Improper Integral Proof

[Seeker555]

Prove that for \(\displaystyle r>1\) the improper integral

\(\displaystyle \displaystyle \int_{1}^{\infty} \frac{1}{(1+x^{5})^{\frac{r}{5}}}dx\) exists.

the boundaries for integration are infinity and one. (i don't know how to write that in code sorry)


And the hint is: Compare with the function \(\displaystyle x\rightarrow \frac{1}{x^r}\).


Any help or answer very much appreciated.

 

[galactus]

Let's check convergence of \(\displaystyle \frac{1}{x^{r}}\).

If r=1, then \(\displaystyle \displaystyle\int_{1}^{\infty}\frac{1}{x}dx=\infty\)

If \(\displaystyle r\neq 1\), then \(\displaystyle \displaystyle\int_{1}^{\infty}\frac{1}{x^{r}}dx\)

\(\displaystyle =\displaystyle\lim_{L\to \infty}\left \frac{x^{1-r}}{1-r}\right|_{1}^{L}\)

\(\displaystyle =\displaystyle\lim_{L\to \infty}\frac{L^{1-r}-1}{1-r}=\left\{\begin{array}{rcl}r<1, \;\ \infty \\ r>1, \;\ \frac{1}{r-1}\end{array}\)


Now, it can be shown that if f and g are continuous and \(\displaystyle 0\leq f(x)\leq g(x)\) for all

\(\displaystyle x\geq a\), then \(\displaystyle \displaystyle \int_{a}^{\infty}f(x)dx\) converges if

\(\displaystyle \displaystyle\int_{a}^{\infty}g(x)dx\) converges.

That is, \(\displaystyle \displaystyle\int_{a}^{\infty}f(x)dx\leq \int_{a}^{\infty}g(x)dx\)

This says, if \(\displaystyle \displaystyle\int_{1}^{\infty}\frac{1}{x^{r}}dx\) converges, than

\(\displaystyle \displaystyle\int_{1}^{\infty}\frac{1}{(1+x^{5})^{\frac{r}{5}}}dx\) converges.

\(\displaystyle \displaystyle \int_{1}^{\infty}\frac{1}{(1+x^{5})^{\frac{r}{5}}}dx\leq \int_{1}^{\infty}\frac{1}{x^{r}}dx\)

 

[Seeker555]

I will study your replies to me. Thank you very much.

 

[Seeker555]

however surely for the first part i'd have thought 1/x and you put x into infinity would make it go to 0 rather than infinity. or atleast that's how it is for the limit...

 

[Seeker555]

shortly realised after posting should have left a notice soz hehe. the onething i'm unsure about is seeing as that it goes to 1/(r-1) whether that proves for all cases that it 'exists' and i'm not sure what the convergence has to do with proving it exists...