3
$\begingroup$

The integral I'd like to evaluate is $\int_0^\infty \frac{\log^2 x \, dx}{(1+x)^2}$. I consider the function $f(z) = \frac{\text{Log}^2 z}{(1+z)^2}$, which has a pole of order 2 at $z=-1$ and has a branch point at $z=0$. I set up the integral $\oint_C f(z) dz$ along the contour $C = C_1 + C_2 + C_3 + C_4 + C_5 + C_6$, which consists of $C_1$ going just above the branch cut from $0$ to infinity, $C_2$, which is a big half-circle, $C_3$, which is a piece along the real axis from infinity to a point infinitesimally close to $z=-1$, $C_4$, which is a small half-circle going around $z=-1$ clockwise, $C_5$, which is another small piece on the real axis, and, finally, $C_6$, which is a quarter-circle around the origin to close the contour.

Now, I take the branch with $\text{Log}(z) = \log(r) + i \theta$, $0 \le\theta<2\pi$. Then the piece along $C_1$ is the integral $I$ I want to calculate. The pieces along $C_2$ and $C_6$ are zero when we take appropriate limits, but I'm not sure what to do with all the other pieces, since we can't really (can we?) write them in a way "$\text{something} \cdot I$", because of the singularity at $z=-1$. I know what I would do in the case of $1+z^2$ in the denominator, but not in this one. Suggestions?

$\endgroup$
  • $\begingroup$ The way you are sketching out the solution gives you, as the contour integral $$-PV \int_0^{\infty} dx \frac{(\log{x}+i \pi)^2}{(1-x)^2} + i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{\log^2{(e^{i \pi}+\epsilon e^{i \phi})}}{(\epsilon e^{i \phi})^2} \\ + \int_0^{\infty} dx \frac{\log^2{x}}{(1+x)^2}$$ This forces you to evaluate a wholly different principal value integral and is sort of leading you astray. I prefer the way I outline using a keyhole contour which, while it requires using a log cubed, gets us there. $\endgroup$ – Ron Gordon Dec 27 '15 at 6:38
2
$\begingroup$

I would consider the integral

$$\oint_C dz \frac{\log^3{z}}{(1+z)^2} $$

where $C$ is a keyhole contour about the real axis of radius $R$. Then as $R \to \infty$ the integral is equal to

$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{(1+x)^2} + 12 \pi^2 \int_0^{\infty} dx \frac{\log{x}}{(1+x)^2} +i 8 \pi^3 \int_0^{\infty} dx \frac{1}{(1+x)^2} = i 2 \pi \operatorname*{Res}_{z=e^{i \pi}} \frac{\log^3{z}}{(1+z)^2}$$

Note that the definition of the keyhole contour implies that the pole at $z=-1$ is necessarily represented as $z=e^{i \pi}$, as $\arg{z} \in [0,2 \pi]$.

To evaluate the other integrals we consider similar contour integrals:

$$\oint_C dz \frac{\log^2{z}}{(1+z)^2} = -i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{(1+x)^2} + 4 \pi^2 \int_0^{\infty} dx \frac{1}{(1+x)^2} = i 2 \pi \operatorname*{Res}_{z=e^{i \pi}} \frac{\log^2{z}}{(1+z)^2}$$

$$\oint_C dz \frac{\log{z}}{(1+z)^2} = -i 2 \pi \int_0^{\infty} dx \frac{1}{(1+x)^2} = i 2 \pi \operatorname*{Res}_{z=e^{i \pi}} \frac{\log{z}}{(1+z)^2}$$

Back substituting the other integral expressions into the first one, we find that

$$-i 6 \pi \int_0^{\infty} dx \frac{\log^2{x}}{(1+x)^2} + 12 \pi^2 \left [ i \pi \operatorname*{Res}_{z=e^{i \pi}} \frac{\log{z}}{(1+z)^2}-\frac12 \operatorname*{Res}_{z=e^{i \pi}} \frac{\log^2{z}}{(1+z)^2}\right ] \\ -i 8 \pi^3 \operatorname*{Res}_{z=e^{i \pi}} \frac{\log{z}}{(1+z)^2} = i 2 \pi \operatorname*{Res}_{z=e^{i \pi}} \frac{\log^3{z}}{(1+z)^2} $$

or

$$\begin{align}\int_0^{\infty} dx \frac{\log^2{x}}{(1+x)^2} &= \operatorname*{Res}_{z=e^{i \pi}} \frac{(2 \pi^2/3 )\log{z}+ i \pi \log^2{z} - (1/3) \log^3{z}}{(1+z)^2}\\ &= \left (\frac{2 \pi^2}{3} \right )\frac1{e^{i \pi}} + (i \pi) 2 \frac{i \pi}{e^{i \pi}} -\frac13 \frac{3 (i \pi)^2}{e^{i \pi}}\end{align}$$

Thus,

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

$\endgroup$
  • $\begingroup$ Thanks! I remembered vaguely that this "adding power" technique should work in case of a log, but it's nice to see exactly how. I'll up-vote once I get 15 points. $\endgroup$ – some1 Dec 27 '15 at 6:47
  • $\begingroup$ @some1: No problem. Please note here how I did not need to evaluate the lower-order integrals directly, but rather expressed them in terms of their residues. The result is a nice, nontrivial expression of the integral of a function times log squared in terms of the residue of the function times a cubic polynomial in log. This works in general and not just for your integral here. $\endgroup$ – Ron Gordon Dec 27 '15 at 6:52
  • $\begingroup$ yeah, I know. By the way, what if we consider an equivalent integral (up to changing the variables): $\int_{-\infty}^{+\infty} \frac{dx x^2 e^x}{(1+e^x)^2}$, so that now we have an infinite number of singularities along the imaginary axis. The usual rectangular contour doesn't work here (you get the identity $4\pi^2 = 4\pi^2$). Is there any reasonable contour we can choose to calculate this integral without changing variables? $\endgroup$ – some1 Dec 27 '15 at 7:00
  • $\begingroup$ @some1: for a rectangular contour I would use a similar trick: consider the contour integral of $z^3$ rather than $z^2$. $\endgroup$ – Ron Gordon Dec 27 '15 at 7:01
  • $\begingroup$ OK, so "adding power" technique works here too. Great, thanks! $\endgroup$ – some1 Dec 27 '15 at 7:02
2
$\begingroup$

We have $$\int_1^{\infty} \dfrac{\log^2(x)}{(1+x)^2}dx = \int_1^0 \dfrac{\log^2(1/x)}{(1+1/x)^2} \left(-\dfrac{dx}{x^2}\right) = \int_0^1 \dfrac{\log^2(x)}{(1+x)^2}dx$$ Hence, our integral becomes \begin{align} I & = \int_0^{\infty}\dfrac{\log^2(x)}{(1+x)^2}dx = 2\int_0^1\dfrac{\log^2(x)}{(1+x)^2}dx = 2\int_0^1 \sum_{k=0}^{\infty}(-1)^k (k+1)x^k \log^2(x)dx\\ & = 2\sum_{k=0}^{\infty}(-1)^k(k+1) \int_0^1 x^k\log^2(x)dx = 2\sum_{k=0}^{\infty}(-1)^k(k+1) \cdot \dfrac2{(k+1)^3}\\ & = 4 \sum_{k=0}^{\infty} \dfrac{(-1)^k}{(k+1)^2} = 4\left(\dfrac1{1^2} - \dfrac1{2^2} + \dfrac1{3^2} - \dfrac1{4^2} \pm \cdots\right) = \dfrac{\pi^2}3 \end{align}

$\endgroup$
  • $\begingroup$ I'm aware of this derivation, but I'd like to use complex analysis. $\endgroup$ – some1 Dec 27 '15 at 6:08
0
$\begingroup$

My suggestion would be, if you are to use complex analysis, you have to worry about the definition of functions. Function f is not defined in -1, so it does not have a pole at that point. First of all, you must "throw" the $x<0,y=0$ to somewhere, say to include a change of variable $x \to ix$ and then the singularity will not be what you have stated and your problem will vanish.

$\endgroup$
  • $\begingroup$ Well, it is defined at $z=-1$, isn't it? $\text{Log}(-1) = \log|-1| + i \pi$. I was sloppy with my notation, I've changed my original post, so $f(z)$ is now defined in terms of the principal $\text{Log}$, which exists at $z=-1$. $\endgroup$ – some1 Dec 27 '15 at 6:16
  • $\begingroup$ even if you define it that way, you have to map like I have described, everyone does it in that way. Your problem will vanish, it will not even be in a region you are interested in(-1 will go to -i, and there you have no interest, and no trouble) $\endgroup$ – nikola Dec 27 '15 at 6:22
  • $\begingroup$ No, I don't have to. See the answer by Ron Gordon above. $\endgroup$ – some1 Dec 27 '15 at 6:29

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.