0
$\begingroup$

I would perhaps to this post add problems of a similar kind.

Problem 1. Prove convergence of the integral $\int_0^{\infty} xe^{-x^6\sin^2x}dx$,

From https://www.desmos.com/calculator/sjizhbtbhp we see the function can be best interpreted as follows: $(\frac{1}{e})^{\sin^2x}$ is a periodic function, with 'peak' 1 at $\sin x=0$, 'bottom' 1/e at $\sin x=1$. The wave is greatly shrinked, or each peak becomes greatly narrowed down by taking $x^6$-th power of it (and the greater x is, the narrower each peak becomes, that's key to the integrability); though the peak value is retained for it's 1. The narrowed wave is amplified by x, but that doesn't change the integrability. We need only to show the sum of all narrow peak near x=n$\pi$ converges (for in the tail (x near $\infty$), anywhere else is almost 0 as the result of $x^6$(huge)-th power of a number less than unity.

One possible way is to realize the behavior and size of $\sin(n\pi+h)$ (h small) is exactly like that of $\sin(0+h)$, so we can (by changing independent variable) move the curve near $n\pi$ to near 0. Similar idea may be applied to another problem $\int_{\pi}^\infty \frac{dx}{x^2(\sin x)^{2/3}}$

Proof: $\sin^2(n\pi+h/2) =\frac{1-\cos{(2n\pi+h)}}{2}=\frac{\frac{h^2}{2!}-\frac{h^4}{4!}+\dots}{2}>\frac{h^2}{2\cdot2!}$

When x is near $n\pi$ but no less than h/2 away from it, $\exp\{-\sin^2(x)x^6\}<\exp\{-\sin^2(n\pi-\pi/2)x^6\}<\exp\{-\frac{h^2}{2\cdot2!} (n\pi+h/2)^6\}$,

Let $h<\frac{1}{n^{2+\lambda}}$, where 0<$\lambda<1/2$, then $\sum_n \int_{n\pi-h/2}^{n\pi+h/2} x\exp\{-\sin^2(x)x^6\}<\sum_n h\cdot 1$, which converges (to less than $\frac{\pi^2}{6}$).

$\sum_n(\int_{n\pi-\pi/2}^{n\pi-h/2}+ \int_{n\pi+h/2}^{n\pi+\pi/2}) \exp\{-\sin^2(x)x^6\}<\sum_n\pi \exp\{-\frac{h^2}{2\cdot2!} (n\pi+h/2)^6\}=\sum_n\pi \exp\{-\frac{(\frac{1}{n^{2+\lambda}})^2}{2\cdot2!} (n\pi+h/2)^6\}=\sum_n\pi \exp\{-Kn^{2-2\lambda}\}$, where K is a positive constant, roughly $\frac{\pi^6}{4}$. The series is less than $\sum_n\pi (\frac{1}{e})^{K{n}}$, a geometric series converging to $\frac{\pi}{1-\frac{1}{e^K}}$(possibly plus a constant $C$ representing sum of the beginning iterms).

So the integral is at least bounded to a small neighbourhood of

$$l<\frac{\pi^2}{6}+\frac{\pi}{1-\frac{1}{e^{[\frac{\pi^6}{4}]}}}+C.\blacksquare$$

My questions:

  1. How can one proceed from that to prove the convergence of the integral?
  2. Is there a neater way to prove it?
$\endgroup$
  • 1
    $\begingroup$ Why $x^6$? .... $\endgroup$ – LSpice Aug 13 at 16:18
  • $\begingroup$ I think this question would be more appropriate for math.stackexchange.com. I believe there is some simplification of the answer possible, but the main idea you have (which is to give upper bounds for the integral on the interval $[(n-\frac 12)\pi,(n+\frac 12)\pi]$) is fine. $\endgroup$ – Anthony Quas Aug 13 at 18:08
  • $\begingroup$ What’s the difference between the two sites? $\endgroup$ – Charlie Chang Aug 14 at 3:35
2
$\begingroup$

Call $I$ the integral, then $$I=\sum_{k} \int_{k\pi}^{(k+1)\pi} xe^{-x^6 \sin^2x}dx=\sum_{k} \int_{0}^{\pi} (z+k\pi)e^{-(z+k\pi)^6 \sin^2z}dz=:\sum_k I_k.$$ Then $$ I_k \le (k+1)\pi \int_0^\pi e^{-k^6 \pi^6 \sin^2 z}dz=2(k+1)\pi\int_0^{\pi/2} e^{-k^6 \pi^6 \sin^2 z}dz \le 2(k+1)\pi\int_0^{\pi/2} e^{- 4 k^6 \pi^4 z^2}dz, $$ since $\sin z \ge (2/\pi)z$ in $[0, \pi/2]$. Finally, for $k \ge 1$ $$ \int_0^{\pi/2} e^{- 4 k^6 \pi^4 z^2}dz \le \int_0^{\infty} e^{- 4 k^6 \pi^4 z^2}dz=C k^{-3} $$ which gives the convergence of $\sum_k I_k$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I see it seems that we connect sin0 and sin\pi/2 and get sin z greater than linear function of z which Simplified the problem $\endgroup$ – Charlie Chang Aug 13 at 17:23
  • $\begingroup$ Yes, only the behavior near the zeros of $\sin$ plays a role. $\endgroup$ – Giorgio Metafune Aug 13 at 17:59
  • 2
    $\begingroup$ If the question isn't right for the site, it's best not to answer it... $\endgroup$ – Anthony Quas Aug 13 at 18:09

Not the answer you're looking for? Browse other questions tagged or ask your own question.