For each positive integer define
Prove that the following limit and integral cannot be interchange:
Proof. First, we have
On the other hand,
where is a nonnegative, increasing function defined for all .
Use this to deduce the inequalities
by taking .
Proof. Since is increasing we know (where denotes the greatest integer less than or equal to ) for all . Define step functions and by
Then and are constant on the open subintervals of the partition
Since is integrable we must have
for every pair of step functions . Hence,
Next, if we take (which is nonnegative and increasing on ) we have
From this we then get two inequalities
Find values for the constants and so that
First, since the integral is continuous (Theorem 3.4 of Apostol) we know
We also have
Hence, we may apply L’Hopital’s rule to the quotient giving us (using the First Fundamental Theorem of Calculus for the numerator)
Since the limit as of the numerator is 0, we must have the limit of the denominator equal to 0 as well (otherwise the whole limit would be 0 instead of 1).
Now, substituting back into our limit and using L’Hopital’s two more times (since again we have the indeterminate form ),
Hence, the requested constants are
Let and be functions, both differentiable in a neighborhood of 0, with and such that
Prove or disprove each the following statements.
- as .
- as .
Proof. Since as we know by the definition of that
Thus, for every there exists a such that
So, for we have
The final line follows since by hypothesis. Therefore,
By definition, we then have
Consider for and for . Then, for ,
For we have .
Since we have as . However, since
does not exist.
Using the Taylor polynomial approximation of find an approximation for the integral
Give an estimate for the error of the approximation. [Define when .]
We know (from this exercise, Section 7.8, Exercise #1) that the Taylor polynomial approximation of is given by
(Note: I cannot get the bounds Apostol asks for. I prove a different set below. I cannot figure out if it is a mistake in the book or not.)
Proof. Using the algebraic identity, valid for ,
Therefore, integrating term by term,
Furthermore, we have
Taking , we then have
From the inequality for this integral we then have
Prove that there exists a number with such that
Proof. For we have
(Since for we know .) Therefore, integrating the terms in the inequality from 0 to 1,
The integral logarithm is defined for by
Prove the following properties of .
- where is a constant depending on . Find the value of for each .
- Prove there exists a constant such that
and find the value of this constant.
- Let . Find an expression for
in terms of .
- Define a function for by
- Proof. We derive this by integrating by parts. Let
Then we have
- Proof. The proof is by induction. Starting with part (a) we have
To evaluate the integral in this expression we integrate by parts with
This gives us
Therefore we have
where . This is the case . Now, assume the formula hold for some integer . Then we have
We then evaluate the integral in this expression using integration by parts, as before, let
Therefore, we have
Plugging this back into the expression we had from the induction hypothesis we obtain
Therefore, the formula holds for the case , and hence, for all integers , where
- Proof. We start with the definition of the integral logarithm,
and make the substitution , . This gives us . Therefore,
where is a constant
- (Note: In the comments, tom correctly suggests an easier way to do this is to use part (c) along with translation and expansion/contraction of the integral. The way I have here works also, but requires an inspired choice of substitution.) We start with the given integral,
and make the substitution
Therefore, using the given fact that , we have
- From part (d) we know that
Then, for the term we consider the integral
where . Similar to part (d) we make the substitution,
This gives us
Therefore, we have
Taking the derivative we then have