4
r
1
= [(4 6)2 + (8 6)2 + ( 3 6)2 + (15 6)2 ]% ' 7.5%.
3
Therefore, the probability that the next return will be between 10% and 20% is
given by
Z 20  ✓ ◆ ✓ ◆
1 (x 6)2 1 20 6 10 6
p e 2·7.5 2
dx = erf p erf p ' 26.6%.
10 2⇡7.5 2 27.5 27.5
Analogously, we can compute the probability that the next return will be in any
bounded interval. On the other hand, to compute the probability that the return
will be greater than a certain percentage or not less than another percentage, we
need to extend the integration to unbounded intervals. Does this make sense?

17.3 Improper integrals

The definition of definite integral of a bounded function in an interval, requires
the interval be bounded. We cannot divide an unbounded interval in n equal parts
and procede with the construction required by the definition. On the other hand,
we can ask what value a definite integral approaches when one of the limits of
integration becomes larger and larger in absolute value, positive or negative. In
more precise terms, we can give a meaning to an integral in an unbounded interval
as the limit of a definite integral.
Definition. The improper integral of a bounded function f : D ⇢ R ! R in
R +1
the unbounded interval [c, +1) ⇢ D, resp. ( 1, c] ⇢ D, denoted by c f (x)dx,

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 17.3 Improper integrals

Rc
resp. 1
f (x)dx, is defined as the limit
Z +1 Z h Z c Z c
f (x)dx = lim f (x)dx, resp. f (x)dx = lim f (x)dx.
c h!+1 c 1 h! 1 h

Being a limit, an improper integral can be convergent or divergent. The conver-

gence of the improper integral over the entire real line is defined by splitting the
integral as
Z +1 Z c Z +1
f (x)dx = f (x)dx + f (x)dx,
1 1 c

with c 2 R and arbitrary constant, and requiring that both improper integrals
obtained converge.
1
Example 17.5. As a first example let us consider the integral of x
in the interval
[1, +1),
Z +1 Z h Z h
1 1
dx = lim dx = lim x 1 dx =
1 x h!+1 1 x h!+1 1
= lim ln x|h1 = lim ln h
h!+1 h!+1

which diverges because ln h ! +1 for h ! +1. Since x1 is positive in [1, +1),

the integral represents the area of an unbounded region and the fact that it does
y
3.0
2.5
2.0
1.5
1.0
0.5
x
2 4 6 8 10
-0.5

not converge is in agreement with our intuition.

Example 17.6. As a second example, we consider the integral of x12 again in the
interval [1, +1),
Z +1 Z h Z h
1 1
2
dx = lim 2
dx = lim x 2 dx =
1 x h!+1 1 x h!+1 1
h ✓ ◆
1 1
= lim = lim +1 =1
h!+1 x 1 h!+1 h

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17 More topics in integration

y
3.0
2.5
2.0
1.5
1.0
0.5

2 4 6 8 10
x
-0.5

which converges to 1. Again, since x12 is positive in [1, +1), the integral represents
the area of an unbounded region. Contrary to our intuition, this area is finite. On
the other hand, x12 approaches the x-axis much faster than x1 (dashed in the figure
for comparison) and we have already had a similar surprise with series, where we
learned that the sum of infinite terms can be finite.

Convergence test Since we are not always able to evaluate integrals, in certain
circumstances it may be useful to be able to at least determine whether a certain
improper integral converges or diverges. A simple test is suggested by the figure
above.
Z +1 Z +1
,! If f (x) g(x) 0 and f (x)dx converges, then g(x)dx converges.
c c

This is because the area between the graph of g and the x-axis is smaller than the
one between the graph of f and the x-axis, and if the latter is finite, the former
must be too. We can also conclude that,
Z +1 Z +1
,! If g(x) f (x) 0 and f (x)dx diverges, then g(x)dx diverges.
c c

This is because the area between the graph of g and the x-axis is larger than the
one between the graph of f and the x-axis, and if the latter diverges, the former
must too. Let us consider some examples.
Example 17.7. Consider the improper integral
Z +1
1
2
dx.
1 x +1
With what we have done so far, we are not able to calculate the indefinite integral
of x21+1 . However, x2 + 1 > x2 so that 1+x
1 1
2 < x2 and

Z +1 Z +1
1 1
2
dx < dx = 1 (see example 17.6).
1 x +1 1 x2
R +1
We can conclude that 1 x21+1 dx converges.

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 17.3 Improper integrals

Example 17.8. Consider the improper integral

Z +1
1
dx.
1 ln x

Since for x > 1, ln x < x we have that ln1x > x1 and

Z +1 Z +1
1 1
dx > dx.
1 ln x 1 x
R +1 R +1
Since in example 17.5 we saw that 1 x1 dx diverges, 1 1
ln x
dx also diverges.
x2
Example 17.9. As we mention already, even if the indefinite integral of e can
not be expressed in terms of elementary function, the improper integral
Z +1 Z 0 Z +1 Z +1
x2 x2 x2 2
e dx = e dx + e dx = 2 e x dx
1 1 0 0

can be evaluated, but the calculation is beyond our current capabilities. However,
we can at least show that it converges. We start by splitting the integral as
Z +1 Z 1 Z +1
x2 x2 2
e dx = e dx + e x dx.
0 0 1

The first integral is finite. To prove that the second one il also finite we observe
2
that for x > 1, x2 > x so that x2 < x and e x < e x . Correspondingly,
Z +1 Z +1 Z h
x2 x h
e dx < e dx = lim e x dx = lim e x1
1 1 h!+1 1 h!+1

h 1 1
= lim e +e =
h!+1 e
R +1 x2
The integral 1
e dx converges.

Cumulative distribution function In probability theory and statistics, the cu-

mulative distribution function of a real distribution function, evaluated at x 2 R,
is the probability that the variable will take a value less than or equal to x.
The cumulative distribution function F : R ! [0, 1], is then defined as the im-
proper integral in the interval ( 1, x] of the probability distribution, and satisfies
lim F (x) = 0 and lim F (x) = 1. If the variable is normally distributed,
x! 1 x!+1
Z x
1 (t µ)2
F (x) = p e 2 2 dt.
2⇡ 1

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17 More topics in integration

By recalling the definition of improper integral, the formula obtained in section

17.2 and that limt! 1 erf t = 1, we can express the cumulative distribution
function for the normal distribution in terms of the error function as
Z x
1 (t µ)2
F (x) = lim p e 2 2 dt =
h! 1 2⇡ h
 ✓ ◆ ✓ ◆  ✓ ◆
1 x µ h µ 1 x µ
= lim erf p erf p = erf p +1 .
h! 1 2 2 2 2 2
This formula is useful in applications.
Example 17.10. Resuming example 17.4 we evaluate the probability that the next
return is less than 5% and also the probability that it will be greater than 10%. The
former is immediately provided by the cumulative distribution function evaluated
at 5 percent,
 ✓ ◆
1 5 6
F (5%) = erf p + 1 ' 45%.
2 27.5
The latter is instead given by one minus the cumulative distribution function eval-
uated at 10 percent, also called complementary cumulative distribution function,
 ✓ ◆
1 10 6
1 F (10%) = 1 erf p + 1 ' 30%,
2 27.5
because it corresponds to certainty, probability one, minus the probability that
next return will be less than 10 precent.

17.4 Exercises
Functions defined by integrals
Exercise 17.1. Is the function
Z x ⇢
1 for x 0
F (x) = f (t)dt, with f (x) = ,
0 1 for x < 0
di↵erentiable? Explain.
answer F (x) = |x| not di↵erentiable at x = 0, the point of discontinuity of f .

Exercise 17.2. Let f be a continuous function in [a, b]. Assume that Zf takes only
x
positive values. That is f (x) > 0 for every x 2 [a, b]. Is F (x) = lnf (t)dt,
a
x 2 [a, b], a di↵erentiable function? Explain.
answer yes, F (x) is di↵erentiable; the integrand lnf (x) is defined for all x 2 [a, b] because f (x) > 0 for x 2 [a, b] and continuous

because the composition of continuous functions is continuous; an integral function with continuous integrand is a di↵erentiable

function of its limits of integration.

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 17.4 Exercises

Exercise 17.3. Let

Z x
F (x) = f (t)dt,
x

dF
where f is a continuous function. Evaluate (x).
dF
dx
answer (x) = f (x) + f ( x).
dx

Exercise 17.4. Write down the equation for the tangent at x0 = 0 to the graph of
Z x
2
F (x) = e t dt.
0

answer y = x.

Exercise 17.5. Find the quadratic approximation to

Z x
F (x) = 2t dt.
0

about x0 = 0.
ln 2 2
answer y = x + x.
2

Exercise 17.6. Let

Z x
f (x) = e t dt.
0

Is f (x) a concave function? Here x > 0. Explain.

answer Yes, because f 00 (x) = e x < 0 for all x 2 (0, +1).

Exercise 17.7. Is the function

Z x
2
f (x) = et dt
x

convex? Here x > 0. Explain.

2
answer Yes, because f 00 (x) = 4xex > 0 for all x > 0.

Exercise 17.8. Find the inflection points of

Z x
2
F (x) = e t dt.
0

answer x = 0.

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17 More topics in integration

Improper integrals
Exercise 17.9. Does the improper integral
Z 1
1
p dx
1 x
exist? Explain.
Z h ⇣ p ⌘ ⇣ p ⌘
1 h
solution The integral does not exist because the limit lim p dx = lim 2 x|1 = lim 2 h 2 does not exit
h!1 1 x h!+1 h!+1
p
because h ! +1 as h ! +1.

Exercise 17.10. Find all ↵ > 0 such that

Z 1
1
dx
1 x↵
exists.
answer ↵ > 1.

Exercise 17.11. Does the improper integral

Z 0
x
2
dx
1 x +1

exist? Explain.
Z 0 ✓ ⇣ ⌘ 0◆ ⇣ ⌘
x 1 2 1 2
solution The integral does not exist because the limit lim dx = lim ln x + 1 = lim ln h + 1
2
h! 1 h x + 1 h! 1 2 h! 1 2
⇣ ⌘ h
does not exist as ln h2 + 1 ! +1 as h ! 1.

ln x
Exercise 17.12. The function f is defined for x > 0 by f (x) = 3 . Examine the
x
convergence of
Z 1
f (x)dx.
1
Z 1 Z 1 Z h !
ln x ln x 1 + 2 ln x h 1
solution f (x)dx is convergent because dx = lim dx = lim = , in fact by l’Hopital’s
1 1 x3 h!+1 1 x3 h!+1 4x2 1 4
1
ln h h 1
rule lim = lim = lim = 0.
h!+1 h2 h!+1 2h h!+1 2h2

Exercise 17.13. Does the improper integral

Z +1
xe x dx
0

exist? Explain.
Z h ✓ ◆ ⇣ ⌘
x x 0 h
solution The integral exists because the limit lim xe dx = lim (1 + x)e = lim 1 (1 + h)e = 1
h!+1 0 h!+1 h h!+1
h 1+h 1
exists; in fact by l’Hopital’s rule lim (1 + h)e = lim = lim = 0.
h!+1 h!+1 eh h!+1 eh

238