# About the proof that $\int_0^\infty\frac{dx}{x^2+6x+8} =\frac12\log2$ via residue formula

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 $\text{Figure (1.1)}$

$\text{Figure (1.1)}$

$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,$ 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,

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

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

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.

• 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 – Zophikel Sep 3 '18 at 17:20
• But your Lemma, should be able to make things a bit more clear sorry for the crappy answer :'(. – Zophikel Sep 3 '18 at 17:46
• @Zophikel I hope it is clear enough. – Szeto Sep 3 '18 at 21:59
• @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. – Szeto Sep 4 '18 at 7:57
• @Zophikel I wrote the lemma. – Szeto Sep 4 '18 at 7:58

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,

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

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