Residue Theorem and friends

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Residue Theorem and friends

A compendium of difficult calculus problems, some of which lean on the residue theoremSkip to contentHomeAbout the AuthorModerate Nerd

Logs on a Dogbone II

Many of you who like to evaluate challenging integrals may have tried your hand at this one:

$$\int_0^1 dx \, \log{x} \, \log{(1-x)} = 2-\frac{\pi^2}{6} $$

Even using real methods, this one is a bit tough. The idea is to expand the $\log{(1-x)}$ term in a Maclurin expansion and reverse the order of integration and summation, which is completely justified. This maneuver replaces the integral with a sum

$$\begin{align} \sum_{k=1}^{\infty} \frac1{k (k+1)^2} &= \sum_{k=1}^{\infty} \left (\frac1{k} – \frac1{k+1} \right ) – \sum_{k=1}^{\infty} \frac1{(k+1)^2} \\ &= 1 – \left ( \frac{\pi^2}{6} – 1 \right )\\ &= 2-\frac{\pi^2}{6} \\ \end{align}$$

While this is a relatively simple and satisfying result, the result I derived in Logs on a Dogbone made me wonder if we could use a similar means to evaluate this integral using complex integration techniques. The answer is unsurprisingly yes, but with the caveat that this means is exceedingly complicated, almost to the point of the success in evaluating this integral using complex techniques may be considered a Pyrrhic victory.

Nevertheless, in some cases a Pyrrhic victory is still a victory! By doing this exercise, we further develop the technique for use with other integrals that may not be evaluated so easily using real methods. So let’s illustrate how this integral is evaluated using Cauchy’s Theorem.

We begin by defining a complex integral for evaluation:

$$\oint_C dz \, \log{z} \, \log^2{(z-1)} $$

where $C$ is

The small arcs have a radius $\epsilon$; the large arcs have radius $R$. Accordingly, the contour integral is equal to

$$-e^{i \pi} \int_{\epsilon}^R dx \, \left (\log{x}+ i \pi \right ) \left (\log{(x+1)}+ i \pi \right )^2 + i \epsilon \int_{\pi}^{0} d\phi \, e^{i \phi} \, \log{\left (\epsilon e^{i \phi} \right )} \, \log^2{\left (\epsilon e^{i \phi} – 1\right )} \\ + \int_{\epsilon}^{1-\epsilon} dx \log{x} \, \left (\log{(1-x)}+ i \pi \right )^2 + i \epsilon \int_{-\pi}^{\pi} d\phi \, e^{i \phi} \, \log{\left (1+\epsilon e^{i \phi} \right )} \, \log^2{\left (\epsilon e^{i \phi} \right )} \\ – \int_{\epsilon}^{1-\epsilon} dx \log{x} \, \left (\log{(1-x)}- i \pi \right )^2 + i \epsilon \int_{2 \pi}^{\pi} d\phi \, e^{i \phi} \, \log{\left (\epsilon e^{i \phi} \right )} \, \log^2{\left (\epsilon e^{i \phi} – 1\right )} \\ + e^{-i \pi} \int_{\epsilon}^R dx \, \left (\log{x}- i \pi \right ) \left (\log{(x+1)}- i \pi \right )^2 + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \log{\left (R e^{i \theta} \right )} \, \log^2{\left (R e^{i \theta} – 1 \right )} \\$$

Before we go on evaluating these integrals, let’s take a little perspective. I defined a contour integral that will give us our definite integral. But why that one? Hopefully, as we work the computations, it will become clear that we anticipate cancellations from the square terms and the integral we seek being made up of mixed terms. The definition of a contour integral used in evaluating a real definite integral takes a lot of practice and a little imagination. The contour of course is based on the fact that we are integrating between two branch points. Keep this in mind as we evaluate the above integrals.

To begin, observe first that the three integrals about the small arcs all cancel as $\epsilon \to 0$. We can also combine like integrals and hence reduce the above eight integrals to three in this limit:

$$ \int_0^R dx \, \left ( \left (\log{x}+ i \pi \right ) \left (\log{(x+1)}+ i \pi \right )^2 – \left (\log{x}- i \pi \right ) \left (\log{(x+1)}- i \pi \right )^2 \right ) \\ + \int_0^1 dx \, \left [ \log{x} \, \left (\log{(1-x)}+ i \pi \right )^2 – \log{x} \, \left (\log{(1-x)}- i \pi \right )^2 \right ] \\ + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \log{\left (R e^{i \theta} \right )} \, \log^2{\left (R e^{i \theta} – 1 \right )}$$

These integrals may be simplified to take the forms

$$ i 4 \pi \int_0^R dx \, \log{x} \, \log{(x+1)} + i 2 \pi \int_0^R dx \, \log^2{(x+1)} – i 2 \pi^3 R \\ + i 4 \pi \int_0^1 dx \, \log{x} \, \log{(1-x)} \\ + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \log{\left (R e^{i \theta} \right )} \, \log^2{\left (R e^{i \theta} – 1 \right )}$$

The value of the integral we seek is revealed in the limit $R \to \infty$. Note that this expression for the contour integral is made up of terms that diverge in that limit; I will demonstrate that those terms will cancel.

To begin, we will need to evaluate the first set of integrals over $[0,R]$. While we may be able to evaluate the second integral exactly, the first integrand has an integrand that does not have an exact antiderivative. As an alternative, we will make use of techniques of evaluating these integrals asymptotically in the limit as $R \to \infty$.

Rewrite the integrals as follows.

$$\begin{align} \int_0^R dx \, \log{x} \, \log{(x+1)} &= R \log{R} \int_0^1 du \, \log{\left ( 1+R u \right )} + R \int_0^1 du \, \log{u} \, \log{\left ( 1+R u \right)} \\ \int_0^R dx \, \log^2{(x+1)} &= R \int_0^1 du \, \log^2{\left ( 1+R u \right )} \end{align}$$

There is a well-known approach to determining asymptotic behavior of integrals with general, smooth and integrable kernels. I will summarize the theory below, which one may find in great detail in N. Bleistein and R. Handelsman, “Asymptotic Expansions of Integrals,” Secs. 4.1-4.4.

Consider the integral

$$I(\lambda) = \int_0^{\infty} dx \, f(t) h(\lambda t) $$

where $\lambda \gt 0$ and $f$ and $h$ satisfy certain convergence properties in respective vertical strips in the complex plane. Then the following Parseval relation is true:

$$I(\lambda) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \lambda^{-s} F(1-s) H(s) $$

where $\operatorname{Re}{s}=c$ is in a vertical strip common to both $f$ and $h$ and $F$ and $H$ are, respectively, the Mellin transforms of $f$ and $h$:

$$\begin{align}F(s) &= \int_0^{\infty} dt \, f(t) \, t^{s-1} \\ H(s) &= \int_0^{\infty} dt \, h(t) \, t^{s-1} \end{align}$$

By using analytical continuation, the line of integration may be moved out to the right to a line $\operatorname{Re}{s}=L$, where $L \gt c$. We can complete a rectangular contour, with the horizontal pieces vanishing as $L \to \infty$. The integral may then be written, by the residue theorem, as

$$I(\lambda) = -\sum_{c \lt \operatorname{Re}{s_j} \lt L} \operatorname*{Res}_{s=s_j}\left [ \lambda^{-s} F(1-s) H(s) \right ] + \frac1{i 2 \pi} \int_{L-i \infty}^{L+i \infty} ds \, \lambda^{-s} F(1-s) H(s) $$

For $L$ large enough, the integral on the RHS is negiglible compared to the residues in the sum. We thus have the following asymptotic relation:

$$\int_0^{\infty} dt \, f(t) h(\lambda t) = -\sum_{c \lt \operatorname{Re}{s_j} \lt L} \operatorname*{Res}_{s=s_j}\left [ \lambda^{-s} F(1-s) H(s) \right ] + O \left ( \lambda^{-L} \right )$$

(For a proof, see the above reference.) And here, at last, we see a surprising appearance of residues!

$$\begin{array}{c|cccc} \text{integral} & f(t) & h(t) & F(1-s) & H(s)\\ \hline 1 & \theta(t) \theta(1-t) & \log{(1+t)} & \frac1{1-s} & \frac{\pi}{s \sin{\pi s}}\\ \hline 2 & \theta(t) \theta(1-t) \log{t} & \log{(1+t)} & – \frac1{(1-s)^2} & \frac{\pi}{s \sin{\pi s}}\\ \hline 3 & \theta(t) \theta(1-t) & \log^2{(1+t)} & \frac1{1-s} & -\frac{2 \pi \left (\gamma + \psi(-s) \right )}{s \sin{\pi s}} \end{array}$$

We will use the above asymptotic relation for the integrals defining our limits as $R \to \infty$. In our case, $\lambda =R$, $\theta$ refers to the Heaviside step function, and $\psi$ is the digamma function.

It turns out that we may analytically continue the Mellin transforms to the half plane $\operatorname{Re}{s} > 0$. Now, note that we need only consider, for the leading asymptotic behavior of these integrals, the pole at $s=0$ and $s=1$. (Poles at $s=2$ and beyond will have subdominant behavior – note that it is not trivial that we needed the residue at $s=1$ – realizing this is usually a matter of trial and error.) Thus, we need to consider three pairs of residues as follows.

$$\begin{align} -\operatorname*{Res}_{s=0}\left [ \frac{\pi R^{-s}}{s (1-s) \sin{\pi s}} \right ] &= \log{R} \, – 1 \\ -\operatorname*{Res}_{s=1}\left [ \frac{\pi R^{-s}}{s (1-s) \sin{\pi s}} \right ] &= \frac1{R} \left (\log{R} + 1 \right ) \\ -\operatorname*{Res}_{s=0}\left [ -\frac{2 \pi R^{-s} \left ( \gamma + \psi(-s) \right )}{s (1-s) \sin{\pi s}} \right ] &= \log^2{R} – 2 \log{R} + 2 \\ -\operatorname*{Res}_{s=1}\left [ -\frac{2 \pi R^{-s} \left ( \gamma + \psi(-s) \right )}{s (1-s) \sin{\pi s}} \right ] &= \frac1{R} \left (\log^2{R} \, – 2 \right ) \\ -\operatorname*{Res}_{s=0}\left [ -\frac{\pi R^{-s}}{s (1-s)^2 \sin{\pi s}} \right ] &= -\log{R} \, + 2 \\ -\operatorname*{Res}_{s=1}\left [ -\frac{\pi R^{-s}}{s (1-s)^2 \sin{\pi s}} \right ] &= -\frac1{R} \left (\frac12 \log^2{R} + \log{R} +\frac{\pi^2}{6} + 1 \right ) \end{align}$$

The above residues were evaluated by expanding the expressions into Laurent expansions; I did all steps by hand. Now we may take these results and plug them into the expression for the contour integral (recalling that there is a term $-i 2 \pi^3 R$ outside of the integrals):

$$\begin{align} & i 4 \pi R \log{R} \int_0^1 du \, \log{(1+R u)} + i 4 \pi R \int_0^1 du \, \log{u} \, \log{(1+R u)} + i 2 \pi R \int_0^1 du \, \log^2{(1+R u)} – i 2 \pi^3 R\\ = & i 4 \pi R \log{R} \left (\log{R} – 1 + \frac1{R} \left (\log{R} + 1 \right ) \right ) + i 4 \pi R \left (-\log{R} + 2 – \frac1{R} \left ( \frac12 \log^2{R} + \log{R} + \frac{\pi^2}{6} + 1 \right ) \right ) \\ + & i 2 \pi R \left ( \log^2{R} – 2 \log{R} + 2 + \frac1{R} \left (\log^2{R} – 2 \right ) \right ) – i 2 \pi^3 R+ O \left ( \frac{\log{R}}{R} \right ) \\ & = i 6 \pi R \log^2{R} – i 12 \pi R \log{R} – i 2 \pi \left ( \pi^2 – 6 \right ) R + i 4 \pi \log^2{R} – i 4 \pi \left ( 2 + \frac{\pi^2}{6} \right ) + O \left ( \frac{\log{R}}{R} \right ) \end{align}$$

Keep in mind that the RHS of the above relations are asymptotic relations, and the big-O represents the leading behavior of the error in the approximation. Note that the $\log{R}/R$ comes from the fact that the poles at $s=2$ and beyond are simple and accordingly produce no additional log terms in $R$. Recall also that these relations are valid in the limit as $R \to \infty$. As expected, the approximate terms we are keeping diverge in this limit. This is no problem, because we have more terms to evaluate that will cancel the diverging terms and leave us with the result we seek.

We now need to evaluate the other integral in this limit as $R \to \infty$:

$$\begin{align} & i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \log{\left (R e^{i \theta} \right )} \, \log^2{\left (R e^{i \theta} – 1 \right )} \\ &= i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \left [\log^3{\left ( R e^{i \theta} \right )} – \frac{2}{R e^{i \theta}} \log^2{\left ( R e^{i \theta} \right )} \right ] + O \left ( \frac{\log{R}}{R} \right )\\ &= i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \left (\log^3{R} + i 3 \theta \log^2{R} – 3 \theta^2 \log{R} – i \theta^3 \right ) – i 2 \int_{-\pi}^{\pi} d\theta \, \left (\log^2{R} + i 2 \theta \log{R} – \theta^2 \ \right ) + O \left ( \frac{\log{R}}{R} \right ) \end{align}$$

The $\log^3{R}$ term in the first integral and the $\log{R}$ term in the second integral vanish. For the other integrals, we need the following results which are easily derived using, e.g., integration by parts.

$$\begin{align} \int_{-\pi}^{\pi} d\theta \, \theta \, e^{i \theta} &= i 2 \pi \\ \int_{-\pi}^{\pi} d\theta \, \theta^2 e^{i \theta} &= -4 \pi \\ \int_{-\pi}^{\pi} d\theta \, \theta^3 e^{i \theta} &= i 2 \pi \left ( \pi^2 – 6 \right ) \end{align} $$

With these results, we may then write down the asymptotic expansion of this integral as $R \to \infty$:

$$\begin{align} & i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \log{\left (R e^{i \theta} \right )} \, \log^2{\left (R e^{i \theta} – 1 \right )} \\ &= ( i R) ( i 3) ( i 2 \pi) \log^2{R} + ( i R) (-3) (-4 \pi) \log{R} + ( i R) ( -i) ( i 2 \pi)(\pi^2-6) – i 2 (2 \pi) \log^2{R} + 0 + i 2 \left (\frac{2 \pi^3}{3} \right ) + O \left ( \frac{\log{R}}{R} \right ) \\ &= – i 6 \pi R \log^2{R} + i 12 \pi R \log{R} + i 2 \pi \left ( \pi^2 – 6 \right ) R – i 4 \pi \log^2{R} + i \frac{4 \pi^3}{3} + O \left ( \frac{\log{R}}{R} \right ) \end{align} $$

And we are finally ready to write down the expression for the contour integral in the limit as $R \to \infty$. Note that all terms dependent on $R$ cancel. We should expect that this happens and, as we will see, the requirement that these terms cancel provides a self-check on the solution. The result for the contour integral is

$$\oint_C dz \, \log{z} \, \log^2{(z-1)} = i 4 \pi \int_0^1 dx \, \log{x} \, \log{(1-x)} – i 4 \pi \left ( 2 – \frac{\pi^2}{6} \right ) $$

And now we invoke Cauchy’s Theorem: because there are no poles within $C$, the contour integral is equal to zero. Accordingly, the value of the integral is as derived at the beginning of this post:

$$\int_0^1 dx \, \log{x} \, \log{(1-x)} = 2-\frac{\pi^2}{6} $$

A few observations:

For such a simple-looking integral, there is a remarkable amount of mathematics in the complex analytical technique developed here. This complexity, however, should not obscure the clear steps: define a contour integral about the contour $C$, express the contour integral in terms of real integrals based on parametrizations of $C$, perform asymptotic analysis as detailed above to estimate the integrals over the branch cuts along the negative real axis and the outer arc, make sure all terms in $R$ cancel, and determine the value of the desired real integral. That all said, I evaluated every expression in this post – every Mellin transform, every residue – by hand. Otherwise, what’s the point? None of it was exceedingly difficult. The worst of it was the MT of the log squared term, and all that involved was taking a second derivative of a beta function. The residues I evaluated at $s=0$ and $s=1$ were at double or triple poles and accordingly I evaluated the residues using Laurent expansions, a much much better approach than trying to derive explicit expressions for residues of rational functions at repeated poles.Speaking of residues, isn’t it fantastic how we end up needing them but in a completely unexpected way? People ask me all the time how I am going to evaluate such-and-such an integral using the residue theorem, and many times I say that the integrand has no poles inside the contour so we will use Cauchy’s Theorem instead. The same might be said for this one but the equation for the asymptotic behavior of the integrals over the branch cut along the negative real axis demands the evaluation of residues of Mellin transforms of functions defined by our integrands. The asymptotic analysis presented here is not new but is detailed in the Bleistein/Handelsman reference. That said, I have not seen it used by practitioners much – I am sure I am simply ignorant of application for which this formulation has been used – but it deserves a much wider audience. The problem is that the formulation depends on concepts like Mellin transforms and residue theory, and instructors tend to shy away from these things in a first (i.e., undergraduate- and beginning-graduate-levels) treatment of asymptotics. Too bad, because this stuff is gorgeous.I believe (but have not proven) that complex integration can be used to evaluate every definite integral taken between branch points of the integrand. I think the technique developed here represents an important step toward demonstrating that statement.Clearly, the complex integration technique demonstrated here is way, way more involved than the real method shown here. In that case, one may wonder what use this technique is at all. I counter that there are several. First, this method opens doors to other cases in which the real approach is not as simple. Second, as I demonstrated in this post, many definite integrals can be approximated using integrals between branch points plus a correction. Finally, this method provides a self-check mechanism that real methods typically do not have, and it is always good to derive a result in more than one way.

ADDENDUM

There is a third way to evaluate this integral. Note that it is in the form of a convolution; the convolution theorem for Laplace transforms is as follows:

Let $f$ and $g$ be functions that map $[0,\infty)$ to $\mathbb{R}$ and have respective Laplace transforms $F$ and $G$. Then the following is true:

$$\int_0^{1} dt \, \log{t} \, \log{(1-t)} = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{s} \, F(s) G(s) $$

In this case, $f(t) = g(t) = \log{t} $. The Laplace transform is $F(s) = G(s) = -\frac{\gamma + \log{s}}{s} $. Accordingly, we would need to evaluate the following integral in the complex plane:

$$\begin{align} \int_0^1 dt \, \log{t} \, \log{(1-t)} &= \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{s} \left (\frac{\gamma + \log{s}}{s} \right )^2 \\ &= \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^s \, \left [\frac{\gamma^2}{s^2} + 2 \frac{\log{s}}{s^2} + \frac{\log^2{s}}{s^2} \right ] \end{align} $$

Now, we have three ILTs to evaluate. The first is simple:

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{s t} \frac1{s^2} = t $$

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January 14, 2017 – 2:18 pm|By Ron|Posted in Uncategorized|Comments (0)

Computing an integral over an absolute value using Cauchy’s theorem

The problem is to compute the following integral: $$\int_{-1}^1 dx \frac{|x-y|^{\alpha}}{(1-x^2)^{(1+\alpha)/2}}$$ I will show how to compute this integral using Cauchy’s theorem. It was remarked that it should not be possible to use Cauchy’s theorem, as Cauchy’s theorem only applies to analytic functions, and an absolute value certainly does not qualify. True. Nevertheless, for the […]

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November 29, 2016 – 2:32 am|By Ron|Posted in Uncategorized|Comments (0)

Very nifty limit

Problem: Find the value of the limit $$\lim_{n \to \infty}n\left(\left(\int_0^1 \frac{1}{1+x^n}\,\mathrm{d}x\right)^n-\frac{1}{2}\right)$$ Solution: I chose this problem because the answer is highly nontrivial and just out of left field. But the process of getting there seems so straightforward; it is not really. Substitute $x=u^{1/n}$ in the integral and get $$I(n) = \int_0^1 \frac{dx}{1+x^n} = \frac1n \int_0^1 […]

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November 16, 2016 – 1:43 am|By Ron|Posted in Uncategorized|Comments (5)

Integral of polynomial times rational function of trig function over multiple periods

Problem: Compute $$\int_0^{12\pi} dx \frac{x}{6+\cos 8x}$$ It looks like the OP is trying to compute the antiderivative and use the fundamental theorem of calculus. With multiple periods, that approach is paved with all sorts of difficulty that is really an artifice related to the functional form of the antiderivative. In truth, there should be no […]

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September 18, 2016 – 10:37 pm|By Ron|Posted in Uncategorized|Comments (1)

A method of evaluating a double integral that nobody taught you in school

Many times we are given integrals to evaluate. The standard way to evaluate is to find a series of transformations that will render the integral into something we know how to evaluate and then proceed. Examples of such transformations are substitutions, parts, replacement of an integrand with another integral, reversing order of integration, and so […]

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May 4, 2016 – 3:19 am|By Ron|Posted in Uncategorized|Comments (0)

Deceptively easy product

The problem is to evaluate $$\prod_{n=2}^{\infty} \left (1+\frac{(-1)^{n-1}}{a_n} \right ) $$ where $$a_n = n! \sum_{k=1}^{n-1} \frac{(-1)^{k-1}}{k!} $$ This is one of those cases where trying out a bunch of numbers really isn’t going to help all that much. The $a_n$ look kind of like $n!$, except off by some. Even so, I can find […]

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May 4, 2016 – 2:55 am|By Ron|Posted in Uncategorized|Comments (0)

A sum I can only imagine being evaluated using the residue theorem

The challenge this time is to evaluate the following sum: $$\sum_{n=1}^{\infty} \frac{1}{n^3\sin{\left (\sqrt{2} \, n\pi \right)}}$$ NB (20 Nov 2016) I just got word that Wolfram fixed the problem described below. See below for details. It should be clear that the sum converges…right? No? Then how do we show this? Numerical experiments are more or […]

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February 17, 2016 – 3:14 am|By Ron|Posted in Uncategorized|Comments (5)← Older posts

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