1. ## Improper integrals

$\displaystyle \int_{2}^{+\infty} \frac{dx}{x\sqrt{x^2-4}}$

i dunno what to do with this integral since both the limits are discontinuous:
so
$\displaystyle \lim_{t \to 2} \frac{1}{2} \sec^{-1}(\frac{x}{2}) \big|_{t}^{2\sqrt{2}} + \lim_{s\to +\infty} \frac{1}{2} \sec^{-1}(\frac{x}{2}) \big|_{2\sqrt{2}}^{s}$

so in evaluating the integral:
$\displaystyle =\frac{1}{2}\sec^{-1}(\sqrt{2}) - \frac{1}{2}\sec^{-1}(1) + \frac{1}{2}\sec^{-1}(\frac{+\infty}{2}) - \frac{1}{2}\sec^{-1}(\sqrt{2})$

$\displaystyle =\frac{1}{2}\sec^{-1}(1)$ which is infinite

my answer seems to be divergent

the answer at the back of the book seems to be $\displaystyle \frac{\pi}{4}$

$\displaystyle \int_{2}^{+\infty} \frac{dx}{x\sqrt{x^2-4}}$

i dunno what to do with this integral since both the limits are discontinuous:
so
$\displaystyle \lim_{t \to 2} \frac{1}{2} \sec^{-1}(\frac{x}{2}) \big|_{t}^{2\sqrt{2}} + \lim_{s\to +\infty} \frac{1}{2} \sec^{-1}(\frac{x}{2}) \big|_{2\sqrt{2}}^{s}$

so in evaluating the integral:
$\displaystyle =\frac{1}{2}\sec^{-1}(\sqrt{2}) - \frac{1}{2}\sec^{-1}(1) + \frac{1}{2}\sec^{-1}(\frac{+\infty}{2}) - \frac{1}{2}\sec^{-1}(\sqrt{2})$

$\displaystyle =\frac{1}{2}\sec^{-1}(1)$ which is infinite

my answer seems to be divergent

the answer at the back of the book seems to be $\displaystyle \frac{\pi}{4}$
Check two things.

First
$\displaystyle sec^{-1}(\sqrt{2}) = \frac{\pi}{4}$

$\displaystyle sec^{-1} ( \infty ) = \frac{\pi}{2}$

$\displaystyle \theta = sec^{-1}(1) \implies sec(\theta) = 1$

$\displaystyle sec(\theta) = 1 \implies cos(\theta) = \frac{1}{1} = 1$

So $\displaystyle \theta = 0$

Thus
$\displaystyle sec^{-1}(1) = 0$.

So
$\displaystyle =\frac{1}{2}\sec^{-1}(\sqrt{2}) - \frac{1}{2}\sec^{-1}(1) + \frac{1}{2}\sec^{-1}(\frac{+\infty}{2}) - \frac{1}{2}\sec^{-1}(\sqrt{2})$

$\displaystyle = \frac{1}{2} \cdot \frac{\pi}{4} - \frac{1}{2} \cdot 0 + \frac{1}{2} \cdot \frac{\pi}{2} - \frac{1}{2} \cdot \frac{\pi}{4} = \frac{\pi}{4}$

And, of course, since the x = 2 limit isn't actually infinite we don't need to go through all that mess with the limit on it.

-Dan

3. I don't know if this is correct.
Calculate $\displaystyle \displaystyle\int_{a}^b\frac{dx}{x\sqrt{x^2-4}}$.
Let $\displaystyle \displaystyle x=\frac{1}{t}\Rightarrow dx=-\frac{1}{t^2}dt$.
Now we have
$\displaystyle \displaystyle -\int_{\frac{1}{a}}^{\frac{1}{b}}\frac{dt}{\sqrt{1-4t^2}}=\left.-\frac{1}{2}\arcsin 2t\right|_{\frac{1}{a}}^{\frac{1}{b}}=-\frac{1}{2}\arcsin\frac{2}{b}+\frac{1}{2}\arcsin\f rac{2}{a}$.
Now take the limits when $\displaystyle b\to\infty$ and $\displaystyle a\to 2$ and we get $\displaystyle \frac{\pi}{4}$