Page 291 - DMTH401_REAL ANALYSIS

Unit 23: Differentiation of Integrals

 f(1) – 2p = 0 Notes
 f(1) = 2p.
2p
q
which is the value of the integral ò e cosq cos(sin )dq .
0
d cos x
2
Example: Find ò cosh t dt .
dx sin x
In this example, we shall simply apply the above given formula, to get
d cos x 2 2 d 2 d
ò cosh t dt = cosh(cos x) (cosx) cosh(sin x) (sin x)+
-
dt sin x dx dx
cos x ¶
2
2
2
ò sin x ¶ x cosh t dt = - cosh(cos x)sin x cosh(sin x)cosx
-
Where the derivative with respect to x of hyperbolic cosine t squared is 0. This is a simple
example on how to use this formula for variable limits.
Self Assessment

Fill in the blanks:

1. The differentiation of integrals is the Lebesgue differentiation theorem, as proved by
Henri Lebesgue in ................................. .
2. The result for ................................. turns out to be a special case of the following result,
which is based on the Besicovitch covering theorem.
3. The problem of ........................................ is that of determining under what circumstances
the mean value integral of a suitable function on a small neighbourhood of a point
approximates the value of the function at that point.
4. The problem of the differentiation of integrals is much harder in an infinite-dimensional
setting. Consider a separable Hilbert space (H, ) equipped with a ............................... .

23.3 Summary

 In mathematics, the problem of differentiation of integrals is that of determining under
what circumstances the mean value integral of a suitable function on a small neighbourhood
of a point approximates the value of the function at that point.
 One result on the differentiation of integrals is the Lebesgue differentiation theorem, as
n
proved by Henri Lebesgue in 1910. Consider n-dimensional Lebesgue measure  on
n
n-dimensional Euclidean space R .
 The result for Lebesgue measure turns out to be a special case of the following result,
which is based on the Besicovitch covering theorem: if  is any locally finite Borel measure
n
on R and f : R  R is locally integrable with respect to , then
n
1

n
lim ò f(y)d (y) f(x) for -almost all points x  R .
=
r 0 (B (x)) r B (x)
r
 The problem of the differentiation of integrals is much harder in an infinite-dimensional
setting. Consider a separable Hilbert space (H, ) equipped with a Gaussian measure .

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