# Integral along a contour is $0$, how?

I recently had an extremely failed attempt at asking the same question, so I am posting the same question more or less to hope that someone can give me feedback.

Consider the integral:

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

$\hskip1in$

Image taken and modified from: Complex Analysis Solution (Please Read for background information).

$R$ is the big radius, $\delta$ is the small radius.

Actually, lets consider $u$ the small radius. Let $\delta = u$

Ultimately the goal is to let $u \to 0$

We can parametrize,

$$z = ue^{i\theta}$$

$$\int_{\delta} f(z)dz = (-)\cdot\int_{0}^{\pi} \frac{(i\theta + \log(u))^2\cdot (uie^{i\theta})}{(ue^{i\theta})^2 + 1} d\theta$$

$$\left | \int_{0}^{\pi} \frac{(i\theta + \log(u))^2\cdot (uie^{i\theta})}{(ue^{i\theta})^2 + 1} d\theta \right | \le \int_{0}^{\pi} \frac{|(i\theta + \log(u))|^2\cdot(u)}{|(ue^{i\theta})^2 + 1 |} d\theta$$

$$|(ue^{i\theta})^2 + 1 | < u^2 + 1$$

$$\frac{1}{u^2 + 1} < \frac{1}{|(ue^{i\theta})^2 + 1 |}$$

Since the maximum value of $\theta$ is $\theta = \pi$

$$|(i\theta + \log(u))| = \sqrt{\log^2(u) - \theta^2} \le \sqrt{\log^2(u) + \pi^2}$$

So:

$$|(i\theta + \log(u))|^2 \le \log^2(u) + \pi^2$$

Then:

$$|(i\theta + \log(u))|^2 \le \log^2(u) + \pi^2$$

For values $u$ near $0$.

$$(u)|(i\theta + \log(u))|^2 \le (\log^2(u) + \pi^2)u \le (\pi^2)u + 5\pi^2$$

Therefore,

$$\frac{|\log(z)|}{|z^2 + 1|} \le \frac{(\pi^2)u + 5\pi^2}{u^2 + 1}$$

Then we take the limit as $u \to 0$ which makes the RHS of the inequality 0.

hence the LHS upperbound is $0$.

So is the contour integral around the small semi circle $\delta = 0$?

How do I do this?

Thanks

• @AaronMaroja, I wanted to explore other ways – Amad27 Dec 24 '14 at 22:14
• But log is discontinuous at $z=0$ the residue theorem wont apply – Amad27 Dec 24 '14 at 22:15
• Oh, that's right. – Aaron Maroja Dec 24 '14 at 22:16
• So I still do need help with this. – Amad27 Dec 24 '14 at 22:16
• – Aaron Maroja Dec 24 '14 at 22:17


According to your picture you have chosen a $\ds{\ln\pars{z}}$-branch as follows: $$\ln\pars{z}=\ln\pars{\verts{z}} + \,{\rm Arg}\pars{z}\ic\,\quad z \not= 0\,,\quad -\,{\pi \over 2} < \,{\rm Arg}\pars{z} < {3\pi \over 2}$$

\begin{align}&\int_{C}{\ln^{2}\pars{z} \over z^{2} + 1}\,\dd x =2\pi\ic\,{\bracks{\ln\pars{1} + \pi\ic/2}^{2} \over 2\ic}=-\,{\pi^{3} \over 4} \end{align}

Moreover,

\begin{align} -\,{\pi^{3} \over 4}&=\lim_{\epsilon\ \to\ 0^{+}}\left[\int_{-\infty}^{-\epsilon}{\bracks{\ln\pars{-x} + \pi\ic}^{2} \over x^{2} + 1}\,\dd x\right. \\[5mm]&\left.\phantom{\lim_{\epsilon\ \to\ 0^{+}}\left[A\right.} +\ \underbrace{% \int_{\pi}^{0}{\bracks{\ln\pars{\epsilon} + \ic\theta}^{2}\over \epsilon^{2}\expo{2\ic\theta} + 1}\,\epsilon\expo{\ic\theta}\ic\,\dd\theta} _{\ds{\dsc{\to\ 0}\ \mbox{when}\ \dsc{\epsilon \to 0^{+}}. \mbox{See below.}}} +\int_{\epsilon}^{\infty}{\bracks{\ln\pars{x} + 0\ic}^{2} \over x^{2} + 1}\,\dd x\right] \\[1cm]&=\int_{0}^{\infty} {\ln^{2}\pars{x} + \bracks{\ln\pars{x} + \pi\ic}^{2} \over x^{2} + 1}\,\dd x \\[5mm]&=2\int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + 1}\,\dd x +2\pi\ic\ \overbrace{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1}\,\dd x} ^{\ds{=\ \dsc{0}}}\ -\ \pi^{2}\ \overbrace{\int_{0}^{\infty}{\dd x \over x^{2} + 1}} ^{\ds{=\ \dsc{\pi \over 2}}} \\[5mm]&=2\int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + 1}\,\dd x -{\pi^{3} \over 2}\quad\imp\quad \color{#66f}{\large\int_{0}^{\infty}{\ln^{2}\pars{x} \over x^{2} + 1}\,\dd x} =\half\pars{{\pi^{3} \over 2} - {\pi^{3} \over 4}} =\color{#66f}{\large{\pi^{3} \over 8}} \end{align}

The integral in the small semicircle satisfies $\ds{\pars{~\mbox{with}\ 0 < \epsilon < 1~}}$:

\begin{align} 0&<\verts{\int_{\pi}^{0}{\bracks{\ln\pars{\epsilon} + \ic\theta}^{2}\over \epsilon^{2}\expo{2\ic\theta} + 1}\,\epsilon\expo{\ic\theta}\ic\,\dd\theta}< \epsilon\int_{0}^{\pi}{\bracks{\ln\pars{\epsilon} + \ic\theta}^{2}\over \verts{\epsilon^{2}\expo{2\ic\theta} + 1}}\,\dd\theta \\[5mm]&<{\epsilon \over 1 - \epsilon^{2}} \bracks{\pi\ln^{2}\pars{\epsilon} + \pi^{2}\verts{\ln\pars{\epsilon}} + {\pi^{3} \over 3}}\quad\to\quad 0\quad\mbox{when}\quad\epsilon\to 0^{+}. \end{align}

• @MathN00b I'll wait for a OP comment to delete it. Thanks. – Felix Marin Dec 24 '14 at 23:17
• @MathN00b I changed the answer. I didn't notice the $^{2}$ in the $\log$. Thanks. – Felix Marin Dec 25 '14 at 0:21
• Thanks for answering, it is very helpful, do you mind if I ask a question? How is: $$\epsilon\cdot\int_{0}^{\pi} \frac{(\ln(\epsilon) + i\theta)^2 d\theta}{|(\epsilon^2 \cdot e^{2i\theta} + 1)|} < \frac{\epsilon}{1-\epsilon^2}\cdot[\pi\cdot\ln^2(\epsilon) + \pi^2|\ln(\epsilon)|] + \frac{\pi^3}{3}]$$ Are you using the ML-Inequality? How did you derive the RHS? It is generally: INT < M*(l) $l$ is arc length of contour, and $M$ is the max: en.wikipedia.org/wiki/Estimation_lemma ? – Amad27 Dec 25 '14 at 8:55
• Wait: Also, how did I choose that branch. Thats weird. The angle I have is ranginging from $0 \le \theta \le \pi$ How is it from $-\pi/2 \le \theta \le 3\pi/2$???? Also: What is the form of the log in the beginning of your answer? – Amad27 Dec 25 '14 at 9:05
• If you please answer the above, I came up with a solution $$\left |\int \right| \le ML(\Gamma)$$ The M-L Inequality. Then I figured $M = \max|f(z)|$ $|\log^2(z)| = |\log^2(\epsilon) - \theta^2| \le \log^2(\epsilon)$ $$\frac{1}{|z^2 + 1|} \le \frac{1}{|\epsilon^2 - 1|}$$ $$\frac{\log^2(\epsilon)}{|\epsilon^2 - 1|} \le M$$ $$L = (1/2)(2\pi\epsilon) = \pi\epsilon$$ $$\left |\oint \right| \le \frac{\pi\epsilon \cdot \log^2(\epsilon)}{|\epsilon^2 - 1|}$$ As $\epsilon \to 0$ Limit $\to 0$ But is this correct? I dont know if thats the actual Max (M)?? Can it be $\le Max$?? – Amad27 Dec 25 '14 at 10:37

We want to show that $$\int_{0}^{\infty}\frac{\ln^2 x}{x^2+1} = \frac{\pi^3}{8} \tag{1}$$

Breaking through, let's take $f(z) = \frac{\ln z}{z^2+1}$ with branch $\Big(|z|> 0 -\frac{\pi}{2} < \arg z < \frac{3\pi}{2}\Big)$ of the multiple-valued function $\ln z / (z^2+1)$. As long as we are isolating $z = i$ we're going to take $\delta < 1 < R$. According to Cauchy's Residue Theorem,

$$\int_{L_1} f(z)dz + \int_{C_R} f(z)dz + \int_{L_2} f(z)dz + \int_{C_\delta} f(z)dz= 2\pi i Res_{z=i}f(z)$$

That is,

$$\int_{L_1} f(z)dz + \int_{L_2} f(z)dz = 2\pi i Res_{z=i}f(z) - \int_{C_R} f(z)dz - \int_{C_\delta} f(z)dz \tag{2}$$

Since $$f(z) = \frac{(\ln r + i\theta)^2}{r^2e^{i0} + 1} \ \ \ \ \ \ (z=re^{i\theta})$$

the parametric representations

$$z = r e^{i0} = r \ \ (\delta\leq r\leq R) \ \ \text{and}\ \ z = re^{i\pi} = -r \ \ (\delta\leq r \leq R)$$

for the legs $L_1$ and $-L_2$ can be used to write the LHS pf equation $(2)$ as

$$\int_{L_1} f(z)dz - \int_{-L_2} f(z)dz = \int_{\delta}^{R} \frac{\ln^2 r}{r^2 + 1}dr + \int_{\delta}^{R} \frac{(\ln r + i\pi)^2}{r^2 + 1}dr$$

Also, since

$$Res_{z=i}f(z) = \frac{p(z)}{\phi'(z)}\ \ \text{where}\ \ p(z) = \ln^2 z \ \ \text{and}\ \ \phi(z) = z^2 + 1$$

then

$$Res_{z=i}f(z) = \frac{\Big(\ln (1) + i\frac{\pi}{2}\Big)^2}{2i}$$

Thus equation $(2)$ becomes

\begin{align}&2\int_{\delta}^{R} \frac{\ln^2 r}{r^2 + 1}dr + 2\pi i\int_{\delta}^{R} \frac{\ln r }{r^2 + 1}dr - \pi^2\int_{\delta}^{R} \frac{1}{r^2 + 1}dr\\ & = 2\pi i \frac{\Big(\ln (1) + i\frac{\pi}{2}\Big)^2}{2i} - \int_{C_R} f(z)dz - \int_{C_\delta} f(z)dz \\ &= -\frac{\pi^3}{4} - \int_{C_R} f(z)dz - \int_{C_\delta} f(z)dz \end{align}

Evaluating integrals

1- $\lim_{\delta \to 0}_{R\to \infty}\int_{\delta}^{R} \frac{\ln r }{r^2 + 1}dr = 0$

2- $\lim_{\delta \to 0}_{R \to \infty}\int_{\delta}^{R} \frac{1}{r^2 + 1}dr = \frac{\pi}{2}$

3- $\lim_{R\to\infty}\int_{C_R} f(z)dz = 0$

4- $\lim_{\delta \to 0}\int_{C_\delta} f(z)dz = 0$

Showing $4$.

Take $z = \delta e^{i\theta}$. Notice that if $\delta < 1$ and $z = \delta e^{i\theta}$, \begin{align}|\log^2 z| &=|(\ln \delta + i\theta )^2| = |\ln^2\delta + 2i\theta\ln\delta - \theta^2|\\&\leq |\ln^2\delta| + 2|i\theta\ln\delta|+\theta^2 \leq \ln^2\delta -2\pi\ln \delta + \pi^2\end{align} and $$|z^2+1| \geq ||z^2| - 1| = 1 - \delta^2$$ then

$$\Bigg|\int_{C_\delta} f(z)dz\Bigg| \leq \int_{C_\delta} |f(z)| |dz| \leq \frac{\ln^2\delta -2\pi\ln \delta + \pi^2}{1 - \delta^2} \pi\delta$$

the RHS of the inequality goes to $0$ as $\delta \to 0$.

There fore we get

$$2\int_{0}^{\infty} \frac{\ln^2 r}{r^2 + 1}dr = \frac{\pi^3}{4} \Rightarrow \int_{0}^{\infty} \frac{\ln^2 r}{r^2 + 1}dr = \frac{\pi^3}{8}$$

• If you have any questions concerning residue and other parts, feel free to ask. – Aaron Maroja Dec 24 '14 at 22:45
• We can't, the demoninator is 1, not indeterminate form. – Amad27 Dec 24 '14 at 22:45
• Well, the limit goes to zero, need to solve it here. – Aaron Maroja Dec 24 '14 at 22:48
• But just because the limit is 0 doesn't mean that the integral is 0 does it? – Amad27 Dec 24 '14 at 22:49
• You're estimating the integral value, and finding that as $\delta \to 0$ it "vanishes", then yes, it means that the integral is zero on the limit. – Aaron Maroja Dec 24 '14 at 22:52