# How to determine whether an integral is convergent or divergent?

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.

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

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

• If the numerator had an x-7 and/or 3x+5, would that affect the result of this question? – dg123 Mar 11 '18 at 20:51
• @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. – Mihail Mar 11 '18 at 22:24
• So pretty much, it'd be the same answer? – dg123 Mar 11 '18 at 23:05
• @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. – Mihail Mar 12 '18 at 10:58