Convergence of $ \int_ 0^\infty x^2\sin(x^4)dx$

Question

Prove $ \int_ 0^\infty x^2\sin(x^4)dx$ converges.

It seems really close to the form of Fresnel integral $ \int_ 0^\infty \sin(x^2)dx$, which made me suspect it works the same.

Furthermore after entering some integrals in wolfarm alpha with some modification it seems that the integral $ \int_ 0^\infty x^k\sin x^{2n} dx$ converges when $k < 2n$, so an explanation why its starts to diverge when $k > 2n$ would be appreciated.

Answer 1

Showing Convergence $$ \begin{align} \int_0^\infty x^2\sin\left(x^4\right)\,\mathrm{d}x &=\frac14\int_0^\infty x^{-1/4}\sin(x)\,\mathrm{d}x\tag{1}\\[4pt] &=\lim_{x\to\infty}\frac{1-\cos(x)}{4x^{1/4}}-\lim_{x\to0}\frac{1-\cos(x)}{4x^{1/4}}+\frac1{16}\int_0^\infty\frac{1-\cos(x)}{x^{5/4}}\,\mathrm{d}x\tag{2}\\[4pt] &=\frac1{16}\int_0^\infty\frac{1-\cos(x)}{x^{5/4}}\,\mathrm{d}x\tag{3} \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto x^{1/4}$
$(2)$: integrate by parts
$(3)$: the limits are zero

The integral converges by comparison to $\frac1{16}\int_0^1\frac{\frac12x^2}{x^{5/4}}\,\mathrm{d}x+\frac1{16}\int_1^\infty\frac2{x^{5/4}}\,\mathrm{d}x$

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Evaluation $$ \require{cancel} \newcommand{\Im}{\operatorname{Im}} \begin{align} \int_0^\infty x^2\sin\left(x^4\right)\,\mathrm{d}x &=\frac14\Im\left(\int_0^\infty z^{-1/4}e^{iz}\,\mathrm{d}z\right)\tag{4}\\ &=\frac14\Im\left(\int_0^{i\infty} z^{-1/4}e^{iz}\,\mathrm{d}z\right)\tag{5}\\ &=\frac14\Im\left(e^{i3\pi/8}\int_0^\infty z^{-1/4}e^{-z}\,\mathrm{d}z\right)\tag{6}\\[3pt] &=\frac14\sin\left(\frac{3\pi}8\right)\Gamma\left(\frac34\right)\tag{7} \end{align} $$ Explanation:
$(4)$: substitute $x=z^{1/4}$ and $\sin(z)=\Im\left(e^{iz}\right)$
$(5)$: the integral along $[0,R]\cup\cancel{Re^{i\pi[0,1/2]}}\cup[iR,0]$ is zero
$(6)$: substitute $z\mapsto iz$
$(7)$: evaluate

Answer 2

There is clearly no convergence issue near $0$ so we need only consider the convergence of $\int_ 1^\infty x^2\sin(x^4)dx$.

Let $M>1$. Note that $$\int_ 1^M x^2\sin(x^4)dx =\frac 14 \int_1^{M^4} \frac{\sin(x)}{x^{1/4}}dx=\frac14\cos(1)-\frac14\frac{\cos(M^4)}{M^{1/4}}-\frac{1}{16}\int_1^{M^4}\frac{\cos(x)}{x^{5/4}}dx$$

• $\dfrac{\cos(M^4)}{M^{1/4}}$ goes to $0$ as $M\to \infty$
• $\int_1^{M^4}\frac{\cos(x)}{x^{5/4}}dx$ is absolutely convergent, hence convergent.

Hence the existence of $\lim_{M\to \infty} \int_ 1^M x^2\sin(x^4)dx$.

Answer 3

It can be done with only one integration by parts: for every $M>1$ one has $$ \int_1^M x^2 \sin(x^4)\, dx = \frac{1}{4} \int_1^M \frac{1}{x}\, [4x^3\sin(x^4)]\, dx = -\frac{\cos M^4}{4M} + \cos 1 -\frac{1}{4} \int_1^M \frac{\cos x^4}{x^2}\, dx. $$ The last integral is absolutely convergent, since $\left|\frac{\cos x^4}{x^2}\right| \leq \frac{1}{x^2}$, and $\left|\frac{\cos M^4}{4M}\right| \leq \frac{1}{4M} \to 0$ as $M\to +\infty$.

Answer 4

Indeed the integral is convergent by the substitution $x\mapsto x^{1/4}$ and Dirichlet's test. Additionally: $$ \int_{0}^{+\infty}x^2\sin(x^4)\,dx = \frac{1}{4}\int_{0}^{+\infty}\frac{\sin(x)}{x^{1/4}}\,dx\stackrel{\mathcal{L}}{=} \frac{1}{4\,\Gamma\left(\frac{1}{4}\right)}\int_{0}^{+\infty}\frac{ds}{s^{3/4}(1+s^2)}\tag{1}$$ leads to: $$ \int_{0}^{+\infty}x^2\sin(x^4)\,dx = \frac{1}{\Gamma\left(\frac{1}{4}\right)}\int_{0}^{+\infty}\frac{du}{1+u^8}=\color{red}{\frac{\pi}{8\, \Gamma\left(\frac{1}{4}\right) \sin\left(\frac{\pi}{8}\right)}}.\tag{2}$$