# Thread: Convergence of an improper integral

1. ## Convergence of an improper integral

I cant figure out how to answer this calculus problem. I'm having trouble actually calculating the integral.

a) (integral from 0 to 1) sin(x+1/x)dx

I haven't been able to answer that, theres a part b that I havent tried yet since the first one has been giving me problems.

b) (integral from 0 to 1) sin^2(x+1/x)

2. Originally Posted by blorpinbloo
I cant figure out how to answer this calculus problem. I'm having trouble actually calculating the integral.

a) (integral from 0 to 1) sin(x+1/x)dx

I haven't been able to answer that, theres a part b that I havent tried yet since the first one has been giving me problems.

b) (integral from 0 to 1) sin^2(x+1/x)
Try the sine of a sum formula.

3. Originally Posted by blorpinbloo
I cant figure out how to answer this calculus problem. I'm having trouble actually calculating the integral.

a) (integral from 0 to 1) sin(x+1/x)dx

I haven't been able to answer that, theres a part b that I havent tried yet since the first one has been giving me problems.

b) (integral from 0 to 1) sin^2(x+1/x)
Specify it:
$x+\frac{1}{x}$ or $\frac{x+1}{x}$ ?!

4. My bad, its x+(1/x)

5. Originally Posted by blorpinbloo
My bad, its x+(1/x)
It can not be expressed in terms of elementary functions.
http://www.wolframalpha.com/input/?i...%281%2Fx%29%5D
Did you post all informations ?

6. Originally Posted by General
It can not be expressed in terms of elementary functions.
Integrate sin[x+(1/x)] - Wolfram|Alpha
Did you post all informations ?
Could this be integrated?

$sin\left(x+\frac{1}{x}\right)=sin(x)cos\left(\frac {1}{x}\right)+sin\left(\frac{1}{x}\right)cos(x)$

I'm not very good at integration, so I can't always tell right away. I would have to work through it.

7. Nevermind that, I didn't notice your wolfram link.

8. Originally Posted by adkinsjr
Could this be integrated?

$sin\left(x+\frac{1}{x}\right)=sin(x)cos\left(\frac {1}{x}\right)+sin\left(\frac{1}{x}\right)cos(x)$

I'm not very good at integration, so I can't always tell right away. I would have to work through it.
No.
Wolfaram\alpha integrator considers all cases.

9. The question was "Show that the following improper integrals both converge." I assumed you had to calculate the integral because that's what the section was about, but I suppose there may be another way to solve it without calculating it.