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I am interested in evaluating the following integral: $$ \int_{0}^{\infty} \frac{|2-2\cos(x)-x\sin(x)|}{x^4}~dx $$ Using Matlab, Numerically it seems that the integral is convergent, but I'm not sure about it. How can we prove that the integral is convergent or not?

Many Thanks in advance.

Warren Moore
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Shahram Khazaie
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  • You say you are interested in evaluating it, then change only ask if it converges. It does (as stated below), but I highly doubt it has a closed form value because the values for $ x $ at which the interior of the absolute value changes sign are not equally distributed $ \mod 2\pi $. – Jon Claus May 18 '13 at 00:00

3 Answers3

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The integral converges. There is potential trouble at $0$ and "at" infinity. So we split the integral into (i) the part from $0$ to $1$, and (ii) the part from $1$ to $\infty$.

(i) Look at the power series expansion of $2-2\cos x-x\sin x$. The first few terms are $2-2\left(1-\frac{x^2}{2!}+\frac{x^4}{4!}\right) -x\left(x-\frac{x^3}{31}\right)$. There is cancellation, and the first non-zero term is the $x^4$ term. So our function behaves well as $x$ approaches $0$ from the right: It approaches $\frac{1}{12}$. If we define your integrand to be $\frac{1}{12}$, the resulting function is continuous on $[0,1]$, and hence integrable.

(ii) For large $x$, the integrand is $\lt \frac{K}{x^3}$ for some constant $K$, and we know that $\int_1^\infty \frac{dx}{x^3}$ converges.

André Nicolas
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  • Thanks for your reply. Knowing the fact that for large values of $x$ the function is smaller than $\frac{K}{x^3}$, can we prove that the integral itself is convergent? because the integral of $\frac{K}{x^3}$ from $0$ to $\infty$ does not exist ... – Shahram Khazaie May 18 '13 at 00:16
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    Note that the first step in the solution was breaking up into two integrals, $0$ to $1$ and $1$ to $\infty$. In (ii), it was the $1$ to $\infty$ part that was being deal with. And we used the fact that $\int_1^\infty \frac{dx}{x^3}$ converges. That's why splitting was important. – André Nicolas May 18 '13 at 00:19
  • Ooops! I totaly forgot that splitting! Thank you very much. It really helped me. – Shahram Khazaie May 18 '13 at 00:27
  • You are welcome. When there is potential "badness" in two or more places, it is ordinarily useful to split the integral, since the analysis is typically different in the two places. – André Nicolas May 18 '13 at 00:32
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let $I=\int_{0}^{\infty}\dfrac{2-2\cos{x}-x\sin{x}}{x^4}dx$

then using integration by parts we have $$I=-\dfrac{1}{3}\dfrac{2-2\cos{x}-x\sin{x}}{x^3}|_{0}^{\infty}+\dfrac{1}{3}\int_{0}^{\infty}\dfrac{\sin{x}-x\cos{x}}{x^3}dx$$

then $$I=\dfrac{1}{3}\int_{0}^{\infty}\dfrac{\sin{x}-x\cos{x}}{x^3}dx$$ and application of integration by parts yields $$I=-\dfrac{1}{6}\lim_{x\to 0^{+}}\dfrac{\sin{x}-x\cos{x}}{x^2}+\dfrac{1}{6}\int_{0}^{\infty}\dfrac{x\sin{x}}{x^2}dx=\dfrac{\pi}{12}$$

math110
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  • Thanks a lot for your reply. But the major problem that I have is about the absolute value on the numerator. As a consequence, the result will be slightly different from $\frac{\pi}{12}$, but I don't know what is its exact value ... – Shahram Khazaie May 18 '13 at 07:24
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It is a good exercise to utilize Laplace properties $$\int_0^\infty f(x)g(x)\mathrm{d}x = \int_0^\infty \mathcal{L}\{f\}(u)\mathcal{L}^{-1}\{g\}(u)\mathrm{d}u,$$ Let's apply consecutively: \begin{align} I&=\int_{0}^{\infty} \frac{|2-2\cos(x)-x\sin(x)|}{x^4}\mathrm dx\\ &=\int_{0}^{\infty}\mathcal{L^{-1}}\left[\frac{1}{x^4}\right](\xi)\mathcal{L}\Bigl[|2-2\cos(x)-x\sin(x)|\Bigr](\xi)\mathrm d\xi\\ &=\int_{0}^{\infty}\frac{\xi^3}{3!}\frac{2}{\xi(\xi^2+1)^2}\mathrm d\xi\\ &=\frac16\int_{0}^{\infty}\frac{2\xi^2}{(\xi^2+1)^2}\mathrm d\xi\\ &=\frac16\int_{0}^{\infty}\mathcal{L^{-1}}\left[\frac{2\xi}{(\xi^2+1)^2}\right](t)\mathcal{L}[\xi](t)\mathrm dt\\ &=\frac16\int_{0}^{\infty}t\sin t\frac{1}{t^2}\mathrm dt\\ &=\frac16\int_{0}^{\infty}\frac{\sin t}{t}\mathrm dt\\ &=\frac16\frac{\pi}{2}=\boxed{\frac{\pi}{12}} \end{align} No complicated formulas from Transform Table is used, but the following trivial one. $$\mathcal{L}[t\sin t]=–\frac{\mathrm d}{\mathrm ds}\left(\frac{1}{s^2+1}\right)=\frac{2s}{(s^2+1)^2}$$

MathArt
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