3
$\begingroup$

Evaluate $$\int_0^\infty \dfrac {\log{x}}{(x^2+1)^2}dx$$

I've been working on this problem for half the day. I'm not getting anywhere.

1) I first changed the integral from negative infinity to positive infinity

2) Then I'm using the fact that

$\int_{-\infty}^\infty \dfrac {P(x)}{Q(x)}dx = 2\pi i $ $\sum$ {residues of $P/Q$ in upper half plane}

3) I'm calculating residues in the upper half plane which are x=+i

4) After I calculate residue and multiply by $2\pi i$, I do not get the answer -$\pi/4$

5) I'm under the impression I have to convert $logx$ to something else.

Any help will be appreciated. Thank you

$\endgroup$
6
  • 3
    $\begingroup$ But, your integral is not of the form $P(z)/Q(z)$. You need to think about the branch-cut of the log etc... $\endgroup$ May 5, 2013 at 0:33
  • 1
    $\begingroup$ The formula you cite in 2) undoubtedly has some hypotheses on $P$ and $Q$, hypotheses which may not be met in the problem at hand --- also, note that that formula involves an integral from $-\infty$ to $\infty$. $\endgroup$ May 5, 2013 at 0:34
  • $\begingroup$ @JamesS. Yikes you are right. The conditions fail! This whole time I thought they were satisfied. What do you mean by branch cut? $\endgroup$
    – User69127
    May 5, 2013 at 0:36
  • $\begingroup$ would it be wise to make a substitution. Say $x=e^t$ $\endgroup$
    – User69127
    May 5, 2013 at 0:42
  • 1
    $\begingroup$ That'll get you nowhere. The Wikipedia page on contour integration has this exact integral as an example. $\endgroup$
    – Javier
    May 5, 2013 at 0:56

3 Answers 3

6
$\begingroup$

This is not the usual contour integral with simple poles. The log term has a branch point at zero and must be treated with care. The usual way to treat integrals with such branch points is to use something called a keyhole contour, which goes up and back a branch cut (here, the positive real axis) and makes use of the multivaluedness of the integrand.

keyhole

In general, the way to attack integrals such as the one you have is to exploit the multivaluedness of the log to extract the integral from $[0,\infty)$ in terms of residues of the integrand. In this case, however, there is already a log in the integrand, so we need to add another factor of log to extract the desired integral. To wit, consider

$$\oint_C dz \frac{\log^2{z}}{(z^2+1)^2}$$

where $C$ is the keyhole contour illustrated above. This integral is equal to the integral over the four segments of $C$:

$$\oint_C dz \frac{\log^2{z}}{(z^2+1)^2} = \left [\int_{C_+} + \int_{C_R} + \int_{C_-} + \int_{C_{\epsilon}} \right] dz \frac{\log^2{z}}{(z^2+1)^2}$$

The integrals over $C_R$ and $C_{\epsilon}$ vanish as $R \to \infty$ and $\epsilon \to 0$, respectively:

$$\int_{C_R} dz \frac{\log^2{z}}{(z^2+1)^2} = i R \int_0^{2 \pi} d\phi\, e^{i \phi} \frac{\log^2{(R e^{i \phi})}}{(1+R^2 e^{i 2 \phi})^2} \sim \frac{\log^2{R}}{R^3} \quad (R \to \infty)$$

$$\int_{C_{\epsilon}} dz \frac{\log^2{z}}{(z^2+1)^2} = i \epsilon \int_{2 \pi}^0 d\phi e^{i \phi} \frac{\log^2{(\epsilon e^{i \phi})}}{(1+\epsilon^2 e^{i 2 \phi})^2} \sim \epsilon \, \log^2{\epsilon} \quad (\epsilon \to 0) $$

This leaves the integrals up and down the real axis, $C_+$ and $C_-$, respectively. The integral over $C_+$ is simply the usual integral over the $x$ axis:

$$\int_{C_+} dz \frac{\log^2{z}}{(z^2+1)^2} = \int_0^{\infty} dx \frac{\log^2{x}}{(x^2+1)^2}$$

(I am assuming that the above limits have been taken.) The integral over $C_-$, however, reflects the fact that $z$ has advanced in argument by $2 \pi$. Normally, with single-valued functions, this doesn't matter. With multi-valued functions, however, this is crucial, as $\log{(x\,e^{i 2 \pi})} = \log{x} + i 2 \pi$. Thus we have

$$\int_{C_-} dz \frac{\log^2{z}}{(z^2+1)^2} = \int_{\infty}^0 dx \frac{(\log{x}+i 2 \pi)^2}{(x^2+1)^2}$$

Putting this altogether:

$$\begin{align}\oint_C dz \frac{\log^2{z}}{(z^2+1)^2} &= \int_0^{\infty} dx \frac{\log^2{x}}{(x^2+1)^2} - \int_0^{\infty} dx \frac{(\log{x}+i 2 \pi)^2}{(x^2+1)^2}\\ &= -i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{(x^2+1)^2} + 4 \pi^2 \int_0^{\infty} dx \frac{1}{(x^2+1)^2} \end{align}$$

This is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand. The poles are at $z = \pm i$ and are double poles. Because these are double poles, the sum of the residues is given by

$$\begin{align}\lim_{z \to i} \frac{d}{dz}\left [ (z-i)^2 \frac{\log^2{z}}{(z^2+1)^2} \right ] \\+ \lim_{z \to -i} \frac{d}{dz}\left [ (z+i)^2 \frac{\log^2{z}}{(z^2+1)^2} \right ]\\ &= \frac{d}{dz}\left [\frac{\log^2{z}}{(z+i)^2} \right]_{z=i}+\frac{d}{dz}\left [\frac{\log^2{z}}{(z-i)^2} \right]_{z=-i}\\ &= \left [ \frac{2 \log (z)}{z (z+i)^2}-\frac{2 \log ^2(z)}{(z+i)^3} \right]_{z=i} + \left [ \frac{2 \log (z)}{z (z-i)^2}-\frac{2 \log ^2(z)}{(z-i)^3} \right]_{z=-i}\\ &= \frac{i\pi}{i (-4)} - \frac{2 (-\pi^2/4)}{-8 i} + \frac{i 3\pi}{(-i) (-4)} - \frac{2 (-9 \pi^2/4)}{8 i}\\ &= \frac{\pi}{2} - i \frac{\pi^2}{2}\end{align}$$

In that next-to-last line, I used $\arg{-i} = 3 \pi/2$; this is crucial to get right so we are consistent with how we defined the contour integral.

We may now write

$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{(x^2+1)^2} + 4 \pi^2 \int_0^{\infty} dx \frac{1}{(x^2+1)^2} = i 2 \pi \left (\frac{\pi}{2} - i \frac{\pi^2}{2}\right) = i \pi^2 +\pi^3$$

To finish this off, we need to evaluate the latter integral:

$$\int_0^{\infty} dx \frac{1}{(x^2+1)^2} = \frac12 \int_{-\infty}^{\infty} dx \frac{1}{(x^2+1)^2}$$

In this case, we can simply use a semicircular contour in the upper half-plane; the integral is (details left to reader):

$$i 2 \pi \frac12 \frac{d}{dz}\left [\frac{1}{(z+i)^2} \right ]_{z=i} = i \pi \frac{-2}{(2 i)^3} = \frac{\pi}{4}$$

Thus the integral we seek is

$$\int_0^{\infty} dx \frac{\log{x}}{(x^2+1)^2} = \frac{( i \pi^2 +\pi^3) - 4 \pi^2 (\pi/4)}{-i 4 \pi}$$

or

$$\int_0^{\infty} dx \frac{\log{x}}{(x^2+1)^2} = -\frac{\pi}{4}$$

$\endgroup$
2
  • $\begingroup$ You're the complex analysis God, Ron. This is amazing work. Thank you $\endgroup$
    – User69127
    May 5, 2013 at 3:59
  • $\begingroup$ @User69127: you are very welcome. I hope that this derivation made the concepts clear. It would help if you worked out a simpler one on your own. $\endgroup$
    – Ron Gordon
    May 5, 2013 at 4:05
2
$\begingroup$

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{\ln\pars{x} \over \pars{x^{2} + 1}^{2}}\,\dd x:\ {\large ?}}$

${\large\tt\mbox{Short Solution:}}$ \begin{align} &\color{#c00000}{\int_{0}^{\infty}{\ln\pars{x} \over \pars{x^{2} + 1}^{2}}\,\dd x} =\left.-\,\partiald{}{\mu}\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + \mu}\,\dd x\, \vphantom{\Huge A^{a}}\right\vert_{\ \mu\ = 1} \\[3mm]&=-\,\partiald{}{\mu}\bracks{% \mu^{-1/2}\int_{0}^{\infty}{\ln\pars{\mu^{1/2}x} \over x^{2} + 1}\,\dd x} _{\ \mu\ = 1} \\[3mm]&=-\,\partiald{}{\mu}\bracks{% \half\,\mu^{-1/2}\ln\pars{\mu}\,{\pi \over 2} +\mu^{-1/2}\ \overbrace{\int_{0}^{\infty}{\ln\pars{x} \over x^{2} + 1} \,\dd x}^{\ds{=\ 0}}}_{\ \mu\ = 1} \end{align} The vanishing integral can be divided in two integrals over $\pars{0,1}$ and over $\pars{1,+\infty}$: They are of equal magnitude but they have different signs. \begin{align} &\color{#00f}{\large\int_{0}^{\infty}{\ln\pars{x} \over \pars{x^{2} + 1}^{2}}\,\dd x} =-\,{\pi \over 4}\bracks{-\,\half\,\mu^{-3/2}\ln\pars{\mu} + \mu^{-3/2}}_{\ \mu\ =\ 1} = \color{#00f}{\large\,-{\pi \over 4}} \end{align}

$$ \vphantom{\large a} $$

${\large\tt\mbox{Long Solution:}}$
With $\ds{t \equiv {1 \over 1 + x^{2}}\quad\imp\quad x = \pars{{1 \over t} - 1}^{1/2}}$: \begin{align} &\color{#c00000}{\int_{0}^{\infty}{\ln\pars{x} \over \pars{x^{2} + 1}^{2}}\,\dd x} =\lim_{\mu \to 0}\partiald{}{\mu} \int_{0}^{\infty}{x^{\mu} \over \pars{x^{2} + 1}^{2}}\,\dd x \\[3mm]&=\lim_{\mu \to 0}\partiald{}{\mu} \int_{1}^{0}t^{2}\pars{{1 \over t} - 1}^{\mu/2}\, \bracks{\half\,\pars{{1 \over t} - 1}^{-1/2}\,\pars{-\,{1 \over t^{2}}}\,\dd t} \\[3mm]&=\half\lim_{\mu \to 0}\partiald{}{\mu}\int_{0}^{1}t^{\pars{1 - \mu}/2} \pars{1 - t}^{\pars{\mu - 1}/2}\,\dd t \\[3mm]&=\half\lim_{\mu \to 0}\partiald{}{\mu}\bracks{% \Gamma\pars{3/2 - \mu/2}\Gamma\pars{\mu/2 + 1/2} \over \Gamma\pars{2}} \\[3mm]&=\half\lim_{\mu \to 0}\partiald{}{\mu}\bracks{% \pars{\half - {\mu \over 2}}\Gamma\pars{\half - {\mu \over 2}} \Gamma\pars{{\mu \over 2} + \half}} \\[3mm]&=\half\lim_{\mu \to 0}\partiald{}{\mu}\bracks{% \pars{\half - {\mu \over 2}}\,{\pi \over \sin\pars{\pi\bracks{\mu/2 + 1/2}}}} \\[3mm]&=\half\,\pi\lim_{\mu \to 0}\partiald{}{\mu}\bracks{% \pars{\half - {\mu \over 2}}\sec\pars{\pi\mu \over 2}} \\[3mm]&=\half\,\pi\ \overbrace{\lim_{\mu \to 0}\bracks{% -\,\half\,\sec\pars{\pi\mu \over 2} + \half\pars{\half - {\mu \over 2}}\pi\sec\pars{\pi\mu \over 2} \tan\pars{\pi\mu \over 2}}}^{\ds{=\ -\,\half}} \end{align}

$$\color{#00f}{\large% \int_{0}^{\infty}{\ln\pars{x} \over \pars{x^{2} + 1}^{2}}\,\dd x = -\,{\pi \over 4}} $$

$\endgroup$
0
$\begingroup$

It is an improper integral. You must take the limit $$\lim_{t-> +\infty} \int_{0}^{t}{\frac{logx}{(x^2 + 1)^2}}dx$$

$\endgroup$
2
  • $\begingroup$ Also as the lower limit goes to zero. $\endgroup$
    – Javier
    May 5, 2013 at 1:06
  • $\begingroup$ Yes, you are right! :-) $\endgroup$
    – EmmanouilG
    May 5, 2013 at 1:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .