3
$\begingroup$

Does the following improper integral converges ? $$\int_{1}^{\infty} x^2 \sin(x^4) dx$$

Tried to find some known improper integral to compare this one to, but didn't find one.

Thanks for helping!

$\endgroup$
2
$\begingroup$

Hint. By the change of variable $x=t^{1/4}$, one gets

$$ \int_{1}^{\infty} x^2 \sin(x^4) dx=\frac14\int_{1}^{\infty}\frac{\sin t}{t^{1/4}} dt $$

The latter integral is convergent by applying the Dirichlet test of convergence for improper integrals.

Alternatively, one may integrate by parts, for $M\geq1$:

$$ \int_{1}^M\frac{\sin t}{t^{1/4}} dt=\left. \frac{-\cos t}{t^{1/4}}\right]_1^M+\frac14\int_{1}^M\frac{\cos t}{t^{5/4}} dt $$

As $M \to \infty$, the first term on the right hand side is finite and the latter integral is clearly absolutely convergent.

$\endgroup$
0
$\begingroup$

It does converge. You can try Wolfram Alpha to get this answer:

$$\int_1^\infty x^2 sin(x^4) dx\approx 0.150932$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy