# 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}

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(I'm assuming you're using a semicircular contour). A somewhat related question: how would one justify that $\lim_{R \to \infty}\left|\int_0^\pi \frac{e^{i3Re^{it}}-3e^{iRe^{it}}}{(Re^{it}-i \epsilon)^5}\right|=0$? – Alyosha Jan 31 '14 at 10:28
So is there no question left? I don't see one. – robjohn Jul 24 '14 at 21:45
@robjohn Not long after I posted this question earlier this year, I realized why my original approach was wrong. So I modified my approach and explained what I did in the original post. Another user bumped this thread today, and when I looked at my post, I realized it could be made a lot clearer. So I edited it. If you have another way to evaluate it using contour integration, please post it. EDIT: I see you already have. :) – Random Variable Jul 24 '14 at 22:33
@RandomVariable: I don't think my approach is any different, now that I look more closely at yours. I should just delete mine. – robjohn Jul 24 '14 at 22:38
@robjohn Isn't the difference that you shifted the contour while I moved the pole? – Random Variable Jul 24 '14 at 22:41

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

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

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

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

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

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I considered doing it like that, but the problem came from a complex analysis textbook. But then again, you could evaluate that last integral by contour integration. Maybe that's what the authors had in mind. I don't know. Thanks. – Random Variable Feb 2 '14 at 15:55
+1 Quite nice and straightforward. – Felix Marin May 3 '14 at 9:25


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}

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

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The answer is accurate (+1), but it would really be nice to see more verbal explanation to your answers. Just a mention that you are integrating by parts 4 times would be helpful. Such a terse answer is sometimes hard to follow. – robjohn Jul 24 '14 at 22:44
@robjohn I just followed your advice. Thanks. – Felix Marin Jul 25 '14 at 2:16
How did the $\sin^3 x$ become $\frac{1}{4} \sin 3x - \frac{3}{4} \sin x$? – Jeel Shah Jul 25 '14 at 7:28

Here is another contour integration approach. Note that the integrand is even, so using the contours $\gamma=[-R,R]\cup Re^{i[0,\pi]}$ as $R\to\infty$ and $\beta=[-R,R]\cup Re^{-i[0,\pi]}$, we have \begin{align} \int_0^\infty\frac{x^3-\sin^3(x)}{x^5}\mathrm{d}x &=\frac12\int_{-\infty}^\infty\frac{x^3-\sin^3(x)}{x^5}\mathrm{d}x\tag{1}\\ &=\lim_{\epsilon\to0^+}\frac12\int_{-\infty-i\epsilon}^{\infty-i\epsilon}\frac{z^3-\sin^3(z)}{z^5}\mathrm{d}z\tag{2}\\ &=\lim_{\epsilon\to0^+}\frac12\int_{\gamma-i\epsilon}\frac{\frac1{8i}(e^{3iz}-3e^{iz})}{z^5}\mathrm{d}z\tag{3}\\ &+\lim_{\epsilon\to0^+}\frac12\int_{\beta-i\epsilon}\frac{z^3+\frac1{8i}(3e^{-iz}-e^{-3iz})}{z^5}\mathrm{d}z\tag{4}\\ &=\lim_{\epsilon\to0^+}\frac1{16i}\int_{\gamma-i\epsilon}\frac{e^{3iz}-3e^{iz}}{z^5}\mathrm{d}z\tag{5}\\[3pt] &=\frac\pi8\frac1{4!}\left((3i)^4-3i^4\right)\tag{6}\\[6pt] &=\frac{13\pi}{32}\tag{7} \end{align} Explanation:
$(1)$: integrand is even, duplicate the domain and divide by $2$
$(2)$: offset the path of integration since there are no singularities and the integrand decays at $\infty$
$(3)$: take some of the terms along the upper contour
$(4)$: take the rest along the lower contour
$(5)$: there are no singularities inside $\beta-i\epsilon$ and the respective integrands decay appropriately on the circular arcs
$(6)$: the residues depend on the $z^4$ terms in the expansions of the exponentials

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Shoot... now that I look at this, I think this may be the same as RandomVariable's approach except that I moved $\frac{z^3}{z^5}$ to the lower contour. – robjohn Jul 24 '14 at 22:34
I've undeleted since Random Variable points out that I did use a different contour (however slightly different). – robjohn Jul 24 '14 at 22:47

Another approach :

Sorry Random Variable, this is not using contour integration technique since I don't know how to approach the integral using that way. $\ddot\smile$

Consider $$\mathcal{I}(\alpha)=\int_0^\infty\frac{(\alpha x)^3-\sin^3\alpha x}{x^5}dx.\tag1$$ Differentiating $(1)$ four times yields \begin{align} \frac{d^4\mathcal{I}}{d\alpha^4}&=\int_0^\infty\frac{\partial^4}{\partial\alpha^4}\left(\frac{(\alpha x)^3-\sin^3\alpha x}{x^5}\right)dx\\ &=\color{green}{\int_0^\infty\left(\frac{81\sin3\alpha x-3\sin\alpha x}{4x}\right)dx}\\ &=\frac{81}{4}\cdot\frac\pi2-\frac{3}{4}\cdot\frac\pi2\\ &=\frac{39\pi}{4},\tag2 \end{align} where $$\int_0^\infty\frac{\sin\alpha x}{x}dx=\frac\pi2\qquad\text{for }\alpha\neq0.$$ Then from $(2)$ we obtain $$\large\color{blue}{\mathcal{I}(\alpha)=\frac{13\pi}{32}a^4},\tag3$$ where $\mathcal{I}(0)=\mathcal{I'}(0)=\mathcal{I''}(0)=\mathcal{I'''}(0)=0$. Thus

$$\mathcal{I}(1)=\int_0^\infty\frac{x^3-\sin^3 x}{x^5}dx=\large\color{blue}{\frac{13\pi}{32}}.$$

P.S.

I think you can easily apply the contour integration technique in line $2$ (green-colored) equation $(2)$.

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What integration technique is this? – Jeel Shah Jul 25 '14 at 10:28
@gekkostate Differentiation under the integral sign a.k.a. Feynman's method. – Tunk-Fey Jul 25 '14 at 11:02
Are you aware of any tutorials which explain this method in detail? For example, how it works and the motivation behind it? Thanks! – Jeel Shah Jul 25 '14 at 11:44
@gekkostate You may refer here: $(1)$ and $(2)$. – Tunk-Fey Jul 25 '14 at 11:50