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Question

Determine whether convergent or divergent.

$$ \int_{-4}^{1} \frac{dz}{(z + 3)^3} $$


Thinking

I'm not sure how best to go about this, whether I'm justified in my result. Basically I'm saying that as I can't find the first limit, the integral is divergent.

I'm not sure if I should be, in some way, trying to combine the two limits (and using L'Hopitals), or if as soon as I've established that one doesn't exist the whole thing can be determined to be divergent (I think this is correct).

If of one of the two limits is divergent, can I conclude that the integral is divergent?


Definition

If $f$ is continuous at all $x$ in the interval $[a, b]$, except maybe at $c$ , where $a < c < b$ , and if $\lim_{x \to c} |f(x)| = + \infty$ , then

$$ \int_{a}^{b} f(x) \mathop{dx} = \lim_{t \to c^- } \int_{a}^{t} f(x) \mathop{dx} + \lim_{s \to c^+ } \int_{s}^{b} f(x) \mathop{dx} $$

if this limit exists, otherwise it is divergent.


Working

The improper integral is

$$ \int \frac{dz}{(z + 3)^3} = - \frac{1}{2(z + 3)^2} + C $$

There's a discontinuity at $z = -3$ , so splitting the integral up as

\begin{equation*} \begin{aligned} \int_{-4}^{1} \frac{dz}{(z + 3)^3} & = \lim_{a \to -3^-} \int_{-4}^{a} \frac{dz}{(z + 3)^3} + \lim_{b \to -3^+} \int_{b}^{1} \frac{dz}{(z + 3)^3} \\ &= - \frac{1}{2} \left( \lim_{a \to -3^-} \left[ \frac{1}{(z + 3)^2} \right]_{-4}^{a} + \lim_{b \to -3^+} \left[ \frac{1}{(z + 3)^2} \right]_{b}^{1} \right) \end{aligned} \end{equation*}

The first limit, $\lim_{a \to -3^-} \left[\frac{1}{(z + 3)^2} \right]_{-4}^{a}$, is found as

\begin{equation*} \begin{aligned} \lim_{a \to -3^-} \left[\frac{1}{(z + 3)^2} \right]_{-4}^{a} &= \lim_{a \to -3^-} \left[\frac{1}{(a + 3)^2} - \frac{1}{(-1)^2} \right] \\ &= \lim_{a \to -3^-} \left[\frac{1}{(a + 3)^2} - 1 \right] \\ &= \lim_{a \to -3^-} \left[\frac{1 - (a + 3)^2}{(a + 3)^2} \right] \\ \end{aligned} \end{equation*}


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  • $\begingroup$ The last limit is of the form $1/0$, so it doesn't converge. $\endgroup$ – Simply Beautiful Art Jul 21 '17 at 13:34
  • $\begingroup$ The integrand function is not locally integrable at $[-4,1] $. $\endgroup$ – hamam_Abdallah Jul 21 '17 at 13:38
  • $\begingroup$ @SimplyBeautifulArt yes, but what I wasn't confident with was saying at this stage that the integral was certainly divergent. I was wondering if there was perhaps a combination of the two limits that would result in a convergent integral, even though one of them was divergent? $\endgroup$ – baxx Jul 21 '17 at 13:38
  • $\begingroup$ It's a matter of definition. But in this specific case both sides diverge so it does not matter. $\endgroup$ – user76568 Jul 21 '17 at 13:40
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    $\begingroup$ @Salahamam_Fatima I'm not sure what you're saying, I see that I can't integrate directly over [-4, 1], but I can by taking limits (?). Or at least, this is sometimes valid (here I don't think so, though I wasn't too confident with my reasoning) $\endgroup$ – baxx Jul 21 '17 at 13:40
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Strictly speaking, this integral is divergent because the integral of the absolute value of the integrand diverges: $$ \int_{-4}^1\frac1{|x+3|^3}\,\mathrm{d}x=\infty $$ However, in the Cauchy Principal Value sense, this integral converges: $$ \begin{align} \operatorname{PV}\int_{-4}^1\frac1{(x+3)^3}\,\mathrm{d}x &=\lim_{\epsilon\to0^+}\left[\int_{-4}^{-3-\epsilon}\frac1{(x+3)^3}\,\mathrm{d}x+\int_{-3+\epsilon}^1\frac1{(x+3)^3}\,\mathrm{d}x\right]\\ &=\lim_{\epsilon\to0^+}\left[\int_{-1}^{-\epsilon}\frac1{x^3}\,\mathrm{d}x+\int_{\epsilon}^4\frac1{x^3}\,\mathrm{d}x\right]\\ &=\lim_{\epsilon\to0^+}\left[\int_{-1}^{-\epsilon}\frac1{x^3}\,\mathrm{d}x+\int_{\epsilon}^1\frac1{x^3}\,\mathrm{d}x+\int_1^4\frac1{x^3}\,\mathrm{d}x\right]\\ &=\int_1^4\frac1{x^3}\,\mathrm{d}x\\ &=-\frac12\left[\frac1{4^2}-\frac1{1^2}\right]\\ &=\frac{15}{32} \end{align} $$ The approach in the question diverges because the integral is not strictly convergent. We need to keep the two halves together, and omit an interval symmetric about the singularity, in order to compute the Principal Value.

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  • $\begingroup$ Your first sentence: It shouldn't be "strictly speaking", it should be "in the Lebesgue sense" For example $\int_0^\infty (\sin x)/x\, dx$ is viewed as a convergent improper integral, although it's not integrable in the Lebesgue sense. $\endgroup$ – zhw. Dec 18 '17 at 0:51
  • $\begingroup$ @zhw.: i don't believe this is integrable in any sense other than the Principal Value sense, whether Lebesgue, Riemann, or probably any since the singularity is degree three. Principal Value works because of the oddness of the singularity and the absence of a second order component. Dividing this function by $\cosh(x)$ would render it non-integrable even in the principal value sense since it would add a second order component to the singularity. $\endgroup$ – robjohn Dec 18 '17 at 4:16

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