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

Question

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*}

---

Answer 1

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.