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Prove some properties of the complex logarithm

Extend the logarithm function to all nonzero complex numbers by defining

Use this formula to prove the following properties of the complex logarithm.

1. , .
2. for an integer.
3. , where is an integer.
4. .

1. Proof. For these we use the definition and compute,

and

2. Proof. Let and . Then,

3. Proof. Again, we compute,

4. Proof. Finally, we have

Find the limit as x to 0 of the given expression

Evaluate the limit.

First, we write the expression using the definition of the exponential,

Now, considering the expression in the exponent and using the expansion of as (page 287 of Apostol) we have as ,

Therefore, we have

Find the limit as x goes to 0 of (x + e2x)1/x

Evaluate the limit.

From the definition of the exponential we have

So, first we use the expansion of as (page 287 of Apostol) to write

Therefore, as we have

Now, since as we can use the expansion (again, page 287) of as to write

Therefore, as we have

So, getting back to the expression we started with,

But, as in the previous exercise (Section 7.11, Exercise #23) we know . Hence,

Find the limit as x goes to 1 of log x / (x2 + x – 2)

Evaluate the limit.

From this exercise (Section 7.11, Exercise #4) we know

Therefore,

Find the limit as x goes to 0 of (ax – 1) / (bx – 1)

Evaluate the limit for .

First we write and . Then we use the expansion of (p. 287), to obtain expansions for and ,

Therefore, we have

Find an expression for log x as a quadratic polynomial in (x-1)

Find constants , and such that

From the Taylor formula for we have

Replacing by we then have

Therefore,

Find an inverse for the function log |x|

Consider the function for . Prove that this function has an inverse, determine the domain of this inverse, and find a formula to compute the inverse .

Proof. From the discussion on page 146 of Apostol we know that a function which is continuous and strictly monotonic on an interval has an inverse on . The function is continuous and strictly monotonic on the negative real axis; therefore, it has an inverse. We know it is continuous since the log function is continuous on the positive real axis, and for all , in particular, for all . Furthermore, we know it is strictly monotonic since

Therefore, has an inverse for all . The domain of this inverse is the range of which is all of

To find a formula for the inverse we set

Therefore, valid for all .

A sketch for the graph of is given by

Prove some properties of the integral logarithm, Li (x)

The integral logarithm is defined for by

Prove the following properties of .

1. .
2. where is a constant depending on . Find the value of for each .
3. Prove there exists a constant such that

and find the value of this constant.

4. Let . Find an expression for

in terms of .

5. Define a function for by

Prove that

1. Proof. We derive this by integrating by parts. Let

Then we have

2. 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

3. Proof. We start with the definition of the integral logarithm,

and make the substitution , . This gives us . Therefore,

where is a constant

4. (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

5. 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

Prove an inequality of exponentials

For all and for any constants such that prove that

Proof. We want to consider the function

If we can show this function is decreasing on the positive real axis then we establish the inequality since this would mean that if then

(So, the trick here is to think of this as a function of the exponent. The and are some positive fixed constants.) To take the derivative of we use logarithmic differentiation,

Multiplying both sides by we then obtain

Now we can conclude that for all since the first term in the product

Since (any real power of a positive number is still positive) and . For the second term we have

since and are positive, but both logarithms are negative. We know these logarithms are negative since

implies

Hence, for all . This means is a decreasing function. Therefore, if then we have

Prove inequalities of the log of 1 + 1/x

For prove that

Before the proof, a picture for this one is probably useful. In the graph below we are going to get the inequality on the right by comparing the area under the graph of from to (blue curve) and the area of the rectangle under the blue dashed line from to . We’ll show that the area of the rectangle is and the area under the curve is .

Proof. First, we note that

Furthermore,

Therefore we have

(In the picture this is the area under the curve between and .)
Since the function is strictly decreasing on the positive real axis (since it’s derivative is is negative everywhere) we know that for any fixed we have for every . Therefore, by the monotone property of the integral,

(Note the inside the integral is a constant here since we have chosen some fixed . For any fixed the point is the point on the curve at the left end of the interval. So, this is saying the integral of the curve from to is less than the integral of the rectangle of height from to .) This gives us the inequality on the right that we wanted,

Now, to prove the inequality on the left we know that if and only if . Thus, the given inequality holds for all if and only if it holds for all . Therefore,

if and only if

Now, consider

for all . Therefore, is increasing on the positive real axis. Since

we have that for all . Hence, we obtain the inequality on the left