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Consider $$f_p(x)=x^p \exp\left(-x^8\sin^2x\right)$$

I have to show that $f_2\in\mathscr L(0,+\infty)$ whilst $f_3\notin \mathscr L(0,+\infty)$. Now, I am looking at the case $p=2$. The problematic points are $x_n=\pi n$ for $n=1,2,3,\ldots$ since the function peaks there, with $f_p(x_n)=n^2\pi^2$. One has to show that the width of the peaks decreases sufficiently fast so that the function is indeed integrable. Now, one can consider the local bounds at $x_n$ $$\left(\frac{x-n\pi}2\right)^2\leq\sin^2x\leq (x-n\pi)^2$$

whence $$\exp \left( { - {x^8}{{\left( {\frac{{x - n\pi }}{2}} \right)}^2}} \right) \geqslant \exp \left( { - {x^8}{{\sin }^2}x} \right) \geqslant \exp \left( { - {x^8}{{(x - n\pi )}^2}} \right)$$

and then we break up each integral at $n\pi\pm\dfrac \pi 2$.

Thus by a shift $x\mapsto x+n\pi$ one end ups looking at the integrals $$ \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\left( {y + n\pi } \right)}^2}\exp \left( { - {{\left( {y + n\pi } \right)}^8}{y^2}} \right)dx}=A_n \\ \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {{{\left( {y + n\pi } \right)}^2}\exp \left( { - {{\left( {y + n\pi } \right)}^8}{{\left( {\frac{y}{2}} \right)}^2}} \right)dx} =B_n$$

and the integral in question will be the first bit in $[0,\pi/2]$ bounded above and below then by the sums $$\sum_{n\geqslant 1} A_n\;\;;\;\;\sum_{n\geqslant 1} B_n$$

How can I bound those integrals from above and below respectively to prove the function is Lebesgue integrable? Laplace's method comes to mind. Is there a better way?

ADD While we're at it: what is the the infimum and supremum of $$A=\left\{p\in\Bbb R:f_p(x)\in\mathscr L(0,+\infty)\right\}\;\;\rm ?$$

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  • $\begingroup$ Laplace's method seems to work. The essential conclusion is that $\int_{-\pi/2}^{\pi/2} \exp\left(-(y+n\pi)^8 y^2\right)\,dy \sim C/n^4$, so multiplying it by $n^2$ or $n^3$ (depending on whether $p = 2$ or $p=3$) will yield the desired result. Would you like me to work out the details as an answer? $\endgroup$ Aug 6 '13 at 22:51
  • $\begingroup$ @AntonioVargas Pretty pretty please. $\endgroup$
    – Pedro Tamaroff
    Aug 7 '13 at 0:45
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The central estimate.

Let's consider the integral

$$ I(n) = \int_{-\pi/2}^{\pi/2} \exp\left[-A(y+\pi n)^8 y^2\right]\,dy, $$

where $A > 0$ is constant. For convenience we'll define $$f_n(y) = -A(y+\pi n)^8 y^2.$$ Note that $f_n(0) = 0$ and $f_n(y) < 0$ for all $y \in [-\pi/2,\pi/2]\setminus\{0\}$ if $n \geq 1$, so that $f_n(y)$ has a maximum at the point $y=0$ and hence the largest contribution to the integral $I(n)$ comes from a neighborhood of $y=0$.

It can be shown that, for $3 \leq m \leq 10$ and $-\pi/2 \leq y \leq \pi/2$,

$$ |y|^m \leq (\pi/2)^7 |y|^3, $$

so if we define

$$ g_n(y) = f_n(y) + A\pi^8 n^8 y^2 $$

then it follows from the triangle inequality that

$$ |g_n(y)| \leq A 9 \binom{8}{4} (\pi/2)^7 n^7 |y|^3 $$

for $-\pi/2 \leq y \leq \pi/2$. In other words,

$$ f_n(y) = -A\pi^8 n^8 y^2 + O\left(n^7 y^3\right). \tag{1} $$

This suggests that we should focus our attention on the interval $y \in \left[-n^{-7/3},n^{-7/3}\right]$, for in this interval we can write

$$ \exp\left[O\left(n^7 y^3\right)\right] = 1 + O\left(n^7 y^3\right). \tag{2} $$

The contribution from the remaining intervals is exponentially small. Indeed,

$$ \begin{align} \int_{n^{-7/3}}^{\pi/2} \exp\left[-A(y+\pi n)^8 y^2\right]\,dy &\leq \left(\frac{\pi}{2}-n^{-7/3}\right) \exp\left[-A\left(n^{-7/3}+\pi n\right)^8 n^{-14/3}\right] \\ &< \frac{\pi}{2} \exp\left(-An^{10/3}\right), \end{align} $$

and similarly for $\left[-\pi/2,-n^{-7/3}\right]$. Combining this with $(1)$ and $(2)$ we have

$$ \begin{align} I(n) &= \int_{-n^{-7/3}}^{n^{-7/3}} \exp\left[-A(y+\pi n)^8 y^2\right]\,dy + O\left(e^{-An^{10/3}}\right) \\ &= \int_{-n^{-7/3}}^{n^{-7/3}} \exp\left[-A\pi^8 n^8 y^2\right]\,dy + O\left(n^7 \int_{-n^{-7/3}}^{n^{-7/3}} |y|^3 \exp\left[-A\pi^8 n^8 y^2\right]\,dy\right) \\ &\qquad + O\left(e^{-An^{10/3}}\right). \tag{3} \end{align} $$

To these integrals we will add on the tails over the intervals $\left(-\infty,-n^{-7/3}\right)$ and $\left(n^{-7/3},\infty\right)$ and in doing so contribute only an exponentially small error.

We can find a constant $B$ so that $|y|^3 \leq B \exp\left[\frac{A}{2}\pi^8 n^8 y^2\right]$ for all $y$, so that

$$ \begin{align} \int_{n^{-7/3}}^{\infty} |y|^3 \exp\left[-A\pi^8 n^8 y^2\right]\,dy &\leq B \int_{n^{-7/3}}^{\infty} \exp\left[-\frac{A}{2}\pi^8 n^8 y^2\right]\,dy \\ &= Bn^{-7/3}\int_1^\infty \exp\left[-\frac{A}{2}\pi^8 n^{10/3} z^2\right]\,dz \\ &< Bn^{-7/3}\int_1^\infty \exp\left[-\frac{A}{2}\pi^8 n^{10/3} z\right]\,dz \\ &= Bn^{-7/3}\,\frac{2\exp\left[-\frac{A}{2}\pi^8 n^{10/3}\right]}{A\pi^8 n^{10/3}} \\ &= B\,\frac{2\exp\left[-\frac{A}{2}\pi^8 n^{10/3}\right]}{A\pi^8 n^{17/3}} \end{align} $$

This method will also show that

$$ \int_{n^{-7/3}}^{\infty} \exp\left[-A\pi^8 n^8 y^2\right]\,dy = O\left(n^{-17/3} \exp\left[-A\pi^8 n^{10/3}\right]\right), $$

and by symmetry these estimates also hold when integrating over the interval $\left(-\infty,-n^{-7/3}\right)$.

Thus

$$ \begin{align} &\int_{-n^{-7/3}}^{n^{-7/3}} |y|^3 \exp\left[-A\pi^8 n^8 y^2\right]\,dy \\ &\qquad = \int_{-\infty}^\infty |y|^3 \exp\left[-A\pi^8 n^8 y^2\right]\,dy + O\left(n^{-17/3} \exp\left[-\frac{A}{2}\pi^8 n^{10/3}\right]\right) \\ &\qquad = 2\int_0^\infty y^3 \exp\left[-A\pi^8 n^8 y^2\right]\,dy + O\left(n^{-17/3} \exp\left[-\frac{A}{2}\pi^8 n^{10/3}\right]\right) \\ &\qquad = \frac{1}{A^2 \pi^{16} n^{16}} + O\left(n^{-17/3} \exp\left[-\frac{A}{2}\pi^8 n^{10/3}\right]\right) \end{align} $$

and

$$ \begin{align} &\int_{-n^{-7/3}}^{n^{-7/3}} \exp\left[-A\pi^8 n^8 y^2\right]\,dy \\ &\qquad = \int_{-\infty}^{\infty} \exp\left[-A\pi^8 n^8 y^2\right]\,dy + O\left(n^{-17/3} \exp\left[-A\pi^8 n^{10/3}\right]\right) \\ &\qquad = \frac{1}{A^{1/2}\pi^{7/2} n^4} + O\left(n^{-17/3} \exp\left[-A\pi^8 n^{10/3}\right]\right). \end{align} $$

Combining this with $(3)$ and absorbing the superfluous exponential error terms yields

$$ I(n) = \frac{1}{A^{1/2}\pi^{7/2} n^4} + O\left(\frac{1}{n^9}\right). $$

(For another application of this method, see this answer.)

Applying this to the problem at hand.

It remains to show that

$$ \int_{-\pi/2}^{\pi/2} (y+\pi n)^p \exp\left[-(y+\pi n)^8 y^2\right]\,dy \sim \pi^p n^p \left.I(n)\right|_{A=1} $$

and

$$ \int_{-\pi/2}^{\pi/2} (y+\pi n)^p \exp\left[-(y+\pi n)^8 \left(\frac{y}{2}\right)^2\right]\,dy \sim \pi^p n^p \left.I(n)\right|_{A=1/4}, $$

which is left as an exercise (Hint: factor out $\pi^p n^p$ first then bound $$1-\epsilon < (1+y/\pi n)^p < 1+\epsilon$$ for $n$ large enough). We may then conclude by your estimates that $f_p \in \mathscr{L}(0,\infty)$ if $-1 < p < 3$ and $f_p \notin \mathscr{L}(0,\infty)$ if $p \geq 3$ or $p \leq -1$.

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    $\begingroup$ I hope there is a simpler way to solve the problem. $\endgroup$ Aug 7 '13 at 3:46
  • $\begingroup$ I need your powers of estimation here! =D $\endgroup$
    – Pedro Tamaroff
    Aug 9 '13 at 2:44

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