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Function $f:(0, \infty)\longrightarrow \mathbb{R}$ is called $\textbf{signomial}$, if $$ f(x)=a_0x^{r_0}+a_1x^{r_1}+\ldots+a_kx^{r_k}, $$ where $k \in \mathbb{N}^*:=\{0,1,2, \ldots\}$, and $a_i, r_i \in \mathbb{R}$, $a_i\neq 0$, $r_0<r_1<\ldots<r_k$, and $x$ is a real variable with $x>0$.

My question is simple in the first glamce, but I cannot get it.

Question: whether function $\displaystyle{\sqrt p \int_0^{\infty}\left(\frac{\sin t}{t}\right)^p}dt$, for $t>0, p\ge 2$ is signomial?

Thank you for your help.

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Can you expand $\sin t$ as a finite polinomial? I assume the $k$ must terminate. If it is not the case, can you expand $\sin t $ as a power series? – Pedro Tamaroff May 7 '12 at 21:49
Since your "function" has no $x$ in it, I guess it is a constant, and thus is certainly signomial. Your integral has no $dt$ or $dp$ or $dq$ or something in it either, making it hard to interpret... – GEdgar May 8 '12 at 0:38
@GEdgar: I was assuming that the integral is over $t$ (since $p$ occurs outside the integral) and that the expression is being considered as a function of $p$. – joriki May 8 '12 at 0:41
I was assuming David should fix the question. – GEdgar May 8 '12 at 0:50
@GEdgar: I completely agree; perhaps I should have said that. – joriki May 8 '12 at 0:55
up vote 0 down vote accepted

This is a non-rigorous derivation of an expansion of the function in inverse powers of $p$. I asked a question here about a rigorous justification for it. It turns out that a) the expansion was known, b) it can be rigorously justified and c) it appears to be only an asymptotic expansion, not a convergent series. However, the conclusion that the function cannot be a signomial remains valid, since the errors of the partial sums of the expansion are bounded such that each term in the expansion would have to be contained in the signomial, which would thus need to have an infinite number of terms.

Let $u=\sqrt pt$. Then

$$ \begin{align} \left(\frac{\sin t}t\right)^p &=\left(1-\frac16t^2+\frac1{120}t^4-\dotso\right)^p \\ &=\left(1-\frac16\frac{u^2}p+\frac1{120}\frac{u^4}{p^2}-\dotso\right)^p \\ &=\left(1+\frac1p\left(-\frac16u^2+\frac1{120}\frac{u^4}p-\dotso\right)\right)^p\;. \end{align} $$


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

(see Wikipedia), we have

$$ \begin{align} \left(\frac{\sin t}t\right)^p &=\mathrm e^{-u^2/6}\left(1+\frac1{120}\frac{u^4}p+\dotso\right)\left(1-\frac1{72}\frac{u^4}p+\dotso\right) \\ &= \mathrm e^{-u^2/6}\left(1-\frac1{180}\frac{u^4}p+\dotso\right)\;, \end{align} $$

where the expansions are in inverse powers of $p$. The expansion cannot terminate, since otherwise the left-hand side would have to exhibit Gaussian decay, which it doesn't. Thus we have

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

with a non-terminating expansion in decreasing powers of $p$. If this were a signomial, the leading term would have to be the leading term of the expansion, then the leading term of the remainder would have to be the leading term of the remainder of the expansion, and so on; thus the expansion cannot be replicated my a finite linear combination of powers of $p$.

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Thank you very much for your answer. – David May 8 '12 at 21:29
@David: You're welcome! – joriki May 8 '12 at 21:29

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