# Improper integral

1. Sep 24, 2006

### Song

basically I'm stuck with this problem........

Integral the upper limit is positive infinite and the lower limit is 1
X/(sqrt(1+x^6) dx......

can someone give me an idea on how to start this?...I really don't know. Thanks.

2. Sep 25, 2006

### dextercioby

Are you sure it's this...?

$$\int_{1}^{\infty} \frac{x}{\sqrt{1+x^{6}}} \ dx \ = ...?$$

Daniel.

3. Sep 25, 2006

### dextercioby

If so, then

$$\int_{1}^{\infty} \frac{x}{\sqrt{1+x^{6}}} \ dx = {}_{2}F_{1}\left(\frac{1}{6},\frac{1}{2}; \frac{7}{6}; -1\right)$$

, with F Gauss' hypergeometric function.

Daniel.

Last edited: Sep 25, 2006
4. Sep 25, 2006

### Song

I don't know how to do that. How do you use the comparison theorem to do this? Thanks.

5. Sep 25, 2006

### dextercioby

What comparison theorem are you referring to ...?

A simple substitution $x^{2}=t$ will bring your integral to a simpler form, however that wouldn't help too much.

Daniel.

6. Sep 25, 2006

### Song

7. Sep 25, 2006

### shmoe

It's always a good idea to state your question in full. I suspect yours is

Prove the integral $$\int_{1}^{\infty} \frac{x}{\sqrt{1+x^{6}}} \ dx$$ is convergent.

Specifically you don't care about finding a value for this integral. When x is large, what simple polynomial is the denominator kinda like? This should suggest a function for your comparison test.

8. Sep 25, 2006

### Song

When X is large, x/(sqrt(1+x^6)) is getting close to the x-axis?
x/(sqrt(1+x^2)) is greater than x/(sqrt(1+x^6))..is my logic right?

9. Sep 25, 2006

### Song

I'm sorry that I didn't state my question clearly at the beginning.

10. Sep 25, 2006

### Song

..........................

11. Sep 25, 2006

### courtrigrad

$$\frac{x}{\sqrt{1+x^{6}}}$$ is about $$\frac{x}{x^{3}} = \frac{1}{x^{2}}$$. Since $$\frac{1}{x^{2}}$$ is convergent ($$p > 1$$), then so is the original integral by the comparison test (the original function is less than the new function)

Last edited: Sep 25, 2006
12. Sep 25, 2006

### Song

but why did you choose x/(x^3)?

13. Sep 25, 2006

### courtrigrad

$$\frac{1}{\sqrt{1+ x^{6}}}= \frac{1}{ \sqrt{x^{6}}} = \frac{1}{x^{3}}$$. The 1 doesnt matter for large enough x.

Last edited: Sep 25, 2006
14. Sep 25, 2006

### shmoe

I don't consider pointing this out to be picky- you never have $$\frac{1}{\sqrt{1+ x^{6}}}= \frac{1}{ \sqrt{x^{6}}}$$.

Much better to write $$\frac{1}{\sqrt{1+ x^{6}}}\sim \frac{1}{ \sqrt{x^{6}}}$$ with a precise asymptotic meaning behind $$\sim$$.

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