# Math Help - Convergence of an integral

1. ## Convergence of an integral

Show whether or not, by bounding (from below of from bottom, depends of the appropriate choice) it, if the integral converges or not :
$\int_1^5 \frac{dx}{\sqrt{x^4-1}}$. I'm sure it converges, so I wanted to bound it from above but couldn't, due to the indetermination of the lower limit of the integral. Can you help me?

2. For $t>1$, note that $0<\frac{1}{\sqrt{t^4-1}}<\frac{1}{\sqrt{t-1}}$.

Hence

$0\ <\ \int_t^5{\frac{dx}{\sqrt{x^4-1}}}\ <\ \int_t^5{\frac{dx}{\sqrt{x-1}}}$

Since $\int_1^5{\frac{dx}{\sqrt{x-1}}}=\lim_{t\to1^+}{\int_t^5{\frac{dx}{\sqrt{x-1}}}}=\lim_{t\to1^+}{\left[2\sqrt{x-1}\right]_t^5}=4$, we can say that $\int_1^5{\frac{dx}{\sqrt{x^4-1}}}$ converges since it is positive and bounded above.

3. Originally Posted by arbolis
Show whether or not, by bounding (from below of from bottom, depends of the appropriate choice) it, if the integral converges or not :
$\int_1^5 \frac{dx}{\sqrt{x^4-1}}$. I'm sure it converges, so I wanted to bound it from above but couldn't, due to the indetermination of the lower limit of the integral. Can you help me?
Just note that

$\sqrt{x^4-1}\geq\sqrt{x-1}~\forall{x}\in{D}$

Where D is the region you are talking about.

EDIT: Wow, I am slow at typing. Jane Benett did what I did and beyond =D