# Improper integral $-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$

If I have to solve $$-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$$ it is right to write: $$-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\geq \int_{3}^{+\infty}\frac{-1}{x^{\frac{1}{3}}}$$ and so the integral is divergent? I have use the fact that $$-\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\geq \frac{-1}{x^{\frac{1}{3}}}$$.

If I have instead $$\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$$, so without minus sign, what can I do to prove the divergence? Is it true that from the following I can deduce the divergence? $$\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx=-\int_{3}^{+\infty}\frac{-\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx\geq -\int_{3}^{+\infty}\frac{-1}{x^{\frac{1}{3}}}\, dx$$ My doubt arises since I have seen here that a similar integral here (Solution verification of the improper integral $\int_1^{+\infty}\frac{\cos^2{t}}{t}\,dt$) is solved in a more complicate way...

• Whether there is a minus sign in front of an integral is irrelevant to its convergence/divergence. Apr 29, 2021 at 6:47
• Ok so my idea is right and can I apply this also when I have minus or plus sign?
– cely
Apr 29, 2021 at 6:47
• In this case, it just means you can't use $\sin x\le1$ as it is an upper bound rather than a lower bound. It is true that $\int_a^\infty dx/x^{1/3}=\infty$ but that tells you nothing about $\int_a^\infty\sin^4x/x^{1/3}\,dx$. Apr 29, 2021 at 6:48
• But when I have a minus sign I can wirite that the original integral is $\geq$ $\int_a^{\infty}\frac{-1}{x^{\frac{1}{3}}}$ so the divergence follows, right?
– cely
Apr 29, 2021 at 6:51
• No. The minus sign does not change anything, as all you're saying now is that $-\int_a^\infty\sin^4x/x^{1/3}\,dx\ge-\int_a^\infty dx/x^{1/3}\ge-\infty$ and this gives no new information. Apr 29, 2021 at 6:53

$$\int_{2n\pi}^{2n\pi+\pi}\frac{\sin^4x}{x^{1/3}}dx\geq\frac{1}{\sqrt{2n\pi+\pi}}\int_{2n\pi}^{2n\pi+\pi}\sin^4xdx\\ =\frac{1}{\sqrt{2n\pi+\pi}}\int_{0}^{\pi}\sin^4xdx\\ =\frac{2}{\sqrt{2n\pi+\pi}}\int_{0}^{\frac\pi2}\sin^4xdx =\frac{1}{\sqrt{2n\pi+\pi}}\cdot\frac{3\pi}{8}.$$ So $$\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx\geq\frac{3\pi}{8}\sum_{n=1}^{\infty}\frac{1}{\sqrt{2n\pi+\pi}}=+\infty.$$

• Ok thanks! And for the minus sign? I would need a lowe or an upper bound?
– cely
Apr 29, 2021 at 7:43
• $\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$ is divergent if and only if $-\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$ is divergent. Apr 29, 2021 at 7:45
• If you know Cauchy's convergent Principle, this is enough to show your improper integral is divergent! Apr 29, 2021 at 7:47
• In fact one can stop at the first line, saying that the integral is a positive constant.
– user65203
Apr 29, 2021 at 7:48
• Ok but If I want to study the divergence with the minus sign I have to find out a function GREATER (and not SMALLER as in case of positive function) and negative that gives me a divergent integral?
– cely
Apr 29, 2021 at 7:48

Hint:

Let us avoid the confusion raised by the minus sign, using the fact that if $$-I$$ diverges, so does $$I$$.

Now

$$\frac{\sin^4x}{x^{1/3}}<\frac1{x^{1/3}}$$ is of no help because it only gives an upper bound on the function and the function could remain much smaller. What you need is a lower bound.

A correct way to solve is to write for instance

$$\frac{\sin^4x}{x^{1/3}}>\frac1{\color{red}4\,x^{1/3}}$$ which holds in a constant fraction of every period, and this will be enough to justify divergence.

• Ok thanks I have not understand why I can't state $$\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx=-\int_{3}^{+\infty}\frac{-\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx\geq -\int_{3}^{+\infty}\frac{-1}{x^{\frac{1}{3}}}\, dx$$ and from which the divergence...I have found out a lower bound, not? Where I am failing?
– cely
Apr 29, 2021 at 7:27
• @cely: this inequality is plain wrong. For God's sake, drop that minus sign.
– user65203
Apr 29, 2021 at 7:28
• mm sorry if I don't understand...why it is wrong?
– cely
Apr 29, 2021 at 7:29
• $$\int_{3}^{+\infty}\frac{\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx=-\int_{3}^{+\infty}\frac{-\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx$$ then $$\frac{-\sin^4{t}}{t^{\frac{1}{3}}}\geq \frac{-1}{t^{\frac{1}{3}}}$$ so $$-\int_{3}^{+\infty}\frac{-\sin^{4}{x}}{x^{\frac{1}{3}}}\, dx\geq -\int_{3}^{+\infty}\frac{-1}{x^{\frac{1}{3}}}\, dx$$
– cely
Apr 29, 2021 at 7:32
• Since $\sin^{4}\leq 1$ then $-\sin^{4}\geq -1$
– cely
Apr 29, 2021 at 7:36