Prove convergence of the integral $\int_0^{\infty} xe^{-x^6\sin^2x}dx$ [closed]

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?
• Why $x^6$? .... Aug 13, 2020 at 16:18
• 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. Aug 13, 2020 at 18:08
• What’s the difference between the two sites? Aug 14, 2020 at 3:35

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 $$\sum_k I_k$$.
• Yes, only the behavior near the zeros of $\sin$ plays a role. Aug 13, 2020 at 17:59