7
$\begingroup$

In the text "Functions of one Complex Variable" by Robert E.Greene and Steven G.Krantz is my understanding of the proof to $\text{Proposition (1.1)}$ correct ?

$\text{Proposition (1.1)}$

$$\int_{0}^{ \infty} \frac{dx}{x^{2} + 6x + 8} = \frac{1}{2} \log(2) \, \, $$

$\text{Proof}$

For the sake and using Complex-Analytic techniques the author considers the following integral.

$$\oint_{\eta_{R}} \frac{\log(z)}{z^{2} + 6z + 8}dz$$

As an exercise, it was left to us by the author that $\log(r)$ is a well defined holomorphic function. To address a trivial proof, one can define $\log(z)$ on $U \equiv \mathbb{C} \setminus \{x : x \geq 0 \}$ by $\{ \log(re^{i \theta}) = (\log(r)) + i \theta$ when $0 < \theta < 2 \pi, r > 0 \}$.

Before proceeding any further, take note that

$$u(r, \theta)=\log(r) \ \ \ \text{ and } \ \ \ v(r, \theta) =\theta.$$

Now it's easy to note that $$ \big( \partial_{r}u \big) =\frac{1}{r}= \frac{1}{r} \cdot 1 = \frac{1}{r} \cdot \left( \partial_{\theta} v\right)\ \ \ \ \ \text{and } \ \ \ \ \big( \partial_{r}u \big) = 0 = \frac{-1}{r}\cdot 0 = \frac{-1}{r} \cdot \big( \partial_{\theta} u \big) $$

So indeed, $log(z)$ is analytic.

But before proceeding further he defines $\eta_{R}$ such that,

$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \eta_{R}^{1}(t) = t + i/\sqrt{2R}, \, \, \, \, 1/\sqrt{2R} \leq t \leq R,$$

$$\eta_{R}^{2}(t)= Re^{it}, \, \, \, \, \theta_{0} \leq t \leq 2 \pi - \theta_{0},$$

where $\theta_{0} = \theta_{0}(R) = \sin^{-1}(1/(R \sqrt{2R}))$

$$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \eta_{R}^{3}(t) = R -t -i/\sqrt{2R}, \, \, \, \, 0 \leq t \leq R-1/\sqrt{2R},$$

$$\eta_{R}^{4}(t) = e^{it}/\sqrt{R}, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \pi/4 \leq t \leq 7 \pi /4.$$

$\text{Remark}$

For those who don't have the book on hand a picture of the Contour employed can be found in $\text{Figure (1.1)}$

$\text{Figure (1.1)}$

$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $Figure (1.1

The author now says that:

$(*)$ $$ \bigg| \lim_{R \rightarrow \infty}\oint_{\eta_{R}^{4}} \frac{\log(z)}{z^{2} + 6z + 8}dz\bigg| \rightarrow 0$$

, and that

$(**)$

$$ \bigg| \lim_{R \rightarrow \infty}\oint_{\eta_{R}^{2}} \frac{\log(z)}{z^{2} + 6z + 8}dz\bigg| \rightarrow 0.$$

A particular device that the author cites to justify convergence over $\eta_{R}^{2}$ and $\eta_{R}^{4}$ consider on faith

$$\bigg(\log \bigg( \frac{x + i \sqrt{2R}}{(x-i/\sqrt{2R}} \bigg) \bigg)\rightarrow -2 \pi i.$$

We will come back to this after dealing with the integrals over $\eta_{R}^{2}$ and $\eta_{R}^{4}$.

One should note that

$$ \sum_{\psi}^{4} \bigg(\oint_{\eta_{R}^{\psi}} \frac{\log(z)}{z^{2} + 6z + 8}dz \bigg).$$

Now over $\eta_{R}^{2}$ we have,

\begin{align*} \bigg| \oint_{\eta_{R}^{2}}\frac{\log(z)}{z^{2} + 6z + 8}dz\bigg|& = \bigg| \int_{-R}^{+Ri} \frac{\log(Re^{it})}{(Re^{it})^{2} + 6(Re^{it}) + 8} iRe^{i \theta} d \theta\bigg|\\&= \int_{-R}^{+Ri} \bigg|\frac{\log(Re^{it})}{(Re^{it})^{2} + 6(Re^{it}) + 8} \bigg| \big| iRe^{i \theta} d \theta \big|\\&= \int_{-R}^{+Ri} \frac{\bigg|\log(Re^{it}) \bigg|}{\bigg|(Re^{it})^{2} + 6(Re^{it}) + 8 \bigg|} \bigg|iRe^{i \theta} \bigg| \bigg|d \theta \bigg| \\& = \int_{\theta_{0}}^{2 \pi - \theta_{0}} \frac{\bigg|\log(Re^{it}) \bigg|}{\bigg|(Re^{it})^{2} + 6(Re^{it}) + 8 \bigg|} \bigg|iRe^{i \theta} \bigg| \bigg|d \theta \bigg| \end{align*}

Now we can establish a precise estimate over $\eta_{R}^{2}$,

$$\bigg| \oint_{\eta_{R}^{2}} \frac{\log(z)}{z^{2} + 6z + 8}dz\bigg| \leq \frac{\ln(R) + \pi }{R^{2} - 13} \pi r \, \, \text{as} \, \, \, R \rightarrow \infty $$

There by proving $(*)$.

A similar process can be done for $\eta_{R}^{4}$, hence:

\begin{align*} \bigg| \oint_{\eta_{R}^{4}} \frac{\log(e^{it}/\sqrt{R})}{(e^{it}/ \sqrt{R})^{2} + (e^{it} / \sqrt{R})(6) +8} dz\bigg|& = \oint_{\eta_{R}^{4}} \bigg| \frac{\log(e^{it}/\sqrt{R})}{(e^{it}/ \sqrt{R})^{2} + (e^{it} / \sqrt{R})(6) +8} iRe^{i \theta} d \theta\bigg|\\&= \oint_{\eta_{R}^{4}} \frac{\bigg|\log(e^{it}/\sqrt{R}) \bigg|}{\bigg|(e^{it}/ \sqrt{R})^{2} + (e^{it} / \sqrt{R})(6) +8 \bigg|} iRe^{i \theta} d \theta \\&= \oint_{\eta_{R}^{4}} \frac{\bigg| \log(e^{it})- \frac{1}{2}\log(R^{}) \bigg|}{ \bigg|\frac{e^{2it}}{\sqrt{2R}} + (e^{it} / \sqrt{R})(6) +8 \bigg|} \bigg| iRe^{i \theta} d \theta \bigg|\\& =\oint_{\frac{\pi}{4}}^{\frac{7 \pi}{4}} \frac{\bigg| it\log(e^{})- \frac{1}{2}\log(R^{}) \bigg|}{ \bigg|\frac{e^{2it}}{\sqrt{2R}} + (e^{it} / \sqrt{R})(6) +8 \bigg|} \bigg| iRe^{i \theta}\bigg| d \theta \bigg|. \end{align*}

Now finally a precise estimate for $\eta_{R}^{4}$

$$\bigg| \oint_{\eta_{R}^{4}} \frac{\log(e^{it}/\sqrt{R})}{(e^{it}/ \sqrt{R})^{2} + (e^{it} / \sqrt{R})(6) +8} dz\bigg| \leq \text{length}(\eta_{R}^{4}) \cdot \sup_{\eta_{R}^{4}}(g) \leq \pi R \frac{O(\log(R))}{\sqrt{R}} \, \text{as} \, R \rightarrow \infty $$

Thus proving $(**)$

After achieving our preliminary results now we have that,

$(***)$

\begin{align*} \bigg( \oint_{\eta_{R}^{1}} g(z) dz + \oint_{\eta_{R}^{3}} g(z) dz \bigg)& = \lim_{R \rightarrow \infty } \bigg( \oint_{\mu_{R}^{1} } \frac{\log(x+ \sqrt{2R})}{(\log(x+ \sqrt{2R}))^{2} + 6(\log(x+ \sqrt{2R})) + 8} - \oint_{\mu_{R}^{3} } \frac{\log(x - i/ \sqrt{2R})}{(\log(x -i /\sqrt{2R}))^{2} + 6(\log(x - i /\sqrt{2R})) + 8} \bigg) \\&= -2 \pi i \lim_{R \rightarrow \infty}\int_{0}^{R} \frac{dt}{t^{2} + 6t + 8} \\& \end{align*}

Using the Residue Theorem it's easy to observe that:

$(****)$

$$ \oint_{\eta_{R}} g(z) dz = 2 \pi i (\operatorname{Res_{g}}(-2) \cdot + Res_{g}(-4) \cdot 1) = - \pi i \log(2)$$

Finally putting $(****)$, $(***)$, $(**)$ and $(*)$ together yields that the,

$$\lim_{R \rightarrow \infty}\int_{0}^{R} \frac{dt}{t^{2} + 6t + 8} = \frac{1}{2}\log(2).$$

$\endgroup$
  • 1
    $\begingroup$ You must have mistyped the proposition. $\endgroup$ – Szeto Sep 3 '18 at 5:55
  • $\begingroup$ Yeah I'll have to edit that that $\endgroup$ – Zophikel Sep 3 '18 at 14:33
1
$\begingroup$

First of all, in your proposition, since $x$ is a dummy variable, it makes no sense to say '$\text{for all }x\in\mathbb R$'.

Besides, it is not quite clear how you obtained $(***)$.

Here, I provide a lemma, which can be applied to derive $(***)$, as well as explaining the motivation to introduce $\log(z)$ at the first place.

Lemma

$$\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln(z-s)dz=-2\pi i\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt$$

Proof:

Let $\hat{k}=i\frac{s}{|s|}$

\begin{align*} &~~~~\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln(z-s)dz \\ &=\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln|z-s|dz +i\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\arg(z-s)dz \\ &=\left(\int_{pe^{i\theta}}^{qe^{i\theta}}+\int^{pe^{i\theta}}_{qe^{i\theta}}\right)f(z)\ln|z-s|dz \\ &~~~~+i\lim_{\Delta\to0^+}\int^{qe^{i\theta}+\Delta\hat{k}}_{pe^{i\theta}+\Delta\hat{k}} f(z)\arg(z-s)dz +i\lim_{\Delta\to0^+}\int^{qe^{i\theta}-\Delta\hat{k}}_{pe^{i\theta}-\Delta\hat{k}} f(z)\arg(z-s)dz\\ \end{align*}

Obviously the first term is zero.

For the second term, by the substitution $z=ue^{i\theta}+\Delta\hat{k}$ \begin{align*} &~~~~ i\lim_{\Delta\to0^+}\int^{qe^{i\theta}+\Delta\hat{k}}_{pe^{i\theta}+\Delta\hat{k}} f(z)\arg(z-s)dz \\ &=i\lim_{\Delta\to0^+}\int^q_p f(ue^{i\theta}+\Delta\hat{k})\arg(ue^{i\theta}+\Delta\hat{k}-s)e^{i\theta}du \\ &=i\int^q_p f(ue^{i\theta})\theta e^{i\theta}du \\ &=i\theta\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt \end{align*}

From the second line to the third line, dominated convergence theorem is applied to exchange limit and integral, and $\lim_{\Delta\to 0^+}\arg(ue^{i\theta}+\Delta\hat{k}-s)=\theta$ is used.

For the third term, by the substitution $z=ue^{i\theta}-\Delta\hat{k}$ \begin{align*} &~~~~ i\lim_{\Delta\to0^+}\int_{qe^{i\theta}-\Delta\hat{k}}^{pe^{i\theta}-\Delta\hat{k}} f(z)\arg(z-s)dz \\ &=i\lim_{\Delta\to0^+}\int_q^p f(ue^{i\theta}-\Delta\hat{k})\arg(ue^{i\theta}-\Delta\hat{k}-s)e^{i\theta}du \\ &=-i\int^q_p f(ue^{i\theta})(2\pi+\theta) e^{i\theta}du \\ &=-i(2\pi+\theta)\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt \end{align*}

Similarly, $\lim_{\Delta\to 0^+}\arg(ue^{i\theta}-\Delta\hat{k}-s)=2\pi+\theta$ is used.

As a result, \begin{align*} &~~~~\lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln(z-s)dz \\ &=0+i\theta\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt-i(2\pi+\theta)\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt\\ &=-2\pi i\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt \end{align*}

Q.E.D.

$\endgroup$
  • $\begingroup$ For the proof of the Lemma you gave did you have to use the Dominated Convergence Theorem you could have just supposed our $f(z)$ is continuous and converged uniformly $\endgroup$ – Zophikel Sep 3 '18 at 17:20
  • $\begingroup$ But your Lemma, should be able to make things a bit more clear sorry for the crappy answer :'(. $\endgroup$ – Zophikel Sep 3 '18 at 17:46
  • 1
    $\begingroup$ @Zophikel I hope it is clear enough. $\endgroup$ – Szeto Sep 3 '18 at 21:59
  • 1
    $\begingroup$ @Zophikel If you know vectors, $\frac{s}{|s|}$ is the unit vector from origin pointing toward $s$. Muitiplying it by $i$ rotates it by 90 degrees counter clockwisely. Then, try visualize $\Delta\hat k$ and you would know why I defined it. $\endgroup$ – Szeto Sep 4 '18 at 7:57
  • 1
    $\begingroup$ @Zophikel I wrote the lemma. $\endgroup$ – Szeto Sep 4 '18 at 7:58
0
$\begingroup$

This post is to address, a complete derivation of $(***)$ which where my understanding of the proof breaks down.

$\text{Proof}$

Now recall the Lemma given by Szeto,

$\text{Szeto's Lemma}$

$(1)$ $$ \lim_{\Delta\to0^+}\left(\int_{\gamma_1}+\int_{\gamma_2}\right)f(z)\ln(z-s)dz=-2\pi i\int_{pe^{i\theta}}^{qe^{i\theta}}f(t)dt.$$

Now what we are aiming to derive, is that

\begin{align*} \bigg( \oint_{\eta_{R}^{1}} g(z) dz + \oint_{\eta_{R}^{3}} g(z) dz \bigg)& \rightarrow - 2 \pi i \int_{0}^{\infty} \frac{dt}{t^{2} + 6t + 8}. \tag{1.1}\\& \end{align*}

Applying $(1)$ to $\text{(1.1)}$ we note that

$$ \bigg( \lim_{R \rightarrow \infty}\oint_{\mu_{R}^{1} } \frac{\log(x+ \sqrt{2R})}{(\log(x+ \sqrt{2R}))^{2} + 6(\log(x+ \sqrt{2R})) + 8}\ln(z-s) \, dz \bigg) + \bigg( \lim_{R \rightarrow \infty}\oint_{\mu_{R}^{3} } \frac{\log(x - i/ \sqrt{2R})}{(\log(x -i /\sqrt{2R}))^{2} + 6(\log(x - i /\sqrt{2R})) + 8} \ln(z-s) dz \bigg).$$

Further analysis of $\eta_{R}^{1}$ revels that, \begin{align*} \lim_{R \rightarrow \infty} \bigg( \int_{1 / \sqrt{2R}}^{R} \frac{\log(x + \sqrt{2R})}{(\log(x + \sqrt{2R}^{2} + 6(\log(x + \sqrt{2R}) + 8}\ln(z-s)dz \bigg)&=& \\& \lim_{ R \rightarrow \infty} \bigg( \int_{0}^{R} \frac{\log(x + \sqrt{2R})}{(\log(x + \sqrt{2R}^{2} + 6(\log(x + \sqrt{2R}) + 8} \ln(z-s)dz\bigg) + \lim_{ R \rightarrow \infty} \bigg( \int_{0}^{\frac{1}{\sqrt{2R}}}\frac{\log(x+ \sqrt{2R})}{(\log(x - i/\sqrt{2R}))^{2} + 6(\log(x - i/ \sqrt{2R})) + 8}\ln(z-s)dz\bigg). \end{align*}

Similarly for $\eta_{R}^{3}$ we have that,

\begin{align*} \bigg(\lim_{R \rightarrow \infty} \int_{R - 1 / \sqrt{2R}}^{0} \frac{\log(x + \sqrt{2R})}{(\log(x + \sqrt{2R}^{2} + 6(\log(x + \sqrt{2R}) + 8}\ln(z-s)dz \bigg)&& \\& \end{align*}

Putting everything together it's easy to note that,

$$ \bigg( \oint_{\eta_{R}^{1}} g(z) dz + \oint_{\eta_{R}^{3}} g(z) dz \bigg) =2 \pi i \int_{0}^{\infty} \frac{dt}{t^{2} + 6t + 8}. $$

$\endgroup$
  • 1
    $\begingroup$ I get mental pictures where this is referred to as "Szeto's Lemma" in literature :D $\endgroup$ – Alvin Lepik Sep 10 '18 at 4:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.