Does integral of 1/sqrt(x^6+1) Converge/Diverge?

[cheffy]

Determine if the
integral of 1/sqrt(x^6+1) from 0 to infinity
converges or diverges.

I can't figure out which technique to use/how to prove this, though I'm pretty sure it converges. I think that I should use something to do with 1/sqrt(x^6).

Thanks!

 

[pka]

Are you sure that the integral is from zero to infinity?
Could it be from one to infinity?

From zero to infinity the value is defined as a beta function which in turn is defined in terms of the gamma functon.

So the answer to your question really depends on the correct interval of intergration and upon what is to be done with it. Are you doing probability and/or statistics?

 

[cheffy]

No, it's from 0 to infinity like I said. And it's a calculus class.

 

[confused_07]

I am no expert (as well all know on this board), but the question I had to do, I had to replace the infinity with a Limit, such as when 'a' approaches infinity. Then you can solve for the limit.....
Maybe that will help....

 

[galactus]

Maybe something to consider is the fact that:

\(\displaystyle \L\\\int_{0}^{\infty}\frac{1}{x^{6}+1}dx=\frac{\pi}{3}\)

Maybe you could consider:

\(\displaystyle \L\\\int_{0}^{\infty}\frac{1}{\sqrt{x^{6}+1}}dx=\int_{0}^{K}\frac{1}

{\sqrt{x^{6}+1}}dx+\int_{K}^{\infty}\frac{1}{\sqrt{x^{6}+1}}dx\)

so, \(\displaystyle \L\\\int_{K}^{\infty}\frac{1}{\sqrt{x^{6}+1}}dx<\int_{K}^{\infty}\frac{1}{\sqrt{x^{6}}}dx\)

Letting K=10 for instance, results in 0.004999999375<0.005

Just a thought.

 

[pka]

For a calculus class, we just want to know if it converges.
So by additivity, break it at 1.
\(\displaystyle \L \begin{array}{l}
0 < x < 1 \Rightarrow \quad \sqrt {x^6 + 1} < \sqrt 2 \Rightarrow \quad \frac{1}{{\sqrt {x^6 + 1} }} > \frac{1}{{\sqrt 2 }} \\
\left[ {0 < c < 1} \right]\left( {\int\limits_c^1 {\frac{1}{{\sqrt 2 }}} < \int\limits_c^1 {\frac{1}{{\sqrt {x^6 + 1} }}} } \right) \\
\end{array}\)
So it converges on [0,1]

Then:\(\displaystyle \L \begin{array}{l}
1 \le x \Rightarrow \quad \sqrt {x^6 + 1} > x^3 \sqrt 2 \Rightarrow \quad \frac{1}{{\sqrt {x^6 + 1} }} < \frac{1}{{x^3 \sqrt 2 }} \\
\int\limits_1^\infty {\frac{1}{{\sqrt {x^6 + 1} }}} < \int\limits_1^\infty {\frac{1}{{x^3 \sqrt 2 }}} \\
\end{array}\)
which shows it converges on 1 to infinity.