Evaluating $\int_{0}^{\infty} \frac{x^{3}- \sin^{3}(x)}{x^{5}} \ dx $ using contour integration

EDIT: Instead of expressing the integral as the imaginary part of another integral, I instead expanded $\sin^{3}(x)$ in terms of complex exponentials and I don't run into problems anymore.

\begin{align} \int_{0}^{\infty} \frac{x^{3}-\sin^{3}(x)}{x^{5}} \ dx &= \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^{3}-\sin^{3}(x)}{x^{5}} \ dx \\ &= \frac{1}{2} \ \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i}(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix})}{x^{5}} \ dx \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \ \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i}(e^{3ix}-3e^{ix}+3e^{-ix}-e^{-3ix})}{(x-i \epsilon)^{5}} \ dx \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{x^{3}+\frac{1}{8i} (e^{3ix}-3e^{ix})}{(x-i \epsilon)^{5}} + \frac{1}{16i} \lim_{\epsilon \to 0^{+}} \int_{-\infty}^{\infty} \frac{3e^{-ix}-e^{-3ix}}{(x-i \epsilon)^{5}} \ dx \end{align}

Then I integrated $ f(z) = \frac{z^{3}+ \frac{1}{8i}(e^{3iz}-3e^{iz})}{(z-i \epsilon)^{5}}$ around the upper half of $|z|=R$ and $ g(z) = \frac{3e^{-iz}-e^{-3iz}}{(z-i \epsilon)^{5}}$ around the lower half of $|z|=R$ and applied Jordan's lemma.

\begin{align} \int_{0}^{\infty} \frac{x^{3}-\sin^{3}x}{x^{5}} \ dx &= \frac{1}{2} \lim_{\epsilon \to 0^{+}}2 \pi i \ \text{Res}[f(z),i \epsilon] + \frac{1}{16i} \lim_{\epsilon \to 0^{+}} 2 \pi i (0) \\ &= \frac{1}{2} \lim_{\epsilon \to 0^{+}} \frac{2 \pi i}{4!} \lim_{z \to i \epsilon} \frac{d^{4}}{dz^{4}} \Big(z^{3}+\frac{1}{8i}e^{3iz}-\frac{3}{8i}e^{iz} \Big) \\ &= \frac{\pi i}{24} \lim_{\epsilon \to 0^{+}} \ \lim_{z \to i \epsilon}\Big( \frac{1}{8i}(3i)^{4}e^{3iz}- \frac{3}{8i} (i)^{4} e^{iz} \Big) \\ &= \frac{\pi i}{24} \lim_{\epsilon \to 0^{+}} \Big( \frac{81}{8i}e^{- 3\epsilon} - \frac{3}{8i}e^{- \epsilon} \Big) \\ &= \frac{\pi i}{24} \Big(\frac{81}{8i}-\frac{3}{8i} \Big) \\ &= \frac{13 \pi}{32} \end{align}

Solution 1:

I like to calculate this integral as follows:

Let us note that

$$\frac{1}{x^5}=\frac{1}{4!}\int_0^\infty t^4e^{-xt}dt$$ So

$$I=\frac{1}{4!}\int_{0}^{\infty}(x^{3}-\sin^{3}x)\int_0^\infty t^4e^{-xt}\;dt\;dx$$

$$=\frac{1}{4!}\int_{0}^{\infty}t^4\int_{0}^{\infty}(x^{3}-\sin^{3}x)e^{-xt}\;dx\;dt$$

$$=\frac{1}{4!} \int_{0}^{\infty}t^4\left [\frac{6}{t^4}-\frac{6}{(t^2+1)(t^2+9)}\right ]dt$$

$$=\frac{1}{4}\int_{0}^{\infty}\frac{10t^2+9}{(t^2+1)(t^2+9)}dt=\frac{13\pi}{32}$$

Solution 2:

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}{x^{3}- \sin^{3}\pars{x} \over x^{5}}\,\dd x:\ {\large ?}}$

Lets $\ds{\color{#00f}{\fermi\pars{x}} \equiv x^{3} - \sin^{3}\pars{x} = \color{#00f}{x^{3} + {1 \over 4}\,\sin\pars{3x} - {3 \over 4}\,\sin\pars{x} }\tag{1}}$.

$$ \mbox{The integral in question becomes}\quad \int_{0}^{\infty}{\fermi\pars{x} \over x^{5}}\,\dd x $$

In order to 'reduce' the $\ds{x^{-5}}$ power to a 'simple' $\ds{x^{-1}}$ power, we integrate by parts repeatedly: \begin{align} \color{#c00000}{\int_{0}^{\infty}{\fermi\pars{x} \over x^{5}}\,\dd x}&={1 \over 4}\int_{0}^{\infty}{\fermi'\pars{x} \over x^{4}}\,\dd x ={1 \over 12}\int_{0}^{\infty}{\fermi''\pars{x} \over x^{3}}\,\dd x={1 \over 24}\int_{0}^{\infty}{\fermi'''\pars{x} \over x^{2}}\,\dd x \\[3mm]&={1 \over 24}\int_{0}^{\infty}{\fermi^{\pars{\tt IV}}\pars{x} \over x} \,\dd x\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\pars{2} \end{align}

From expression $\pars{1}$ we can evaluate $\ds{\fermi^{\pars{\tt IV}}\pars{x}}$: $$ \fermi^{\pars{\tt IV}}\pars{x} ={81 \over 4}\,\sin\pars{3x} - {3 \over 4}\,\sin\pars{x} $$

which is replaced in $\pars{2}$: \begin{align} \color{#c00000}{\int_{0}^{\infty}{\fermi\pars{x} \over x^{5}}\,\dd x}& ={27 \over 32}\int_{0}^{\infty}{\sin\pars{3x} \over x}\,\dd x -{1 \over 32}\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x \\[3mm]&={27 \over 32}\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x -{1 \over 32}\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x \\[3mm]&=\underbrace{\pars{{27 \over 32} - {1 \over 32}}}_{\ds{13 \over 16}}\ \underbrace{\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x}_{\ds{\pi \over 2}} \end{align}

$$\color{#00f}{\large% \int_{0}^{\infty}{x^{3}- \sin^{3}\pars{x} \over x^{5}}\,\dd x ={13 \over 32}\,\pi} \approx {\tt 1.2763} $$