Convergence of $\int_0^\infty {\frac{\sin(x)(x+4)}{\sqrt{x^3(x+1)^2}}}$.

Question

I am trying to check the convergence of $$ \int_0^\infty {\frac{\sin(x)(x+4)}{\sqrt{x^3(x+1)^2}}}\,dx. $$

I divided it into two cases, from 0 to 1 and from 1 to $\infty$. I could see, using modulus because of the $\sin$, that $$ \int_0^1 {\frac{\sin(x)(x+4)}{\sqrt{x^3(x+1)^2}}}\,dx $$ converges.

But now I can't seem to find an upper bound that converges or a lower one that diverges to use the comparison criteria for the other part. I tried defining the function as a quotient of two functions and take their limits, but the $\sin$ seems to bother in that procedure.

Am I not seeing some easy way of bounding the function? If not, is there anything that I could do without overcomplicating the excersise?

Thanks a lot.

Answer 1

• On $0$ we have $$\frac{\sin(x)(x+4)}{\sqrt{x^3(x+1)^2}}\sim_0\frac4{\sqrt x}$$ hence the integral $$\int_0^1\frac{\sin(x)(x+4)}{\sqrt{x^3(x+1)^2}}dx$$ is convergent.
• On $+\infty$ we have $$\frac{|\sin(x)|(x+4)}{\sqrt{x^3(x+1)^2}}\le \frac{(x+4)}{\sqrt{x^3(x+1)^2}}\sim_\infty\frac1{x\sqrt x}$$ hence the integral $$\int_1^\infty \frac{\sin(x)(x+4)}{\sqrt{x^3(x+1)^2}}dx$$ is convergent. Conclude.

Answer 2

For convergence on the interval $[1,\infty)$ I believe you can use the fact that $${\frac{\sin(x)(x+4)}{\sqrt{x^3(x+1)^2}}} \leq {\frac{(x+4)}{\sqrt{x^3(x+1)^2}}} \\ \leq {\frac{(x+4x)}{\sqrt{x^3(x+1)^2}}} \\ = \ {\frac{5x}{\sqrt{x^3(x+1)^2}}} \\ \leq {\frac{5x}{\sqrt{x^3(x+0)^2}}} \\ = {\frac{(5x)}{\sqrt{x^3\cdot x^2}}} \\ = {\frac{(5x)}{x^2\sqrt{x}}} \\ = \frac{5}{x\sqrt{x}}$$ So your integral should have a finite upper bound, as $$5\int_1^\infty \frac{1}{x\sqrt{x}}$$ converges.