“Integral Milking”

I begin this post with a plea: please don’t be too harsh with this post for being off topic or vague. It’s a question about something I find myself doing as a mathematician, and wonder whether others do it as well. It is a soft question about recreational mathematics – in reality, I’m shooting for more of a conversation.

I know that a lot of users on this site (e.g. Cleo, Jack D’Aurizio, and so on) are really good at figuring out crafty ways of solving recreational definite integrals, like
When questions like this pop up on MSE, the OP provides an integral to evaluate, and the answerers can evaluate it using awesome tricks including (but certainly not limited to):

  • Clever substitution
  • Exploitation of symmetry in the integrand
  • Integration by parts
  • Expanding the integrand as a series
  • Differentiating a well-know integral-defined function, like the Gamma or Beta functions
  • Taking Laplace and Inverse Laplace transforms

But when I play around with integrals on my own, I don’t always have a particular problem to work on. Instead, I start with a known integral, like
π0cos(mx)cos(nx)dx=π2δmn,  m,nZ+
and “milk” it, for lack of a better word, to see how many other obscure, rare, or aesthetically pleasing integrals I can derive from it using some of the above techniques. For example, using the above integral, one might divide both sides by m, getting
π0cos(mx)mcos(nx)dx=π2mδmn,  m,n,kZ+
Then, summing both sides from m=1 to , and exploiting a well-known Fourier Series, obtain
π0cos(nx)ln(22cos(x))dx=πn,  nZ+
or, after a bit of algebra, the aesthetically pleasing result
π/20cos(2nx)ln(sin(x))dx=π4n,  nZ+
After pulling a trick like this, I look through all of my notebooks and integral tables for other known integrals on which I can get away with the same trick, just to see what integrals I can “milk” out of them in the same way. This is just an example – even using the same starting integral, countless others can be obtained by using other Fourier Series, Power Series, integral identities, etc. For example, some integrals derived from the very same starting integral include
and the astounding identity
Everyone seems to be curious about the proof of this last identity. A proof can be found in my answer here.

I just pick a starting integral, and using every technique I know as many times as possible, try to come up with the most exotic integrals as I can, rather than picking a specific integral and trying to solve it.

Of course, integrals generated this way would be poor (or at least extremely difficult) candidates for contest problems or puzzles to evaluate given the integral, since they are derived “backwards,” and determining the derivation given the integral is likely much harder than pursuing the vague goal of a “nice-looking integral” with no objective objective (ha ha).

QUESTION: Do you (residents of MSE who regularly answer/pose recreational definite integral questions) do this same activity, in which you try to generate, rather than solve, cool integrals? If so, what are some integrals you have come up with in this way? What strategies do you use? Does anyone care to opine on the value (or perhaps lack of value) of seeking integrals in this way?



Yes, definitely. For example, I found that
m\int_0^{\infty} y^{\alpha} e^{-y}(1-e^{-y})^{m-1} \, dy = \Gamma(\alpha+1) \sum_{k \geq 1} (-1)^{k-1} \binom{m}{k} \frac{1}{k^{\alpha}}
(and related results for particular values of \alpha) while mucking about with some integrals. Months later, I was reading a paper about a particular regularisation scheme (loop regularisation) useful in particle physics, and was rather surprised to recognise the sum on the right! I was then able to use the integral to prove that such sums have a particular asymptotic that was required for the theory to actually work as intended, which the original author had verified numerically but not proved. The resulting paper’s on arXiv here.

Never let it be said that mucking about with integrals is a pointless pursuit!

Source : Link , Question Author : Franklin Pezzuti Dyer , Answer Author : Chappers

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