For this question, I'm not sure if I'm doing it right, can anyone please help me out?

Determine whether the following integral is convergent or divergent.

$$\int_5^6 \frac 1 {(x-3)\sqrt {x-5}} \, dx$$

$$ \frac{1}{(x-3)\sqrt {x-5}}\le \frac{1}{x-3}$$

Since $\int_5^6 \frac 1 {x-3} \,dx$ converges, $\int_5^6 \frac 1 {(x-3)\sqrt {x-5}} \, dx$ must also converge.

  • $\begingroup$ So then how would we go about that then? $\endgroup$ – dg123 Mar 11 '18 at 17:59
  • $\begingroup$ That doesn't work: The problematic point is the behavior of $\sqrt{x-5},$ not of $x-3. \qquad$ $\endgroup$ – Michael Hardy Mar 11 '18 at 18:01
  • $\begingroup$ @AndrewLi : A mere discontinuity will not cause any difficulty; rather the issue is the vertical asymptote. $\qquad$ $\endgroup$ – Michael Hardy Mar 11 '18 at 18:02

The inequality is not valid. Rather, \begin{align*} 0<\dfrac{1}{(x-3)\sqrt{x-5}}<\dfrac{1}{2\sqrt{x-5}},~~~~x\in(5,6], \end{align*} and \begin{align*} \int_{5}^{6}\dfrac{1}{\sqrt{x-5}}&=\lim_{\eta\rightarrow 5^{+}}\int_{\eta}^{6}\dfrac{1}{\sqrt{x-5}}dx\\ &=\lim_{\eta\rightarrow 5^{+}}2\sqrt{x-5}\bigg|_{\eta}^{6}\\ &=\lim_{\eta\rightarrow 5^{+}}\left(2-2\sqrt{\eta-5}\right)\\ &=2\\ &<\infty. \end{align*}

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  • $\begingroup$ Why did you multiply the root by two? $\endgroup$ – dg123 Mar 11 '18 at 18:03
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    $\begingroup$ For $x\in(5,6]$, then $x-3>2$, and so $\dfrac{1}{x-3}<\dfrac{1}{2}$. $\endgroup$ – user284331 Mar 11 '18 at 18:04
  • $\begingroup$ Where did the 1 over two go in the integration? $\endgroup$ – dg123 Mar 11 '18 at 18:07
  • $\begingroup$ I don't understand your question? $\endgroup$ – user284331 Mar 11 '18 at 18:08
  • $\begingroup$ You forgot the constant in your integration at the beginning? $\endgroup$ – dg123 Mar 11 '18 at 18:09

Your inequality does not hold because the left handside is unbounded but the right handside is bounded on the interval $[5,6]$.

Actually you have $ \frac{1}{(x-3)\sqrt {x-5}}$ $\le \frac{1}{2\sqrt{x-5}}$ Then substitute $y=x-5$. Now, you obtain $$I\leq \int_5^6 \dfrac{dx}{2\sqrt{x-5}}=\frac12\int_0^1\dfrac{dy}{y^{1/2}}$$

Here the right hand side converges by the p-test, so it implies the convergence of your original integral $I$.

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  • $\begingroup$ If the numerator had an x-7 and/or 3x+5, would that affect the result of this question? $\endgroup$ – dg123 Mar 11 '18 at 20:51
  • $\begingroup$ @dg123 no, again the method would be to write your integrand as $\dfrac{f(x)}{\sqrt{x-5}}$ where f(x) is well-defined on your compact interval and therefore obtain some maximum value $M$. And again you obtain $I\leq M\int_5^6\dfrac{dx}{\sqrt{x-5}}$ so that you have to proceed with a similar substitution. $\endgroup$ – Mihail Mar 11 '18 at 22:24
  • $\begingroup$ So pretty much, it'd be the same answer? $\endgroup$ – dg123 Mar 11 '18 at 23:05
  • $\begingroup$ @dg123 yes. On the contrary, if you multiply with $f(x)$ having singularities on $[5,6]$ there might be some issues: Take $f(x)=\dfrac{1}{x-a}$, where are $a\in \{5,6\}$. Now the integral diverges by the p-test. $\endgroup$ – Mihail Mar 12 '18 at 10:58

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