1. ## Converging improper integrals

I've been asked to determine for what values of n does the integral of x^n.sin(1/x) .dx from 1 to infinity, converge. I've been given the Maclaurin series expansion of sin(x).

I'm not sure how to tackle this problem can anyone offer any advice please?

For what values of n is this integral converging?

2. ## Re: Converging improper integrals

Try plugging in 1/x to the Maclaurin series and multiplying out by x^n

3. ## Re: Converging improper integrals

I've tried its not a nice results, Maclaurin series expansion of sin(x) is given as a hint. Could I possibly replace all the x's in the series with 1/x to get the series expansion for sin(1/x) and then sub the first few terms of this expansion into the integral. Then I could multiply out the series and x^n?

4. ## Re: Converging improper integrals

Lol, I just changed my response to that exactly

5. ## Re: Converging improper integrals

Originally Posted by CRD98
I've tried its not a nice results, Maclaurin series expansion of sin(x) is given as a hint. Could I possibly replace all the x's in the series with 1/x to get the series expansion for sin(1/x) and then sub the first few terms of this expansion into the integral. Then I could multiply out the series and x^n?
yes and note that $\displaystyle \int_1^\infty ~\dfrac{1}{x^n}~dx$ converges for $n>1$

6. ## Re: Converging improper integrals

Originally Posted by CRD98
I've tried its not a nice results, Maclaurin series expansion of sin(x) is given as a hint. Could I possibly replace all the x's in the series with 1/x to get the series expansion for sin(1/x) and then sub the first few terms of this expansion into the integral. Then I could multiply out the series and x^n?
You can also plug in the indefinite integral into Wolframalpha and figure out for what values of n the antiderivative converges as x approaches infinity and the antiderivative exists at 1.

http://m.wolframalpha.com/input/?i=i...%5B1%2Fx%5D%5D

7. ## Re: Converging improper integrals

Originally Posted by romsek
yes and note that $\displaystyle \int_1^\infty ~\dfrac{1}{x^n}~dx$ converges for $n>1$
So in my case the integral will converge for values for n < 0 , as this will give an integrand similar to the one you posted?