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Prove some properties of the complex logarithm

by RoRi

Extend the logarithm function to all nonzero complex numbers z by defining

    \[ \operatorname{Log} z = \log |z| + i \arg (z). \]

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

  1. \operatorname{Log} (-1) = \pi i, \displaystyle{\operatorname{Log} (i) = \frac{\pi i}{2}}.
  2. \operatorname{Log} (z_1 z_2) = \operatorname{Log} z_1 + \operatorname{Log} z_2 + 2n \pi i for n an integer.
  3. \displaystyle{\operatorname{Log} \left( \frac{z_1}{z_2} \right) = \operatorname{Log} z_1 - \operatorname{Log} z_2 + 2n \pi i}, where n is an integer.
  4. e^{\operatorname{Log} z} = z.
  1. Proof. For these we use the definition and compute,

        \[ \operatorname{Log} (-1) = \log |-1| + i \arg (-1) = 0 + \pi i = \pi i, \]

    and

        \[ \operatorname{Log} (i) = \log |i| + i \arg(i) = 0 + \frac{\pi i}{2} = \frac{1}{2} \pi i. \qquad \blacksquare\]

  2. Proof. Let z_1 = r_1 e^{i \theta_1} and z_2 = r_2 e^{i \theta_2}. Then,

        \begin{align*} \operatorname{Log} (z_1 z_2) &= \log |z_1 z_2| + i \arg(z_1 z_2) \\ &= \log (|z_1||z_2|) + i \arg (\theta_1 + \theta_2 + 2n \pi) \\ &= \log |z_1| + i \theta_1 + \log |z_2| + i \theta_2 + 2n i \pi \\ &= \operatorname{Log} z_1 + \operatorname{Log} z_2 + 2n i \pi. \qquad \blacksquare \end{align*}

  3. Proof. Again, we compute,

        \begin{align*} \operatorname{Log} \left( \frac{z_1}{z_2} \right) &= \log \left| \frac{z_1}{z_2} \right| + i \arg \left( \frac{z_1}{z_2} \right) \\[9pt] &= \log \frac{|z_1|}{|z_2|} + i (\theta_1 - \theta_2 + 2n i \pi ) \\[9pt] &= \log |z_1| + i \theta_1 - \log|z_2| - i \theta_2 + 2n i \pi \\[9pt] &= \operatorname{Log} z_1 - \operatorname{Log} z_2 + 2n i \pi. \qquad \blacksquare \end{align*}

  4. Proof. Finally, we have

        \[ e^{\operatorname{Log} z} = e^{\log |z| + i \arg(z)} = e^{\log |z|}e^{i \arg (z)} = e^{\log r} e^{i \theta} = re^{i \theta} = z. \qquad \blacksquare\]

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

by RoRi

Evaluate the limit.

    \[ \lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}}. \]

From the definition of the exponential we have

    \[ \left( x + e^{2x} \right)^{\frac{1}{x}} = e^{ \frac{1}{x} \log \left( x + e^{2x} \right)}. \]

So, first we use the expansion of e^x as x \to 0 (page 287 of Apostol) to write

    \[ e^{2x} = 1 + 2x + o(x) \qquad \text{as} \quad x \to 0. \]

Therefore, as x \to 0 we have

    \[ \log (x + e^{2x}) = \log (x + 1 + 2x + o(x)) = \log (1 + 3x + o(x)). \]

Now, since 3x + o(x) \to 0 as x \to 0 we can use the expansion (again, page 287) of \log (1+x) as x \to 0 to write

    \[ \log (1 + 3x + o(x)) = 3x+o(x) + o(3x+o(x)) = 3x + o(x) \qquad \text{as} \quad x \to 0. \]

Therefore, as x \to 0 we have

    \[ \frac{1}{x} \log (x+e^{2x}) = \frac{1}{x}(3x + o(x)) = 3 + \frac{o(x)}{x}. \]

So, getting back to the expression we started with,

    \[ \lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}} = \lim_{x \to 0} e^{3 + \frac{o(x)}{x}} = e^3 \lim_{x \to 0} e^{\frac{o(x)}{x}}. \]

But, as in the previous exercise (Section 7.11, Exercise #23) we know \lim_{x \to 0} e^{\frac{o(x)}{x}} = 1. Hence,

    \[ \lim_{x \to 0} \left( x + e^{2x} \right)^{\frac{1}{x}} = e^3. \]

Find an inverse for the function log |x|

by RoRi

Consider the function f(x) = \log |x| for x < 0. Prove that this function has an inverse, determine the domain of this inverse, and find a formula to compute the inverse g(y).

Proof. From the discussion on page 146 of Apostol we know that a function which is continuous and strictly monotonic on an interval [a,b] has an inverse on [a,b]. The function f(x) = \log |x| 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 |x| > 0 for all x, in particular, for all x < 0. Furthermore, we know it is strictly monotonic since

    \[ f'(x) = \frac{1}{x} < 0 \qquad \text{for all } x < 0. \]

Therefore, f(x) = \log |x| has an inverse for all x < 0. The domain of this inverse is the range of \log |x| which is all of \mathbb{R}. \qquad \blacksquare

To find a formula for the inverse we set

    \[ y = \log |x| \quad \implies \quad e^y = |x| \quad \implies \quad x = -e^y. \]

Therefore, g(y) = -e^y valid for all y \in \mathbb{R}.

A sketch for the graph of g is given by

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

by RoRi

The integral logarithm \operatorname{Li}(x) is defined for x \geq 2 by

    \[ \operatorname{Li}(x) = \int_2^x \frac{dt}{\log t}. \]

Prove the following properties of \operatorname{Li}(x).

  1. \displaystyle{ \operatorname{Li}(x) = \frac{x}{\log x} + \int_2^x \left(\frac{dt}{\log^2 t}\right) - \frac{2}{\log 2}}.
  2. \displaystyle{ \operatorname{Li}(x) = \frac{x}{\log x} + \sum_{k=1}^{n-1} \left(\frac{k!x}{\log^{k+1} x}\right) + n! \int_2^x \left( \frac{dt}{\log^{n+1} t} \right) + C_n} where C_n is a constant depending on n. Find the value of C_n for each n.
  3. Prove there exists a constant b such that

        \[ \operatorname{Li}(x) = \int_b^{\log x} \frac{e^t}{t} \, dt \]

    and find the value of this constant.

  4. Let c = 1 + \frac{1}{2} \log 2. Find an expression for

        \[ \int_c^x \frac{e^{2t}}{t-1} \, dt \]

    in terms of \operatorname{Li}(x).

  5. Define a function for x > 3 by

        \[ f(x) = e^4 \operatorname{Li}(e^{2x-4}) - e^2 \operatorname{Li}(e^{2x-2}). \]

    Prove that

        \[ f'(x) = \frac{e^{2x}}{x^2-3x+2}. \]

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

        \begin{align*} u &= \frac{1}{\log t} & du &= \frac{-1}{t \log^2 t} \, dt \\ dv &= dt & v &= t. \end{align*}

    Then we have

        \begin{align*} \operatorname{Li}(x) &= \int_2^x \frac{dt}{\log t} \\[9pt] &= \frac{t}{\log t} \Bigr \rvert_2^x + \int_2^x \frac{1}{\log^2 t} \, dt \\[9pt] &= \frac{x}{\log x} + \int_2^x \frac{dt}{\log^2 t} - \frac{2}{\log 2}. \qquad \blacksquare \end{align*}

  2. Proof. The proof is by induction. Starting with part (a) we have

        \[ \operatorname{Li}(x) = \frac{x}{\log x} + \int_2^x \frac{dt}{\log^2 t} - \frac{2}{\log 2}. \]

    To evaluate the integral in this expression we integrate by parts with

        \begin{align*} u &= \frac{1}{\log^2 t} & du &= \frac{-2}{t \log^3 t} \, dt \\ dv &= dt & v &= t. \end{align*}

    This gives us

        \begin{align*} \int_2^x \frac{dt}{\log^2 t} &= \frac{t}{\log^2 t} \Bigr \rvert_2^x + 2 \int_2^x \frac{dt}{\log^3 t} \\[9pt] &= \frac{x}{\log^2 x} + 2 \int_2^x \frac{dt}{\log^3 t} - \frac{2}{\log^2 2}. \end{align*}

    Therefore we have

        \begin{align*} \operatorname{Li}(x) &= \frac{x}{\log x} + \int_2^x \frac{dt}{\log^2 t} - \frac{2}{\log 2} \\[9pt] &= \frac{x}{\log x} + \frac{x}{\log^2 x} + 2 \int_2^x \frac{dt}{\log^3 t} - \left( \frac{2}{\log 2} + \frac{2}{\log^2 2} \right) \\[9pt] &= \frac{x}{\log x} + \sum_{k=1}^{n-1} \frac{n! x}{\log^{k+1} x} + n! \int_2^x \frac{dt}{\log^{n+1} t} + C_n. \end{align*}

    where C_n = -2 \sum_{k=1}^n \frac{(k-1)!}{\log^k 2}. This is the case n = 2. Now, assume the formula hold for some integer m \geq 2. Then we have

        \[ \operatorname{Li}(x) &= \frac{x}{\log x} + \sum_{k=1}^{m-1} \frac{k! x}{\log^{k+1} x} + m! \int_2^x \frac{dt}{\log^{m+1} t} - 2 \sum_{k=1}^m \frac{(k-1)!}{\log^k 2}.\]

    We then evaluate the integral in this expression using integration by parts, as before, let

        \begin{align*} u &= \frac{1}{\log^{m+1} t} & du &= \frac{-(m+1)}{t \log^{m+2} t} \\ dv &= dt & v &= t. \end{align*}

    Therefore, we have

        \begin{align*} \int_2^x \frac{dt}{\log^{m+1} t} &= \frac{t}{\log^{m+1} t} \Bigr \rvert_2^x + (m+1) \int_2^x \frac{dt}{\log^{m+2} t} \\[9pt] &= \frac{x}{\log^{m+1} x} + (m+1) \int_2^x \frac{dt}{\log^{m+2} t} - \frac{2}{\log^{m+1} 2}. \end{align*}

    Plugging this back into the expression we had from the induction hypothesis we obtain

        \begin{align*} \operatorname{Li}(x) &= \frac{x}{\log x} + \sum_{k=1}^{m-1} \frac{k! x}{\log^{k+1} x} + m! \int_2^x \frac{dt}{\log^{m+1} t} - 2 \sum_{k=1}^m \frac{(k-1)!}{\log^k 2} \\[10pt] &= \frac{x}{\log x} + \sum_{k=1}^{m-1} \frac{k! x}{\log^{k+1} x} + m! \left( \frac{x}{\log^{m+1} x} + (m+1) \int_2^x \frac{dt}{\log^{m+2} t} - \frac{2}{\log^{m+1} 2} \right) \\ & \qquad - 2 \sum_{k=1}^m \frac{(k-1)!}{\log^k 2} \\[10pt] &= \frac{x}{\log x} + \left( \sum_{k=1}^{m-1} \frac{k! x}{\log^{k+1} x} + \frac{m! x}{\log^{m+1}} \right) + (m+1)! \int_2^x \frac{dt}{\log^{m+2} t} \\ & \qquad - 2 \left( \frac{m!}{\log^{m+1} 2} + \sum_{k=1}^m \frac{(k-1)!}{\log^k 2} \right) \\[10pt] &= \frac{x}{\log x} + \sum_{k=1}^m \frac{k! x}{\log^{k+1} x} + (m+1)! \int_2^x \frac{dt}{\log^{m+2} t} - 2 \sum_{k=1}^{m+1} \frac{(k-1)!}{\log^k 2} \\[10pt] &= \frac{x}{\log x} + \sum_{k=1}^m \frac{k! x}{\log^{k+1} x} + (m+1)! \int_2^x \frac{dt}{\log^{m+2} t} + C_{m+1}. \end{align*}

    Therefore, the formula holds for the case m+1, and hence, for all integers n \geq 2, where

        \[ C_n = -2 \sum_{k=1}^n \frac{(k-1)!}{\log^k x}. \qquad \blacksquare \]

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

        \[ \operatorname{Li}(x) = \int_2^x \frac{dt}{\log t} \]

    and make the substitution s = \log t, ds = \frac{dt}{t}. This gives us dt = t \, ds = e^s \, ds. Therefore,

        \begin{align*} \operatorname{Li}(x) &= \int_2^x \frac{dt}{\log t} \\[9pt] &= \int_{\log 2}^{\log x} \frac{e^s}{s} \, ds \\[9pt] &= \int_b^{\log x} \frac{e^t}{t} \, dt \end{align*}

    where b = \log 2 is a constant. \qquad \blacksquare

  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,

        \[ \int_c^x \frac{e^{2t}}{t-1} \, dt \]

    and make the substitution

        \begin{align*} s &= e^{2t-2} & \implies && t &= 1 + \frac{1}{2} \log s \\ ds &= 2e^{2t-2}\, dt & \implies && dt &= \frac{1}{2} e^{2-2t} \, ds = \frac{ds}{2s}. \end{align*}

    Therefore, using the given fact that c = 1 + \frac{1}{2} \log 2, we have

        \begin{align*} \int_c^x \frac{e^{2t}}{t-1} \, dt &= \int \limits_{e^{2(1+\frac{1}{2}\log 2) - 2}}^{e^{2x-2}} \frac{e^2 s}{\frac{1}{2} \log s} \frac{ds}{2s} \\[10pt] &= e^2 \int_2^{e^{2x-2}} \frac{ds}{\log s} \\[10pt] &= e^2 \operatorname{Li}(e^{2x-2}). \end{align*}

  5. From part (d) we know that

        \[ e^2 \operatorname{Li}(e^{2x-2}) = \int_c^x \frac{e^{2t}}{(t-1)} \, dt. \]

    Then, for the term e^4 \operatorname{Li}(e^{2x-4}) we consider the integral

        \[ \int_d^x \frac{e^{2t}}{t-2} \, dt, \]

    where d = 2 + \frac{1}{2} \log 2. Similar to part (d) we make the substitution,

        \begin{align*} s &= e^{2t-4} & \implies && t &= 2 + \frac{1}{2} \log s \\ ds &= 2e^{2t-4} \, dt & \implies && dt &= \frac{1}{2}e^{4-2t} \, ds = \frac{ds}{2s}. \end{align*}

    This gives us

        \begin{align*} \int_d^x \frac{e^{2t}}{t-2} &= \int \limits_{e^{2(2+\frac{1}{2} \log 2) - 2}}^{e^{2x-4}} \frac{e^4 s}{\frac{1}{2} \log s} \frac{ds}{2s} \\[9pt] &= e^4 \int_2^{e^{2x-4}} \frac{ds}{\log s} \\[9pt] &= e^4 \operatorname{Li}(e^{2x-4}). \end{align*}

    Therefore, we have

        \[ f(x) = e^4 \operatorname{Li}(e^{2x-4}) - e^2 \operatorname{Li}(e^{2x-2}) = \int_d^x \frac{e^{2t}}{t-2} \,dt - \int_c^x \frac{e^{2t}}{t-1} \, dt. \]

    Taking the derivative we then have

        \begin{align*} f'(t) &= \frac{e^{2x}}{x-2} - \frac{e^{2x}}{x-1} \\ &= \frac{e^{2x}(x-1) - e^{2x}(x-2)}{(x-2)(x-1)} \\ &= \frac{e^{2x}}{x^2 - 3x + 2}. \qquad \blacksquare \end{align*}

Prove an inequality of exponentials

by RoRi

For all x, y > 0 and for any constants a,b such that 0 < a < b prove that

    \[ \big( x^b + y^b \big)^{\frac{1}{b}} < \big( x^a + y^a \big)^{\frac{1}{a}}. \]

Proof. We want to consider the function

    \[ f(t) = (x^t + y^t)^{\frac{1}{t}}. \]

If we can show this function is decreasing on the positive real axis then we establish the inequality since this would mean that if 0 < a < b then

    \[ f(b) < f(a) \quad \implies \quad (x^b+y^b)^{\frac{1}{b}} < (x^a+y^a)^{\frac{1}{a}}.\]

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

    \begin{align*} &&f(t) &= (x^t+y^t)^{\frac{1}{t}} \\[10pt] \implies &&\log f(t) &= \frac{1}{t} \log (x^t+y^t) \\[10pt] \implies &&\frac{d}{dt} (\log f(t)) &= \frac{d}{dt} \left( \frac{1}{t} \log (x^t+y^t) \right) \\[10pt] \implies &&\frac{f'(t)}{f(t)} &= -\frac{1}{t^2} \log (x^t+y^t) + \frac{1}{t} \left( \frac{1}{x^t+y^t} \right) \left( x^t \log x + y^t \log y \right) \\[10pt] \implies &&\frac{f'(t)}{f(t)} &= \frac{-\log(x^t+y^t)}{t^2} + \frac{x^t \log x + y^t \log y}{t (x^t+y^t)}. \end{align*}

Multiplying both sides by f(t) we then obtain

    \begin{align*} f'(t) &= \left( \frac{-\log(x^t+y^t)}{t^2} + \frac{x^t \log x + y^t \log y}{t(x^t+y^t)} \right) (x^t+y^t)^{\frac{1}{t}} \\[10pt] &= (x^t + y^t)^{\frac{1}{t}} \left( \frac{t(x^t \log x + y^t \log y) - (x^t+y^t)\log(x^t+y^t)}{t^2 (x^t+y^t)} \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t} - 1}}{t^2} \right) \left( x^t t \log x + y^t t \log y - (x^t + y^t)\log(x^t+y^t) \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t}-1}}{t^2} \right) \left( x^t \log x^t + y^t \log y^t - x^t \log (x^t+y^t) - y^t \log (x^t+y^t) \right) \\[10pt] &= \left( \frac{(x^t+y^t)^{\frac{1}{t} - 1}}{t^2} \right) \left( x^t \log \left(\frac{x^t}{x^t+y^t}\right) + y^t \log \left( \frac{y^t}{x^t+y^t} \right) \right). \end{align*}

Now we can conclude that f'(t) < 0 for all t > 0 since the first term in the product

    \[ \frac{(x^t+y^t)^{\frac{1}{t}-1}}{t^2} > 0. \]

Since x, y > 0 (any real power of a positive number is still positive) and t^2 > 0. For the second term we have

    \[ x^t \log \left( \frac{x^t}{x^t+y^t} \right) + y^t \log \left( \frac{y^t}{x^t+y^t} \right) < 0 \]

since x^t and y^t are positive, but both logarithms are negative. We know these logarithms are negative since

    \[ \frac{x^t}{x^t+y^t} < 1 \quad \text{and} \quad \frac{y^t}{x^t+y^t} < 1 \]

implies

    \[ \log \left( \frac{x^t}{x^t+y^t} \right) < 0 \quad \text{and} \quad \log \left( \frac{y^t}{x^t+y^t} \right) < 0. \]

Hence, f'(t) < 0 for all t > 0. This means f(t) is a decreasing function. Therefore, if 0 < a < b then we have

    \[ f(b) < f(a) \quad \implies \quad \big( x^b + y^b \big)^{\frac{1}{b}} < \big( x^a + y^a \big)^{\frac{1}{a}}. \qquad \blacksquare \]

Prove inequalities of the log of 1 + 1/x

by RoRi

For x > 0 prove that

    \[ \frac{1}{x + \frac{1}{2}} < \log \left( 1 + \frac{1}{x} \right) < \frac{1}{x}. \]

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 \frac{1}{x} from x to x+1 (blue curve) and the area of the rectangle under the blue dashed line from x to x+1. We’ll show that the area of the rectangle is 1/x and the area under the curve is \log \left( 1 + \frac{1}{x} \right).

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Proof. First, we note that

    \[ \log \left( 1 + \frac{1}{x} \right) = \log \left( \frac{x+1}{x} \right) = \log (x+1) - \log x. \]

Furthermore,

    \[ \int_x^{x+1} \frac{1}{t} \, dt = \log (x+1) - \log x. \]

Therefore we have

    \[ \log \left( 1 + \frac{1}{x} \right) = \int_x^{x+1} \frac{1}{t} \, dt. \]

(In the picture this is the area under the curve between x and x+1.)
Since the function \frac{1}{t} is strictly decreasing on the positive real axis (since it’s derivative is -\frac{1}{t^2} is negative everywhere) we know that for any fixed x> 0 we have \frac{1}{x} > \frac{1}{t} for every t \in (x, x+1). Therefore, by the monotone property of the integral,

    \[ \int_x^{x+1} \frac{1}{t} \, dt < \int_x^{x+1} \frac{1}{x} \, dt = \frac{1}{x} \int_x^{x+1} dt = \frac{1}{x}. \]

(Note the \frac{1}{x} inside the integral is a constant here since we have chosen some fixed x>0. For any fixed x the point \frac{1}{x} is the point on the curve f(x) at the left end of the interval. So, this is saying the integral of the curve \frac{1}{t} from x to x+1 is less than the integral of the rectangle of height \frac{1}{x} from x to x+1.) This gives us the inequality on the right that we wanted,

    \[ \log \left( 1 + \frac{1}{x} \right) < \frac{1}{x}. \]

Now, to prove the inequality on the left we know that x > 0 if and only if \frac{1}{x} > 0. Thus, the given inequality holds for all x>0 if and only if it holds for all \frac{1}{x} > 0. Therefore,

    \[ \frac{1}{x+\frac{1}{2}} < \log \left( 1+ \frac{1}{x} \right) \]

if and only if

    \[ \frac{1}{\frac{1}{x} + \frac{1}{2}} < \log \left( 1 + \frac{1}{1/x} \right) \implies \frac{x}{1+\frac{1}{2} x} < \log (1+x). \]

Now, consider

    \begin{align*} f(x) &= \log(1+x) - \frac{x}{1+\frac{1}{2}x} \\[9pt] f'(x) &= \frac{1}{1+x} - \frac{1+\frac{1}{2}x - \frac{1}{2}x}{(1+\frac{1}{2}x)^2} \\[9pt] &= \frac{1}{1+x} - \frac{1}{1+x+\frac{1}{4}x^2} \\[9pt] &> 0 \end{align*}

for all x > 0. Therefore, f(x) is increasing on the positive real axis. Since

    \[ f(0) = \log (1+0) - \frac{0}{1+\frac{1}{2}\cdot0} = 0, \]

we have that f(x) > 0 for all x > 0. Hence, we obtain the inequality on the left

    \[ \frac{1}{x+\frac{1}{2}} < \log \left( 1+\frac{1}{x} \right). \qquad \blacksquare\]