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So by using tabular integration: $$\int_{2\pi}^\infty \frac{\sin(x)}{x} \, dx$$ $$= \left. -\frac{1}{x}\cos(x) - \frac{1}{x^2}\sin(x) + \frac{2}{x^3}\cos(x) + \frac{2\cdot3}{x^4}\sin(x) - \frac{2\cdot3\cdot4}{x^5}\cos(x) - \frac{2\cdot3\cdot4\cdot5}{x^6}\sin(x) + \cdots \right|_{2\pi}^\infty$$

adding all the cosine terms gives:

$$\left.-\frac{1}{x}\cos(x)+ \frac{2}{x^3}\cos(x)- \frac{2\cdot3\cdot4}{x^5} \cos(x) + \cdots = \sum_{n=1}^\infty \frac{(-1)^n (2n-2)!}{x^{2n-1}} \cos(x) \right|_{2\pi}^\infty$$

and similarly for the sine terms:

$$\left.-\frac{1}{x^2}\sin(x)+ \frac{2\cdot3}{x^4}\sin(x)- \frac{2\cdot3\cdot4\cdot5}{x^6}\sin(x)+\cdots =\sum_{n=1}^\infty \frac{(-1)^n \cdot (2n-1)!}{x^{2n}} \sin(x) \right|_{2\pi}^\infty$$

So putting everything back together we have:

$$\left.\int_{2\pi}^\infty \frac{\sin(x)}{x} \, dx = \sum_{n=1}^\infty \frac{(-1)^n (2n-2)!}{x^{2n-1}} \cos(x)+ \sum_{n=1}^\infty \frac{(-1)^n (2n-1)!}{x^{2n}} \sin(x) \right|_{2\pi}^\infty$$

I think that as $x$ approaches $\infty$ the fractions in both sums approach $0$, so the top limit does not yield anything from the sums. Likewise, $\sin(2\pi) = 0$ and $\cos(2\pi) = 1$, so all that we are left with is:

$$\int_{2\pi}^\infty \frac{\sin(x)}{x} \, dx = -\sum_{n=1}^\infty \frac{(-1)^n (2n-2)!}{(2\pi)^{2n-1}}$$

However, putting this sum into wolfram I get that the sum is divergent, which doesn't seem right to me since $ f(x) = \frac{\sin(x)}{x}$ decreases to $0$ quite fast.

Is my logic inconsistent here? Have I done the integration incorrectly, or used a technique improperly/ inappropriately?

Michael Hardy
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user373763
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  • I have very little doubt that this integral converges, to about $0.1526...$. – Jared Oct 06 '16 at 03:52
  • I think you need to be more careful about plugging in $\infty$. It's not clear to me that that part vanishes since for any positive $x$, those sums do not converge. – Alexis Olson Oct 06 '16 at 03:54
  • @AlexisOlson Oh I see what you are saying, even though the fraction within the sum goes to 0, that doesn't mean that the sum itself will go to 0 as x approaches $\infty$. – user373763 Oct 06 '16 at 04:10
  • I've done a little testing with wolfram alpha, and I've found that $\int_0^{\infty} \frac{\sin(x)}{x} dx$ is $\frac{\pi}{2}$, which would be very difficult to see from this sum. So, long story short, this is probably not an appropriate way to evaluate this integral it seems. – user373763 Oct 06 '16 at 04:55

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$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\int_{2\pi}^{\infty}{\sin\pars{x} \over x}\,\dd x}\ &\ =\ \overbrace{\int_{0}^{\infty}{\sin\pars{x} \over x}\,\dd x}^{\ds{\pi \over 2}}\ -\ \overbrace{\int_{0}^{2\pi}{\sin\pars{x} \over x}\,\dd x}^{\ds{\mrm{Si}\pars{2\pi}}}\ =\ \color{#f00}{{\pi \over 2} - \mrm{Si}\pars{2\pi}} \end{align}

$\ds{\mrm{Si}}$ is the Sine Integral Function.

The $\,\mrm{Si}\pars{z}$ series expansion is given by: $$ \mrm{Si}\pars{z} = \sum_{n = 0}^{\infty}\pars{-1}^{n}\, {z^{2n + 1} \over \pars{2n + 1}!\pars{2n + 1}} $$

Felix Marin
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  • Thank you for this, I was not aware of this function until today! It seems I was steering way off course with my attempt. As an aside, I have also ran into this math exchange answer: http://math.stackexchange.com/questions/5248/solving-the-integral-int-0-infty-frac-sinxx-dx-frac-pi2 – user373763 Oct 06 '16 at 05:17
  • A small addendum: the given integral equals $\int_{0}^{+\infty}\frac{\sin x}{x+2\pi},dx$ that by exploiting the Laplace transform can be written as $\int_{0}^{+\infty}\frac{e^{-2\pi s}}{1+s^2},ds$, probably more suited form numerical approximations. In this form, for instance, it is trivial that the integral is positive but bounded by $\frac{1}{2\pi}$. – Jack D'Aurizio Oct 06 '16 at 16:20
  • @JackD'Aurizio Thanks. It's a nice point of view. – Felix Marin Oct 06 '16 at 20:26
  • Im sorry, but i'm having trouble seeing how you can tell $\int_{0}^{\infty} frac{\e^{-2\pi s}}{1+s^2} ds$ is bounded by $\frac{1}{2\pi}$. Could you elaborate/explain? – user373763 Oct 12 '16 at 15:40
  • sorry I meant : $\int_{2\pi}^{\infty} \frac{e^{-2\pi s}}{1+s^2} ds$, mathematics stack exchange won't let me edit my own comment for some reason. – user373763 Oct 12 '16 at 15:49
  • @JackD'Aurizio I guess the $\texttt{@user373763}$ last $\underline{two}$ comments are related to your comment. – Felix Marin Oct 12 '16 at 21:56
  • @user373763: simply, $$0\leq \int_{0}^{+\infty}\frac{e^{-2\pi s}}{1+s^2},ds < \int_{0}^{+\infty}e^{-2\pi s},ds = \frac{1}{2\pi}.$$ – Jack D'Aurizio Oct 12 '16 at 21:58
  • @FelixMarin: thank you for notifying me. – Jack D'Aurizio Oct 12 '16 at 21:58