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I just remember an observation I made a loong time ago, while learning to use Maple.

Let $f_0$ be the cardinal sine. It is well known that:

$$\int_0^{+\infty} f_0 (t) \ dt = \frac{\pi}{2}.$$

Since the indefinite integral converges, we can define:

$$f_1 (x) := - \int_x^{+ \infty} f_0 (t) \ dt,$$

so that $f_1 (0) = -\pi/2$. But let's do this again! For all $n \geq 0$ and all $x \in \mathbb{R}$, let:

$$f_{n+1} (x) := - \int_x^{+ \infty} f_n (t) \ dt.$$

From numerical experiments, it looks like $f_n$ is well-defined, at least for small values of $n$ (until 5 or so). In addition, the sequence $(f_n (0))_{n \geq 0}$ is remarkable : it begins by $(1, -\pi/2, 1, 0, ...)$. I think there were other $\pi$ later in the sequence, but I don't manage to get these terms at the moment. So, a couple of questions:

  • Is $f_n$ well-defined for all $n$?

  • If so, what is the sequence $(f_n (0))_{n \geq 0}$?

D. Thomine
  • 10,870
  • Kinda off-topic, but would you have a brother called Olivier? If so, I went to university with him :-) – Sylvain Julien Jan 27 '14 at 22:14
  • All terms in the sequence you propose can be expressed in terms of the sine-integral function ${\rm Si}(x)$, polynomials in $x$, and constants. – rajb245 Jan 27 '14 at 22:42
  • @ Sylvain Julien : Yes, I do. Nice to know you studied with him :) – D. Thomine Feb 16 '14 at 22:14
  • @ rabj245 : I see what you mean, and I can formally get these polynomials. However, there are a few holes, due to the fact that the functions are not integrable. I am not sure that the boundary terms I get (while trying to make the formal integration a bit less formal) are negligible. Would you have a reference on the subject, or a nice trick to bypass the woes of improper integrals? – D. Thomine Feb 16 '14 at 22:16

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