This question relates to this answer I gave to a question about the integral

$$\int_0^\infty\left(\frac{\sin t}t\right)^p\mathrm dt\;.$$

I derived an expansion in inverse powers of $p$ and then realized that I don't know how to justify it rigorously or how to determine its radius of convergence. I substituted $u=\sqrt pt$ and applied

$$\left(1+\frac xn\right)^n=\mathrm e^x\left(1-\frac{x^2}{2n}+\frac{x^3(8+3x)}{24n^2}+\dotso\right)$$


$$ \left(\frac{\sin t}t\right)^p=\left(1+\frac1p\left(-\frac16u^2+\frac1{120}\frac{u^4}p-\dotso\right)\right)^p$$

to obtain

$$ \begin{align} \sqrt p\int_0^\infty\left(\frac{\sin t}t\right)^p\mathrm dt &= \int_0^\infty\mathrm e^{-u^2/6}\left(1-\frac1{180}\frac{u^4}p+\dotso\right)\mathrm du \\ &= \sqrt{\frac{3\pi}2}\left(1-\frac{3}{20}\frac1p+\dotso\right)\;. \end{align} $$

(For more details, see the answer.) With the help of Wolfram|Alpha, I worked out further terms:

$$\sqrt p\int_0^\infty\left(\frac{\sin t}t\right)^p\mathrm dt = \sqrt{\frac{3\pi}2}\left(1-\frac{3}{20}\frac1p-\frac{13}{1120}\frac1{p^2}+\frac{27}{3200}\frac1{p^3}+\frac{52791}{3942400}\frac1{p^4}+\dotso\right)\;. $$

I believe all the intermediate series are well-defined and convergent; the problem is that I don't know how to justify interchanging integration and summation at the end. I doubt that the dominated convergence theorem can be applied, since the integrand is $\operatorname{sinc}^p(u/\sqrt p)$, whose expansion I would expect to oscillate about as badly as the power series for $\sin x$, whose partial sums are unbounded.

The last three coefficients given above are roughly of the same order of magnitude, which might indicate that if the series converges at all, it might converge for $p\gtrsim1$. However, while plugging in $p=10$ gives the right result up to six decimal places, the result for $p=2$ is off in the first decimal place, much worse than might be expected from the given terms. That makes me wonder whether this is perhaps just an asymptotic expansion.

So my questions are:

  • How can I justify interchanging integration and summation in the last step?
  • Is the series that I obtained convergent? Or is it just an asymptotic expansion?
  • If it converges, how might I determine the radius of convergence?
  • 2
    $\begingroup$ Have you seen formula 37 here by any chance? $\endgroup$ – J. M. is a poor mathematician May 8 '12 at 16:21
  • $\begingroup$ @J.M.: Thanks, no, I hadn't -- that's rather fascinating. I had worked out formula 38, though, but didn't see how to use it to answer the question (neither mine nor the linked one). Do you think either of these could be of help? $\endgroup$ – joriki May 8 '12 at 16:28
  • 2
    $\begingroup$ I haven't tried it out; maybe I should. You might want to see these two articles as well (unfortunately the AMS site is fritzing on me at the moment so I can't check these two out). $\endgroup$ – J. M. is a poor mathematician May 8 '12 at 16:40
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    $\begingroup$ Interestingly enough, formula 2 in the article in the first link of my previous comment has your expansion to a few more terms than you have, and is attributed to Goddard. $\endgroup$ – J. M. is a poor mathematician May 8 '12 at 18:03
  • $\begingroup$ @J.M.: Thanks, very interesting. It says "Presumably, this is a divergent series (though, since the coefficients are not known in analytical form, it is not easy to see how this could be proved), and Goddard claims it is asymptotic." It also has a discussion of the error term. $\endgroup$ – joriki May 8 '12 at 19:27

Let's first take care of the "tail" of this integral. We have $|\sin(t)/t| \le 1/t$ for $t > 0$, so $$\int_2^\infty |\sin(t)/t|^p \ dt \le \int_{2}^\infty t^{-p}\ dt = \frac{2^{1-p}}{p-1} = O(p^{-k}) \ \text{as} \ p \to \infty$$ for any fixed $k$. Now for the integral from $0$ to $2$, we write $ \sin(t)/t = e^{-U^2(t)/6}$ where $U(t)$ is analytic in $|t| \le 2$ (in fact the closest singularities to the origin are at $t = \pm \pi$) and increasing on $[0,2]$ with $$U(t) = t+{\frac {1}{60}}{t}^{3}+{\frac {139}{151200}}{t}^{5}+{\frac {83}{ 1296000}}{t}^{7}+O \left( {t}^{9} \right) $$ We then have $$\int_0^2 (\sin(t)/t)^p\ dt = \int_0^2 e^{-p U^2(t)/6}\ dt$$ Apply the change of variables $U(t)=u$, where $t = V(u)$ is the inverse function defined in a neighbourhood of $[0,U(2)]$: $$ \int_0^2 (\sin(t)/t)^p\ dt = \int_0^{U(2)} e^{-p u^2/6} V'(u)\ du$$ where $$ V(u) = u-{\frac {1}{60}}{u}^{3}-{\frac {13}{151200}}{u}^{5}+{\frac {1}{336000}}{u}^{7}+O \left( {u}^{9} \right) $$ We can approximate $V'(u)$ by its Maclaurin polynomial $P_k(u)$ of degree $k$, with error bounded by some $K u^{k+1}$ for $u \in [0,U(u)]$. The corresponding error in our integral is then bounded by $\int_0^\infty K e^{-p u^2/6} u^{k+1}\ du = O(p^{-k/2-1})$. And we can approximate $\int_0^{U(2)} e^{-p u^2/6} P_k(u)\ du$ by $\int_0^\infty e^{-pu^2/6} P_k(u)\ du$ with error $O(p^{-k})$ (in fact $O(e^{-\epsilon p})$ for some $\epsilon > 0$). So finally we conclude that an asymptotic expansion of our integral is given by $$\sum_{j=1}^\infty c_{2j} \int_0^\infty e^{-p u^2/6} u^{2j}\ du = \sum_{j=1}^\infty \frac{c_{2j}}{2} \Gamma(j+1/2) \left(\frac{6}{p}\right)^{j+1/2}$$ where $c_{2j}$ are the Maclaurin series coefficients of $V'(u)$.

Here are a few more terms:

$$\eqalign{ \int_0^\infty \left(\frac{\sin(t)}{t}\right)^p\ dt = \sqrt {{\frac {6 \pi }{p}}} &\left( \frac{1}{2} -{\frac {3}{40}}\,{p}^{-1}-{ \frac {13}{2240}}\,{p}^{-2}+{\frac {27}{6400}}\,{p}^{-3}+{\frac {52791 }{7884800}}\,{p}^{-4}\right.\cr & +{\frac {482427}{133120000}}\,{p}^{-5}-{\frac { 124996631}{20070400000}}\,{p}^{-6}-{\frac {5270328789}{272957440000}} \,{p}^{-7}\cr &\left. -{\frac {7479063506161}{536923340800000}}\,{p}^{-8}+{\frac { 6921977624613}{113036492800000}}\,{p}^{-9} + \ldots\right)\cr}$$

Yes, this should be an asymptotic series.

  • $\begingroup$ The derivation is quite similar to what Goddard did in this article, which was referred to in this article (linked to in the comments). $\endgroup$ – J. M. is a poor mathematician May 8 '12 at 18:48
  • $\begingroup$ Thanks for this! It's interesting that these two approaches lead to the same result. Yours comes with a rigorous justification -- can you see how mine might be saved? And when you say "this should be an asymptotic series", is that your impression from the coefficients, or is there a theoretical indication for that? $\endgroup$ – joriki May 8 '12 at 19:32
  • $\begingroup$ Note that if $p$ is a half-integer, the integral will have a nonzero imaginary part due to the intervals where $\sin(t)$ is negative. The sum of a convergent series with real coefficients would not... $\endgroup$ – Robert Israel May 8 '12 at 20:57
  • $\begingroup$ Of course, that the series is asymptotic to, rather than convergent to, the integral is separate from the question of whether the series has a nonzero radius of convergence. I would think that the series for $V'(u)$ has a finite radius of convergence, perhaps around $4$ or $5$ from looking at the first $20$ coefficients, and then the factor $\Gamma(j+1/2)$ would make your series have radius of convergence $0$. $\endgroup$ – Robert Israel May 9 '12 at 0:20

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