A little perspective on reaching 100K Rep on Math.SE

I stumbled onto math.stackexchange.com) on December 16, 2012. Before that, I was occasionally on the prowl for Putnam exam prep questions, various university math contests, and stuff like the IBM monthly puzzle.  At this point 1132 days later, I forget what I was looking for in the first place – likely a new problem to solve.  But stumble I did, and it has changed the way I think about doing and teaching math.

So I reached some sort of milestone by attaining a 100K rep score on Math.SE.  I guess some readers here may know what I mean by a “rep score,” and others have no clue what that is.  So let me explain a little.  Stack Exchange is a network of Question & Answer sites in which people can ask questions about any of a plethora of topics (e.g., computer programming, physics, Judaism, home repair, personal finance, patents,…, and mathematics).    Within each topic, experts answer questions posed by anyone.  If the person who posed the question (the “OP”) likes your answer best of all of the answers provided, the OP may “accept” your answer and you get +15 added to your rep score.  If people in the community like your answer, you get +10 added for each “up vote.”  One may also get an answer downvoted and lose -2 rep score.  One may also get a question upvoted (+5) or downvoted (-1).  There are other ways to gain and lose rep which I will discuss in a bit, but the above are the main ways people gain or lose rep.  Oh, and your rep scores are separate for each topic.

It is pretty easy to review your history on Stack Exchange to understand all that you accomplished.  An actually, looking at my very first post, I now know why I came upon Math.SE: I was stuck on an integral and could not find anyone else who had done it.  This is how I came across Math.SE.  Fortunately, I knew LaTeX and so typing the problem into the interface was amazingly easy.  Best of all, unlike the other LaTeX editors I had worked with before, you could see what you were actually typing in the preview pane below.

Looking back at this first post, I am amazed at how easily I gave up on this integral.  It is a sign of how quickly I learned from other talented people who make the site great.  Over the ensuing months, I would learn more and more until I became one of the experts.  I then got to watch other new people adopt the skills I learned and become experts themselves.  It has been incredibly satisfying.

I also can see that I provided my first answer the very next day, December 17th.  Well, it wasn’t very much of an answer – it was me telling the OP that in my vast experience that I knew of no solution to his/her problem.  Despite its utter uselessness, the answer got 2 upvotes.  I would provide more useful answers subsequently, some of which didn’t even get 2 upvotes.  Then again, some of them got a lot more than 2.

On January 1, 2013, I asked my second question which was completely original.  The question was based on a problem I had done out of The Green Book of Mathematical Problems by Hardy and Williams (#63) involving the minimization of the value of an integral.  I was able to evaluate the integral and then a generalization of it, but I realized that the result possessed a symmetry that wasn’t present in the integral itself.  Normally, that would be a red flag about the result, but it wasn’t – the result was correct.  I spent untold hours manipulating the integral to find the source of the symmetry…and I was genuinely stumped.  So I posed the question to the Math.SE community – and to this day, they remain stumped with me.  They remain stumped not because the question got no attention – the question got a “great question” gold badge and gets views and upvotes to this day 3 years later – but because many people have not been able to figure it out.  I am optimistic that someone will.

Throughout my first year, I became completely addicted to Math.SE.  I became obsessed with observing integration and general problem-solving techniques that I had not known before.  Most importantly, I gained a new appreciation of the residue theorem and Cauchy’s theorem in general.  Of course, after two courses on complex analysis At UMass – including a graduate-level one with the great Sam Holland – one would think that I had the stuff down cold.  Not really.  I was shaky with regard to branch point singularities and in particular how to define branch cuts to maintain the analyticity of an integrand.  I realized that Math.SE was the perfect place to build up my knowledge base and learn how to evaluate a whole new class of integrals I could not evaluate before.  I spent nearly all of my free time working out these integrals, observing good ways and bad ways of attacking them.  After a while, it seemed that I was ubiquitous in the contour-integration tag because I was determined to hone (and show off) my newfound skills.

Here are some significant answers in that first year which influenced my problem-solving strategies:

$$\int_0^1 \frac{dx}{\log{x}} \frac{1-x}{1+x} $$

Here the strategy was clear, but proving that the steps I took to obtain the correct answer were justified were, as I learned the hard way, well above my then-pay grade.  I kept the comments intact as a lesson for me and anyone else who thinks that evaluating integrals is merely about manipulating symbols.  I would refer to this experience numerous times later.  Also note that just because an answer has lots of upvotes does not mean that it was 100% correct from the get-go.

$$\int_0^1 dx \, \left (\frac1{\log{x}} + \frac1{1-x} \right )^2 $$

I was able to transform this challenging integral in terms of a sum that I could not evaluate, but others on the site could.  This was one of the most positive experiences I had on the site, where three of us worked together as a team to evaluate the integral.  I wish I could see more of this.

$$\int_0^{\pi/2} dx \, x^2 \sqrt{\tan{x}} \sin{2 x} $$

This integral looked impossible.  I was happy just to manipulate it into a form in which Mathematica could produce something intelligible.  The exact result, happily, was expressible in terms of $\pi$’s and logs.  A year later, another user performed an analytical evaluation of the reduced integral I derived based on Cauchy’s theorem.

  • 27 Jan 2013: Evaluating

$$\int_0^{\infty} dx \frac{x-\sin{x}}{(x^2+\pi^2) x^3} $$

Here is a nontrivial application of Parseval’s theorem, a technique I found used surprisingly sparsely on the site, but one hammered home to us in Prof. Holland’s Applied Analysis courses.

  • 30 Jan 2013: Closed form, without complex analysis, of

$$\int_0^{\infty} dx \frac{\cosh{ax} \, \cosh{b x}}{\cosh{\pi x}} $$

This was a bit if a challenge – in fact, the integral is also a challenge using complex methods.  But here, the easily-generated series turned into what looked like a mess.  That is until the summand was manipulated into a form in which the symmetry was apparent and the sums then evaluated in closed form.  There would not be many cases in which this symmetry would show up unfortunately.

$$\int_0^{\pi/3} dx \, \log^2{\left [\frac{\sin{x}}{\sin{(x+\pi/3)}} \right ]} $$

The evaluation of this integral was what I would consider my first masterpiece on Math.SE.  Actually, looking back, the contour integral is pretty bland stuff.  But the substitution that gets us there is pure magic.  At this point, I was really hooked.

$$\sum_{n=1}^{\infty} \frac{\sinh{\pi}}{\cosh{2 n \pi} – \cosh{\pi}} $$

This was a bit hairier than I expected, but produced a beautiful result that I used over and over again.

$$\frac1{i 2 \pi} \int_{a-i \infty}^{a+i \infty} ds \, s^{-1/2} e^{s t} = \frac1{\sqrt{\pi t}} $$

This was the first of many Inverse Laplace transforms I did involving transforms that had nasty branch point singularities.  These became my favorite problems to work out.

  • 4 Mar 2013: Evaluate

$$\int_0^1 dx \, h_n(x) $$


2x&\left(0\leq x\leq \frac{1}{3}\right)\\
\frac{1}{2}x+\frac{1}{2} &\left(\frac{1}{3}<x\leq 1\right)
and $h_2(x)=h(h(x)), h_3(x)=h_2(h(x)),\cdots, h_{n+1}(x)=h_n(h(x))$.

This problem was both a triumph and an incredibly frustrating experience.  The problem itself was really cool and the ultimate answer was simple yet nontrivial.  However, the OP accepted an incorrect answer whose author admitted that he didn’t know what he was doing.  This is where I learned that Math.SE can be, like any democracy, an inherently chaotic place.  Fortunately, a little sunlight that was shone on the problem righted the ship.  Lesson: things do work well enough if you have patience.

$$\sum_{k=-\infty}^{\infty} \left (\frac{\sin{k}}{k} \right )^2 = \pi$$

This was actually a nontrivial application of Parseval’s theorem.  This is one of those solutions that looks very simple to understand but was a little tough to formulate.  I find that people here appreciate such solutions and up vote frequently – this was my most-upvoted solution for a few months.  I also found that when a solution gets a lot of upvotes, many other solutions follow.

  • 1 Apr 13 and 4 Apr 13: Inverse Laplace transform of $e^{-\sqrt{s}}$ and a solution to the heat equation on the half-line.

Kind the same problem, sort of.  I would use the contour diagram I generated here in Microsoft Visio for many subsequent problems until I figured out an easy way to generate them in Mathematica.

  • 17 Apr 2013: Why do we choose the coefficients of the Fourier Series that way?

Basic question, but I think Prof. Holland would be proud of the way I explained it.

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

The evaluation of this integral purely by Cauchy’s theorem was a breakthrough of sorts for me.  The colocation of a branch point and pole makes this one a challenge in the complex plane.  I learned here to embrace the divergences and have faith that they eventually would cancel.  I have been able to apply similar techniques to many subsequent problems.


The answer of course is yes.  Actually, the analysis is pretty straightforward if you know Gamma functions cold.  I do not.  But somehow this problem worked itself out beautifully and allowed for a simple explanation.  Maybe that explains the upvotes a lack of other attempts.  This was my most-upvoted answer for almost 6 months.

  • 31 May 2013: Compute the Fourier transform of $\operatorname{sinc}^3{\pi t}$.

Always satisfying to see the same result come from both Cauchy’s theorem and the convolution theorem.

  • 10 Jun 2013: Is there a closed form for this integral:

$$\int_0^1 dt \, t^{-3/2} (1-t)^{-1/2} e^{-a^2/t – b^2/(1-t)} $$

This horrible-looking integral became pretty tame when I looked back to my work computing inverse Laplace transforms.  Then all we needed to do was recognize the convolution and apply the convolution theorem – the integral became quite easy from there.

$$\sum_{j=2}^\infty \prod_{k=1}^j \frac{2 k}{j+k-1} = \pi$$

This is the first of many problems where the Maclurin expansion for $\arcsin{x}/\sqrt{1-x^2}$ came in handy.

$$\displaystyle\int_{0}^{\infty}e^{-ax}\sin(ax)\left(\cot(x)+\coth(x)\right)dx=\frac{\pi}{2}\cdot \frac{\sinh(\pi a)}{\cosh(\pi a)-\cos(\pi a)}, \;\ a\in \mathbb{N}$$

This integral looked innocent enough, but the choice of integrand and contour were far from trivial.  The subsequent analysis was as difficult as any I had done in Math.SE up to this point.

$$\frac{\log{(1+x^3)}}{(1+x^2)^2} $$

It doesn’t get much messier than this, with long, complex expressions whose imaginary parts delicately canceled.  But still, the application of complex techniques to transform the log integrals into much simpler integrals worked like a charm here.  Very satisfying.

  • 2 Aug 2013: A solution to the advection-diffusion equation on a semi-infinite domain.

The application of Cauchy’s theorem resulted in a Parseval-like integral but with three Fourier transforms.  And yet somehow I was able to stick it out and reduce it to error functions.

$$\int_0^{\infty}\frac{e^{-a x^2(x^2-\pi^2)}\cos(2\pi a x^3)}{\cosh x}dx=\frac{\pi}{2}e^{-\pi^4 a/16}$$

This one was done using Cauchy’s theorem and a strange choice of integrand.  If I hadn’t been already aware, I was learning that expertise in using Cauchy’s theorem to evaluate integrals requires a lot of creativity.

$$\int_0^\infty\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt{x}\right)\ e^{-x\sqrt{2}}}{x}dx $$

The answer of course is yes.  This one was an unexpected triumph for Parseval and it was amazing how simple the resulting integral ended up being.  The hard part ended up being simplifying the logs and square roots after evaluating the integral.

$$\int\limits_0^\pi{\frac{{{x^2}}}{{\sqrt 5-2\cos x}}}\operatorname d\!x =\frac{{{\pi^3}}}{{15}}+2\pi \ln^2 \left({\frac{{1+\sqrt 5 }}{2}}\right)$$

I chose to evaluate this using the residue theorem.  The result was a new contour that I had never seen before: a keyhole unit circle.  Amazing how it all worked out, but it did require manipulating dilogs with which I was not facile.  So this problem held a lot of new lessons for me that I would apply to many problems later.

$$\int_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx = \frac{\pi}{2 \log^2{2}}$$

This one was a lot easier than it looked.  However, it was easy to predict that this answer would generate a lot of upvotes because it was a complicated-looking problem with a simple solution.  It did.

$$\int_0^1\frac{x^2-2\,x+2\ln(1+x)}{x^3\,\sqrt{1-x^2}}\mathrm dx=\frac{\pi^2}8-\frac12$$

If I had to pick a single transformative month for me at Math.SE, it was November 2013.  Here I learned that the arcsine function makes for a convenient integrand.  This was also a great demonstration of the, um, power of power series.

$$\int_{-1}^1\frac1x\sqrt{\frac{1+x}{1-x}}\ln\left(\frac{2\,x^2+2\,x+1}{2\,x^2-2\,x+1}\right)\ \mathrm dx $$

I don’t even know where to begin on this one.  I spent about 12 hours and half a pad of paper getting it right.  In the meantime, some user just posted the result without showing any work, infuriating many in the community, including me. (I admit overreaction.) Then, after I posted, the upvotes poured in over next several days…and never stopped.  I found out that two separate deluges followed Reddit discussions pointing out this solution.

As a solution, this is a favorite of mine, but it really is dependent on the symmetry of the posed problem.  What I think made this a popular solution was the fact that, if we know the residue theorem, there are whole new classes of fact patterns that allow for evaluating integrals in closed form.  So the manipulations of the integrand were engineered to achieve a form for the integral that allowed for evaluation in the complex plane.  Then, even when the resulting integrand required the solution of an 8th-degree equation, it turned out this equation had roots expressible in simple, closed form.  It couldn’t have worked out better.

$$\large I=\int_0^\infty{_1F_2}\left(\begin{array}{c}\tfrac12\\1,\tfrac32\end{array}\middle|-x\right)\frac{dx}{1+4\,x}$$

Before this, I avoided hypergeometric-looking problems.  With this one, I realized that the were not all that hard, at least for lower-order cases.  I also recall getting an initial downvote after posting the final answer and getting very upset – I think I was receiving serial downvotes around then.  I learned that downvotes are a fact of life in Math.SE and it is not worth getting upset over.

I think that’s enough to illustrate.  I have now posted 2,672 answers since that first day.  I look forward to posting many more.





Systematic treatment of a deceptively messy Cauchy principal value integral

The problem here is to evaluate the following: $$PV \int_0^{\infty} dx \frac{\log^2{x}}{(x-1)^2(x-4) \sqrt{x}} $$ This can be done using complex analysis, but it is a pretty involved affair, deceptively so. After struggling with the problem of how to present the solution, I am going to lay out a systematic approach that ignores the motivation behind […]

Cauchy principal value of a convolution

The problem here is to compute the following convolution-type integral: $$\int_{-1}^1 dx \frac{\sqrt{1-x^2}}{\lambda-x}$$ When $-1 \lt \lambda \lt 1$, this integral is infinite, but its Cauchy principal value may be defined. This integral is interesting because of the branch points. And so, away we go… Consider the following contour integral: $$\oint_C dz \frac{\sqrt{z^2-1}}{\lambda-z} $$ where […]

Computing the Convolution of Two Pulses: Graphical vs Analytical

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Integral with two branch cuts II

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Integral with two branch cuts

The problem is to evaluate the following integral: $$\int_{-1}^1 dx \frac{\log{(x+a)}}{(x+b) \sqrt{1-x^2}} $$ where $a \gt 1$ and $|b| \lt 1$. It should be obvious to those who spend time around these integrals that this integral does not converge as stated. However, we only have a simple pole at $x=-b$ so that we can compute […]

Fascinating Fourier Transform

Sometimes I come across a Fourier integral that I have no idea how to attack. And then I find that I can convert it to another, more familiar integral using complex analysis. So here’s an example of such a satisfying situation. The problem is to evaluate $$\int_{-\infty}^{\infty} dx \, (1+i a^2 x)^{-1/2} (1+i b^2 x)^{-1/2} […]

Cauchy Principal Value

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Inverse Laplace Transforms and Delta Functions

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Adventures in integration, University Edition

I got a request from an old friend whom I didn’t even know existed. He is the son of my grand-advisor, if that makes any sense. He teaches, among others, a course in Real Analysis at a university in Australia. I know I must like this guy because he says stuff like this: Currently, within the School […]

Inverse Laplace Transform with a coinciding pole and branch point

Recently, the following Laplace transform was asked to be inverted: $$F(s) = s^{-a-1} e^{-s^a} $$ where $a \in (0,1)$. This is a tough problem for two reasons. One is that there is very little chance of there being an analytical result for arbitrary values of $a$. The other, however, is more subtle: there is a […]

The art of using the Residue Theorem in evaluating definite integrals

As many of you know, using the Residue Theorem to evaluate a definite integral involves not only choosing a contour over which to integrate a function, but also choosing a function as the integrand. Many times, this is an easy task when integrating, say, rational functions over the real line. Sometimes there are less trivial […]

A Tale of Two Integrals

Two integrals look almost the same, and even to those fairly well-versed in the art, are the same. But alas, as we shall see. Consider the integral $$I_1 = \int_0^{\pi} dx \frac{x \sin{x}}{1+\cos^2{x}} $$ This may be evaluated by subbing $x \mapsto \pi-x$ as follows: $$\begin{align} I_1 &= \int_0^{\pi} dx \frac{(\pi-x) \sin{x}}{1+\cos^2{x}} \\&= \pi \int_0^{\pi} […]

Weird integral whose simplicity is only apparent in the complex plane

Every so often we come across an integral that seems absolutely impossible on its face, but is easily attacked – in fact, is custom designed – for the residue theorem. I wonder why a first year complex analysis class doesn’t show off this case as Exhibit A in why the residue theorem is so useful. […]

Expansions of $e^x$

A very basic question was asked recently: What is a better approximation to $e^x$, the usual Taylor approximation, or a similar approximation involving $1/e^{-x}$? More precisely, given an integer $m$, which is a better approximation to $e^x$: $$f_1(x) = \sum_{k=0}^m \frac{x^k}{k!} $$ or $$f_2(x) = \frac1{\displaystyle \sum_{k=0}^m \frac{(-1)^k x^k}{k!}} $$ The answer is amazingly simple: […]

Real evaluation of an improper log integral

The problem posed in M.SE concerns real methods of evaluating $$\int_0^{\infty} dx\frac{\log(x)}{\cosh(x) \sec(x)- \tan(x)} $$ The place I started is the nifty result, proven here, that $$\frac{\sin{x}}{\cosh{t} – \cos{x}} = 2 \sum_{k=1}^{\infty} e^{-k t} \sin{k x} $$ Of course, the integral actually looks like $$\int_0^{\infty} dx \frac{\cos{x}}{\cosh{x} – \sin{x}} \log{x} $$ so we need to […]

Generalizing an already tough integral

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Mathematica v9.0.1 states that this integral does not converge.

The problem is to evaluate the following integral: $$\int_0^{\infty} dx \frac{\log(1+x)}{\left(1+x^2\right)\,\left(1+x^3\right)}$$ This one turned out to be messy but straightforward. What did surprise is the way in which I would need to employ the residue theorem. Clearly, the integral is more amenable to real methods than a contour integration. What happens, though, is that the […]

An unusually alternating sum

Problem: evaluate the following sum… $$\sum_{k=0}^{\infty}\dfrac{(-1)^{\frac{k(k+1)}{2}}}{(2k+1)^2}$$ This is unusual because the $-1$ is raised to the $k (k+1)/2$ power, rather than the usual $k$th power. On the surface, this problem may look hopeless, but really, it is all about determining the pattern of odd and even numbers from the sequence $k (k+1)/2$, which turns out […]

Algebraically difficult integral

Well, some integrals are not all that hard to evaluate in principle. The one I am posting here should be an open and shut application of the residue theorem, using the unit circle as a contour. The form of the integrand, however, should give a little pause. It turns out that actually computing residues on […]

An integral involving a quadratic phase

This one was first posted on the site Integrals and Series and was brought to my attention on M.SE by Cody. My solution involves a contour integration, although the approach is far from trivial. Yet again, the solution boils down to finding a contour and a function to integrate over the contour. The problem involves […]

An integral that illustrates the beauty of contour integration methods

I attack many of the integrals in this blog using contour integration methods, some of which are obvious and some of which take a little more imagination. Here is one that illustrates the beauty an power of such methods as much as any other integral here. The problem is to derive the integral representation $$\sin […]