Convergence of $\int_1^2 \frac 1{\sqrt{x^4 - 1}}dx$ I want to show that $\int_1^2 \frac 1{\sqrt{x^4 - 1}}dx$ is convergent.
I tried to use the limit convergence test by comparing with $\frac{1}{x^2}$ but that doesn't work. Any hints on how to proceed with this?
 A: You can set $x^2=sect$ and perform this substitution. When you do that, you in up with an integral of the form$\int\frac{dt}{\sqrt{cost}}$. The key is in the upper and lower bound. For both $x=1$ and $x=2$ you can conclude that on the corresponding $t$ interval, the function is continuous. in other words, it is no longer an improper integral, the integral has an answer (though through elementary techniques we can' figure that out) thus convergent.
A: You can use the fact that $\frac{1}{\sqrt{x^4-1}} < \frac{1}{\sqrt{x-1}}$ for all $x>1$. This is because $x^4-1>x-1$ for all $x>1$.
Therefore, we have $\int_1^2 \frac{1}{\sqrt{x^4-1}} < \int_1^2 \frac{1}{\sqrt{x-1}} = 2\sqrt{x-1} |_1^2 = 2$
A: The problem is at the endpoint $1$.
You want to compare with $\int_1^2 \frac{1}{\sqrt{x-1}}\; dx$.
Possibly helpful is
$$  x^4-1 = (x-1)(x^3 + x^2 + x + 1)$$
A: Check that the integral $$I=\int_{a}^{b} \frac{dx}{(x-a)^p}$$ is finte if $p<1$. So
$$J=\int_{1}^{2} \frac{dx}{(x-1)^{1/2} (x+1)^{1/2} \sqrt{x^2+1}},$$
converges as $p=1/2<1$
A: The only problem could be at the lower bound.
But, by Taylor
$$ \frac 1{\sqrt{x^4 - 1}}=\frac{1}{2 \sqrt{x-1}}-\frac{3 \sqrt{x-1}}{8}+O\left((x-1)^{3/2}\right)$$ Then, no problem, and you will get a number (the answer is quite complicated).
But, if you continue the expansion
$$ \frac 1{\sqrt{x^4 - 1}}=\frac{1}{2 \sqrt{x-1}}+\sum_{n=0}^\infty(-1)^n a_n  (x-1)^{n+\frac 12}$$ the first coeffiicients form the sequence
$$\left\{\frac{3}{8},\frac{11}{64},\frac{7}{256},\frac{141}{4096},\frac{627}{16
   384},\frac{2465}{131072},\frac{81}{524288},\frac{142019}{16777216},\cdots\right\}$$ You will have
$$\int_1^t \frac 1{\sqrt{x^4 - 1}}=\sqrt{t-1}+2\sum_{n=1}^\infty(-1)^n \frac{ a_n }{2 n+3}(t-1)^{n+\frac{3}{2}}$$
Using the above coefficients and $t=2$, the approximation will give
$$\frac{22479314089}{27808235520}\approx 0.8084$$ while the excat value is $0.8078$
