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I'm a little stuck with this one. I've found the singularities to be at $\pm 2i$ and $0$ (branch point). So far, using a branch cut at $2\pi$ I've found that $$\int_{0}^{\infty}\frac{\ln(2r)}{4+r^2}dr+\int_{0}^{\infty}\frac{\ln(2r)+2\pi i}{4+r^2}dr = 2I+\int_{0}^{\infty}\frac{2\pi i}{4+r^2}dr$$ And $$\int_{0}^{\infty}\frac{2\pi i}{4+r^2}dr = \text{Res}(r=2i) \\ = -2\pi^2\lim_{r\to2i}\frac{r-2i}{(r-2i)(r+2i)} = \frac{i}{2}\pi^2$$ My problem is that the answer I found is completely imaginary, and I'm not sure how that's possible given that the original function is real. Any help is appreciated.

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  • $\begingroup$ The original integral does not make sense (as a real integral) because $2x$ does not have a real logarithm when $x\le0$. Should the integral be from $0$ to $\infty$? $\endgroup$
    – David
    Mar 10 '17 at 2:18
  • $\begingroup$ You're right! I accidentally wrote down the integral wrong here, it's changed to the right bounds now $\endgroup$
    – John Page
    Mar 10 '17 at 2:29
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    $\begingroup$ If you are integrating around a key-hole contour then the first and second integrals should cancel. $\endgroup$ Mar 10 '17 at 2:31
  • $\begingroup$ Not entirely sure what you mean here. You say "the answer I found is completely imaginary", do you mean the answer $i\pi^2/2$? As far as I can tell from your working, that's the answer to a different integral and therefore no problem. $\endgroup$
    – David
    Mar 10 '17 at 2:33
  • $\begingroup$ My problem is I don't know how to get the final answer from here, or whether this answer is correct. $\endgroup$
    – John Page
    Mar 10 '17 at 2:34
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Consider

$$f(z) = \frac{\log(2z)}{z^2+4}$$

By using a key-hole integral with branch-cut on positive axis we should get

$$\int^\infty_0 \frac{\log(|2x|)}{x^2+4}\,dx + \int_{\infty}^{0} \frac{\log(|2x|)+2\pi i}{x^2+4}\,dx =2\pi i \sum \mathrm{Res}(f,z_0)$$

We see that the first and second integrals will cancel. Now to avoid that consider

$$f(z) = \frac{\log(2z)^2}{z^2+4}$$

By using a key-hole integral with branch-cut on positive axis we should get

$$\int^\infty_0 \frac{\log^2(|2x|)}{x^2+4}\,dx + \int^0_{\infty} \frac{(\log(|2x|)+2\pi i)^2}{x^2+4}\,dx =2\pi i \sum \mathrm{Res}(f,z_0)$$

$$\int^\infty_0 \frac{\log^2(|2x|)}{x^2+4}\,dx - \int_0^{\infty} \frac{(\log(|2x|)+2\pi i)^2}{x^2+4}\,dx =2\pi i \sum \mathrm{Res}(f,z_0)$$

Now you can see that $\log^2(2x)$ will be cancelled and we are left out with $\log(2x)$.


Another approach

Integrate around a big half-circle indented at 0 where the branch cut is chosen on the imaginary axis then for

$$f(z) = \frac{\log(2z)}{z^2+4}$$

we have only one pole at $z = 2i$

$$\int_{-\infty}^0 \frac{\log|2x|+\pi i}{x^2+4}\,dx +\int^{\infty}_0 \frac{\log|2x|}{x^2+4}\,dx = 2\pi i \,\mathrm{Res}(f,2i)$$

$$2\int^\infty_0 \frac{\log(2x)}{x^2+4}\,dx + \pi i\int^\infty_0\frac{1}{x^2+4}\,dx = 2\pi i \,\mathrm{Res}(f,2i) $$

Note that

$$\mathrm{Res}(f,2i) = \lim_{z \to 2i} (z-2i) \frac{\log(2z)}{(z-2i)(z+2i)} = \frac{\log(4i)}{4i} = \frac{\log(4)+(\pi i)/2}{4i}$$

$$2\int^\infty_0 \frac{\log(2x)}{x^2+4}\,dx + \pi i\int^\infty_0\frac{1}{x^2+4}\,dx = \pi \frac{\log(4)+(\pi i)/2}{2} $$

By comparison we have

$$\int^\infty_0 \frac{\log(2x)}{x^2+4}\,dx = \frac{\pi}{2} \log(2)$$

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METHODOLOGY $1$: COMPLEX ANALYSIS

As already established in the answer left by @ZaidAlyafeai, the classical approach begins by analyzing the contour integral

$$I=\oint_C \frac{\log^2(2z)}{z^2+4}\,dz$$

where $C$ is the "keyhole" contour with the keyhole taken along the positive real axis. In that case $0\le \arg(z)<2\pi$. Then, we have

$$\begin{align} \color{blue}{\int_0^\infty \frac{\log^2(2x)}{x^2+4}\,dx-\int_0^\infty\frac{(\log(2x)+i2\pi)^2}{x^2+4}\,dx}&=\color{red}{2\pi i \text{Res}\left(\frac{\log^2(2z)}{z^2+4}, z=\pm i2\right)}\\\\ &=\color{red}{2\pi i\left(\frac{\log^2(4i)}{4i}+\frac{\log^2(-4i)}{-4i}\right)}\\\\ &=\color{red}{\frac{\pi}{2}\left((\log(4)+i\pi/2)^2-(\log(4)+i3\pi/2)^2\right)}\\\\ \color{blue}{4\pi^2\int_0^\infty\frac{1}{x^2+4}\,dx-i4\pi\int_0^\infty\frac{\log(2x)}{x^2+4}\,dx}&=\color{red}{\frac{\pi}{2}\left(2\pi^2-i2\pi\log(4)\right)}\tag 1 \end{align}$$

where we find by setting real and imaginary parts of $(1)$ equal

$$\int_0^\infty \frac{1}{x^2+4}\,dx=\frac{\pi}{4}$$

and

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty\frac{\log(2x)}{x^2+4}\,dx=\frac{\pi\log(2)}{2}} \tag 2$$


METHODOLOGY $2$: REAL ANALYSIS

I thought it might be instructive to present an approach that relies on real analysis only.

To proceed, we begin by enforcing the substitution $x\to 4/x$. Then, we have

$$\begin{align} \int_0^\infty \frac{\log(2x)}{x^2+4}\,dx&=\int_0^\infty \frac{\log(8/x)}{(4/x)^2+4}\,\frac{4}{x^2}\,dx\\\\ &=4\log(2)\int_0^\infty\frac{1}{x^2+4}\,dx-\int_0^\infty \frac{\log(2x)}{x^2+4}\\\\ 2\int_0^\infty \frac{\log(2x)}{x^2+4}\,dx&=4\log(2)\int_0^\infty\frac{1}{x^2+4}\,dx\\\\ \int_0^\infty \frac{\log(2x)}{x^2+4}\,dx&=2\log(2)\int_0^\infty\frac{1}{x^2+4}\,dx\\\\ &=\frac{\pi\log(2)}{2} \end{align}$$

agreeing with the result in $(2)$ as expected!

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  • $\begingroup$ @ZaidAlyafeai Thank you my friend! And yours was a very well-written solution! (+1) also. -Mark $\endgroup$
    – Mark Viola
    Mar 10 '17 at 6:50
  • $\begingroup$ Without doubt you are one of the best here on posting instructive, well-written and formatted answers. keep it up. $\endgroup$ Mar 10 '17 at 6:52
  • $\begingroup$ @ZaidAlyafeai Thank you Zaid! $\endgroup$
    – Mark Viola
    Mar 10 '17 at 6:55

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