Show that improper integral converges.

Question

$$\int_{0}^{\infty}{x^{10}\sin(x^{20})} \mathrm{dx}$$

In preparation for the calculus test, I have such a calc to determine whether it converges. I'm a bit stuck and don't know which way to go. I know it converges, but I don't see how to show it. I've thought about boundary criteria/comparison test, but I have little probelm using it here. Can someone give me some hint? Thank you in advance!

Answer 1

We can avoid substitutions. By applying the integration by parts we get $$20\int\limits_1^\infty x^{10}\sin(x^{20})\,dx =20\int\limits_1^\infty x^{-9}x^{19}\sin(x^{20})\,dx \\ =-{x^{-9}\cos(x^{20})}\Big\vert_1^\infty -9\int\limits_1^\infty x^{-10}\cos(x^{20})\,dx$$ The last integral us absolutely convergent.

Answer 2

Outline:

• $\int_0^1x^{10}\sin(x^{20})\,dx$ is obviously fine.
• Next, consider $\int_1^{R}x^{10}\sin(x^{20})\,dx$. We’ll eventually take the limit $R\to\infty$ after proving it exists, but consider this for now. Make the change of variables $t=x^{20}$. You will end up with an integral which looks like $\int_1^r\frac{\sin t}{t^{\alpha}}\,dt$ for some $r$ and some number $\alpha$. Make sure you figure out exactly what they are (I could tell you, but you should do it yourself).

Now, here’s a standard theorem:

For any $\alpha>0$, the improper integral $\int_1^{\infty}\frac{\sin t}{t^{\alpha}}\,dt$ converges.

If $\alpha>1$, this is obvious by comparison with $\frac{1}{t^{\alpha}}$. If $0< \alpha\leq 1$, then we do an integration by parts: for any $r>0$, \begin{align} \int_1^r\frac{\sin t}{t^{\alpha}}\,dt&=\left[\frac{-\cos t}{t^{\alpha}}\right]_1^r-\int_1^r\left(\frac{-\alpha}{t^{1+\alpha}}\right)(-\cos t)\,dt\\ &=\cos(1)-\frac{\cos(r)}{r^{\alpha}}-\alpha\int_1^r\frac{\cos t}{t^{1+\alpha}}\,dt. \end{align} Notice that as $r\to\infty$, the second term approaches zero, while the third term has a limit, because the integral is absolutely convergent on $(1,\infty)$ (by comparison with $\frac{1}{t^{1+\alpha}}$), and absolute convergence of the integral implies convergence of the improper integral. Thus, we have shown the RHS has a limit as $r\to\infty$, and therefore, the LHS has a limit as well, so \begin{align} \int_1^{\infty}\frac{\sin t}{t^{\alpha}}\,dt&=\cos(1)-\int_1^{\infty}\frac{\cos t}{t^{1+\alpha}}\,dt. \end{align} So, the point is that via integration by parts, we bump up the power $t^{\alpha}$ to $t^{1+\alpha}$, and this in the denominator allows us to use our beloved comparison test (and absolute convergence test).

I’ll leave it to you to put this all together.

Answer 3

The integral $\displaystyle\int_0^1 x^{10} \sin(x^{20}) dx$ clearly converges. So let's study the integral $\displaystyle\int_1^{+\infty} x^{10} \sin(x^{20}) dx$.

Let $A>1$. Substituting $t=x^{20}$, one has $$I(A):=\int_1^A x^{10} \sin(x^{20}) dx = \int_1^{A^{20}} \sqrt{t} \sin(t) \dfrac{dt}{20t^{19/20}} = \dfrac{1}{20}\int_1^{A^{20}} t^{-9/20} \sin(t) dt $$

Integrating by parts, we get \begin{align*} I(A)& =\dfrac{1}{20} \left( \left[-\cos(t) t^{-9/20} \right]_1^{A^{20}}-\dfrac{9}{20}\int_1^{A^{20}} t^{-29/20}\cos(t)dt\right)\\ & = \dfrac{1}{20} \left(\cos(1) -\cos(A^{20}) A^{-9}-\dfrac{9}{20}\int_1^{A^{20}} t^{-29/20}\cos(t)dt\right) \end{align*}

But $$\lim_{A \rightarrow +\infty} \cos(A^{20}) A^{-9} = 0$$

and since for every $t > 1$, $|t^{-29/20}\cos(t)| \leq t^{-29/20}$, then the integral $\displaystyle \int_1^{+\infty} t^{-29/20}\cos(t)dt$ is absolutely convergent, so $$\int_1^{A^{20}} t^{-29/20}\cos(t)dt$$

has a finite limit when $A$ tends to $+\infty$.

So $I(A)$ has a finite limit when $A$ tends to $+\infty$, so the integral $\displaystyle\int_0^{+\infty} x^{10} \sin(x^{20}) dx$ converges.