The solutions are in! As is so often the case, solvers came up with a number of different approaches; the one which follows broadly represents the type of solution that came up most often. I’ll mention some others in the remarks below, some of it connected with the lore of Richard Feynman.

*Solution by Nilay Vaish*: The answer to POW-10 is . Put

and integrate by parts: we have

The first term vanishes by a simple application of L’hôpital’s rule. We now have

where the second equation follows from , and the general elementary fact that Adding the last two displayed equations, we obtain

This gives

after a simple substitution. The last integral splits up as two integrals:

but these two integrals are equal, using the identity together with the general integration fact cited above. Hence the two sides of (3) equal

recalling equation (1) above. Substituting this for the last integral in equation (2), we arrive at

whence we derive the value of the desired integral .

**Remarks:**

1. A number of solvers exploited variations on the theme of the solution above, which could be summarized as involving symmetry considerations together with a double angle formula. For example, Philipp Lampe and Paul Shearer in their solutions made essential use of the identity

in conjunction with the complementarity , and the general integration fact cited above.

2. Arin Chaudhuri (and Vishal in private email) pointed out to me that the evaluation of the integral

is actually fairly well-known: it appears for example in Complex Analysis by Ahlfors (3rd edition, p. 160) to illustrate contour integration of a complex analytic function via the calculus of residues, and no doubt occurs in any number of other places.

3. Indeed, Simon Tyler in his solution referred to this as an integral of Clausen type, and gave a clever method for evaluating it: we have

which works out to

The last integral can be expanded as a series

where the summands for odd vanish; the other terms can be collected and then resummed to yield

and by substituting this for the integral in (*), the original integral is easily evaluated.

4. Arin C. later described still another method which he says he got from an exercise in a book by Apostol — it’s close in spirit to the one I myself had in mind, called “differentiation under the integral sign”, famously referred to in *Surely You’re Joking, Mr. Feynman!*. As Feynman recounts, he never really learned fancy methods like complex contour integration, but he did pick up this method of differentiating under the integral sign from an old calculus book. In typical Feynman style, he writes,

“It turns out that’s not taught very much in the universities; they don’t emphasize it. But I caught on how to use that method, and I used that one damn tool again and again. So because I was self-taught using that book [Wilson's Advanced Calculus], I had peculiar methods of doing integrals. The result was, when guys at MIT or Princeton had trouble doing a certain integral, it was because they couldn’t do it with the standard methods they had learned in school. If it was contour integration, they would have found it; if it was a simple series expansion, they would have found it. Then I come along and try differentiating under the integral sign, and often it worked. So I got a great reputation for doing integrals, only because my box of tools was different from everybody else’s, and they had tried all their tools on it before giving the problem to me.”

So, what is this method? Wikipedia has a pretty good article on it; the idea is to view a given definite integral as an instance of a smoothly parametrized family of integrals, where the extra parameter is inserted at some judicious spot, in such a way that has an easy-to-integrate derivative. Then one figures out , and evaluates it at the which yields the original definite integral.

The best way to understand it is through an example. Pretend you’re Feynman, and someone hands you

[I suppose there may be a contour integration method to handle that one, but never mind -- you're Feynman now, and not supposed to know how to do that!] After fiddling around a little, you find that by inserting a parameter in the exponent,

this has a very simple derivative:

i.e., by differentiating under the integral sign, you manage to kill off that annoying factor in the denominator:

We therefore have

and notice that from the original definition of , we have . Thus , and the problem integral evaluates to . **Bam!**

It takes a little experience to judge where or how to stick in the extra parameter to make this method work, but the Wikipedia article has some good practice problems, including the integral of POW-10. For this problem they recommend considering

and I’ll let you the reader take it from there.

The point is not that this method is much more elegant than others, but that a little practice with it should be enough to convince one that it’s incredibly powerful, and often succeeds where other methods fail. (Not always, of course, and Feynman had to concede defeat when in response to a challenge, his pal Paul Olum concocted some hellacious integral which he said could *only* be done via contour integration.)

Doron Zeilberger is also fond of this method (as witnessed by this paper). Something about it makes me wonder whether it’s secretly connected with the idea of “creative telescoping” and the powerful methods (e.g., see here) developed by Gosper, Wilf, Zeilberger, and others to establish identities of hypergeometric type. But I haven’t had time to consider this carefully.

Also solved by Arin Chaudhuri, Philipp Lampe (University of Bonn), Paul Shearer (University of Michigan), and Simon Tyler. Thanks to all who wrote in!

## 16 comments

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October 12, 2008 at 8:44 am

Américo TavaresCongratulation to all solvers!

I failed to prove the answer, although trying during some hours the integration by parts and contour integration.

Now that others solved it and the explanations are here it looks like as if it were simple.

Very interesting and instructive. I hope to be able to apply these techniques in the future.

October 12, 2008 at 9:15 am

Américo Tavaresseveral hours instead of some hours

October 12, 2008 at 1:41 pm

Todd TrimbleSimple, maybe, but I still think some of those solutions are pretty clever! Which in a sense can be annoying — the purpose of a

calculusshould be to mechanize the process of making a calculation (in the same sense that high school students can compute certain areas and volumes; it doesn’t take an Archimedes or any especial cleverness).At the end of the most recent post at the Secret Blogging Seminar, there’s a great quote by Knuth, “The aim of mathematics is to reduce inspiration with perspiration.”

October 12, 2008 at 2:45 pm

Américo TavaresPretty clever indeed. You just said what I wanted to have said, if I had the mathematical and the English knowledge to do it.

October 13, 2008 at 2:55 pm

Arin CThe ultimate goal of mathematics is to eliminate any need for intelligent thought.

Alfred N. Whitehead

October 13, 2008 at 3:53 pm

Todd TrimbleExactly. Another Whitehead quote: “Civilization advances by extending the number of important operations which we can perform without thinking of them.”

October 16, 2008 at 3:45 pm

Paul ShearerI don’t particularly care for the method I used, it was just the first thing that worked. Although I do wonder if this sort of symmetry argument has any connections with differentiating under the integral sign or contour integration.

I’m also curious just how much one can do with differentiation under the integral sign. Certainly you could generate a managerie of seemingly impossible integrals just by taking a familiar one, inserting a parameter, differentiating with respect to it, then evaluating the derivative at a given point. Perhaps it would be interesting to investigate the set of functions generated by this process. Just how big is it? Can it be characterized in any way? Could some kind of search algorithm attempt to systematically determine whether an integral can be evaluated by differentiation under the integral? As a first try, it might insert a parameters in a given spot, differentiate with respect to it, and algebraically manipulate the results until it finds something it already knows how to integrate. If it fails, it tries to insert the parameter in a different spot. Who knows, maybe Mathematica already does something to this effect…

October 16, 2008 at 7:40 pm

Todd TrimbleThose are great questions, Paul, and I have a feeling they lead quickly into some pretty deep waters and unsettled questions. One question is what counts as an “answer” to evaluating a definite integral; e.g., should we allow roots of Bessel functions or something as possible answers, or do we want just numbers in “closed form”, whatever that might mean.

Some years back, Timothy Chow wrote a very nice article on what counts as a number in closed form, which appeared in the American Mathematical Monthly. Some idea of the depth of this topic can be gleaned from the interesting connections Chow makes with Schanuel’s conjecture.

Just a few months ago, Chow was also putting questions to sci.math.research on just how powerful differentiation under the integral sign is, but unfortunately this didn’t generate much discussion. My own gut feeling is that the search space for deciding when this method “works” is infinitely large, and that there are undecidability issues lurking nearby. If anyone in the world is equipped to tackle such questions, it’s probably Chow, who has a substantial background and interest in formal logic. Perhaps in a few years we’ll know more!

December 3, 2008 at 12:34 am

Cálculo automático de um integral difícil que me resistiu aos métodos matemáticos usuais « problemas | teoremas[...] o método extremamente engenhoso que Nilay Vaish, autor da solução do integral do problema POW-10 (problem of the week 10, de Todd and Vishal’s blog), usou para lá chegar — obteve uma equação linear em relação a esse integral [...]

December 5, 2008 at 12:26 am

Américo TavaresIn Example 3.3 of INTEGRATION: THE FEYNMAN WAY by an ANONYMOUS one finds also the same advice .

I proceed as follows

The difficult part now is how to integrate which is not detailed in the article.

After knowing the result, I worked backwards. By splitting the integrand intowe can evaluate

Hence

(due to the integration constant is )

and

The original integral is therefore

Does anyone know another way of evaluating ?

December 5, 2008 at 12:48 am

Américo TavaresSmall typo

December 5, 2008 at 3:24 am

Todd TrimbleAmérico,

To integrate

I think I wound up substituting for , so that we get

and this can be handled with partial fraction techniques, and so on. Not the most elegant solution, but as we discussed above, the idea is to try to make the solutions mechanical. What what success, it’s not clear!

December 5, 2008 at 3:12 pm

Américo TavaresDear Todd

Thanks for reminding me that “the idea is to try to make the solutions mechanical”!

I should have thought of the standard substitution . I checked the partial fractions technique you mentioned which gives

Of course the integral has the same value as before

December 5, 2008 at 6:01 pm

Integração pelo método de diferenciação em relação a um parâmetro « problemas | teoremas[...] a substituição recomendada por Todd Trimble transforma-se este integral [...]

July 25, 2009 at 8:39 pm

A quick integral « Delta Epsilons[...] Todd Trimble posted solutions to similar integrals, which use the result of this post as a lemma, here. [...]

March 8, 2010 at 10:21 pm

Américo TavaresHi!

I invite you and your readers to

Prove or disprove:.Posted here (My Problem of the month #4).

Américo