Improper integral for a one sided limit (Using CT)

Question

Question: Determine if the integral is conv. or div.

Integral : $$\int_{5}^{6} \frac{1}{(x-3)(x-5)^{1/2}} dx$$

Firstly I needed to figure if the $f(x)$ was divergent or convergent. And so, I tried subbing in the limit values, and found that x = 5 becomes 1/0 which is indeterminent form. So I added a limit to the integral, A, where A approaches $5^+$ and subbed it again. This gave me infinity, which implies divergence. However, I graphed $f(x) = 1/(x-3)(x-5)^{1/2} $ and it looked like: http://puu.sh/zE9yH/ef90963004.png . This is where I'm confused.... If $x = 5^+$, $f(x) = inf $, and $x= 6, f(x) = constant$, where there is no V.A. in between, then isn't f(x) convergent? This is just the rough work of the equation, I cannot move onto the next step where I solve the integral of g(x) > f(x) or f(x) < g(x) without knowing this...

Can anybody answer my question? Thanks!

Or if somebody could provide a detailed solution I would really appreciate it. Thanks!

Answer 1

Note that $$ 0\leq \int_5^6\frac{1}{(x-3)\sqrt{x-5}}\mathrm dx\leq\frac{1}{2}\int_5^6(x-5)^{-1/2}\mathrm dx $$ Now $$ \frac{1}{2}\int_5^6(x-5)^{-1/2}\mathrm dx=\sqrt{x-5}\bigg\vert_5^6=1-\lim_{x\to 5^+} \sqrt{x-5}=1 $$

Answer 2

Observe that in a neighbourhood of $5$ the function's singularity behaves like $\frac{1}{\sqrt{x-5}}$. More precisely $$ \lim_{x\to5}\frac{1/[(x-3)\sqrt{x-5]}}{1/\sqrt{x-5}}=2 $$ Since $$ \int_{5}^6\frac{1}{\sqrt{x-5}}\,dx=\int_{0}^1\frac{1}{\sqrt{x}}\,dx<\infty $$ It follows that your integral is finite too by the limit comparison test.

Answer 3

Perhaps this is skirting the issue of the discontinuity by using a substitution which makes it go away, but if we let $u^2=x-5$ then $2u\,du=dx$ and the integral becomes

$$ 2\int_0^1\frac{du}{u^2+2} $$

which is elementary.