here is a solution that my friend gave me.
using the hint we have
the first expression converges to . for the third expression we can change the lower limit to 1, since this will not effect the convergence. so we have
which converges.
To handle the second expression we apply integration by parts once again and get.
to evaluate the first term we can either use l'hopitals rule or taylor series, your pick, either way you find that it does converge. For the last 2 we use fact that
which converges.
Have you graphed the integrand? For large x the function creates "wave-packets" : oscillations within a sine envelope where, on the inside of the envelope, the wavelength is very small. For all intents the function "fills in" all the empty spaces of the sine envelope, creating a net zero area effect.
I don't see why this integral shouldn't coverge, though obviously my argument above is hardly a proof.
-Dan
Aha! I think I caught a mistake with my trained eye.
If, the continous functions
And, converges then so does (Of course I mean the improper intergral).
But your inequality does not satisfy the one above, namely because it can be negative.
However, I have a way to fix this error.
Using the integral series test we need to test,
and see whether it converges. It does since its absolutely convergent, since .