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Showing $$\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx=0$$ We can show this by re-writing $I$ as

$$ \implies I=6\int_{0}^{\infty}\frac{\frac{1-\cos(2x)}{2x}-\frac{1-\cos(3x)}{3x}}{x}\,\mathrm dx, $$ which is Frullani Integral. $$J=\int_{0}^{\infty} \frac{f(ax)-f(bx)}{x} dx=[f(\infty)-f(0)]\ln(a/b).$$ Here, $f(x)=\frac{1-\cos(x)}{x},$ hence $I=0.$

So the question is how to show (1), otherwise?

Mittens
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Z Ahmed
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  • Laplace transforms should work here by appending an $e^{-ax}$ at the end of the integrand and defining it as $I(a)$. Equivalently, differentiation under the integral sign (Leibniz rule) should work. – Prasun Biswas Jun 22 '21 at 03:54
  • Just a thought: $1-\cos(2x)=2\sin^2(x)$, and the corresponding integral is well-known. Then you can add and subtract $1/x^2$ for the other integrals, use integration by parts, and it should work out. Am on my phone or I'd type it out. :/ – Integrand Jun 22 '21 at 03:55

3 Answers3

8

Alternatively, integrate by parts

\begin{align} &\int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx\\ = &\int_{0}^{\infty} \frac{6\sin 2x-6\sin 3x}{x} dx\\ =& \>6 \int_{0}^{\infty} \frac{\sin 2x}{2x} d(2x) - 6 \int_{0}^{\infty} \frac{\sin 3x}{3x} d(3x) \\ =&\>0 \end{align}

Quanto
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  • If there was an International Integration Challenge, I'm pretty sure that you'd get the gold medal. Your integrating skill is incredible. +1. – A-Level Student Jun 22 '21 at 09:17
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    This is a great answer. However, I think that $\lim_{x\to 0}-\frac{1-3\cos(2x)+2\cos(3x)}{x} = 0$ is a detail that should not be omitted; just for the sake of completion. – Alias K Jun 22 '21 at 09:32
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Here is an approach that uses the following identity for Laplace transforms $$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (s)\mathcal{L}^{-1} \{g(x)\} (s) \, ds.$$ Recently this identity has been referred as the Maz identity for Laplace transforms.

Setting $f(x) = 1 - 3\cos 2x + 3 \cos 3x$ and $g(x) = \frac{1}{x^2}$, it is easy to see that \begin{align*} \mathcal{L}\{f(x)\} &= \mathcal{L}\{1 - 3\cos 2x + 2 \cos 3x\} = \mathcal{L}\{1\} - 3 \mathcal{L}\{\cos 2x \} + 2 \mathcal{L}\{\cos 3x \}\\ &= \frac{1}{s} - \frac{3s}{s^2 + 4} + \frac{2s}{s^2 + 9}, \end{align*} and $$\mathcal{L}^{-1} \{g(x)\} = \mathcal{L}^{-1} \left \{\frac{1}{x^2} \right \} = s.$$ So from the Maz identity one has \begin{align*} \int_0^\infty \frac{1 - 3\cos 2x + 2 \cos 3x}{x^2} \, dx &= \int_0^\infty \left [\frac{1}{s} - \frac{3s}{s^2 + 4} + \frac{2s}{s^2 + 9} \right ] s \, ds\\ &= \int_0^\infty \left [1 - 3 \frac{s^2}{s^2 + 4} + 2 \frac{s^2}{s^2 + 9} \right ] \, ds\\ &= \int_0^\infty \left [1 - 3 \frac{(s^2 + 4) - 4}{s^2 + 4} + 2 \frac{(s^2 + 9) - 9}{s^2 + 9} \right ] \, ds\\ &= \int_0^\infty \left [1 - 3 + \frac{12}{s^2 + 4} + 2 - \frac{18}{s^2 + 9} \right ] \, ds\\ &= 12 \int_0^\infty \frac{ds}{s^2 + 4} - 18 \int_0^\infty \frac{ds}{s^2 + 9}\\ &= 6 \arctan \left (\frac{s}{2} \right ) \Big{|}_0^\infty - 6 \arctan \left (\frac{s}{3} \right ) \Big{|}_0^\infty\\ &= 6 \cdot \frac{\pi}{2} - 6 \cdot \frac{\pi}{2} = 0, \end{align*} as required to show.

omegadot
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The result can be found almost immediately from the work in the question statement. Note that
\begin{align*} \int_{0}^{\infty} \frac{1-3\cos 2x+2\cos 3x}{x^2} dx &= 6 \underbrace{\int_0^\infty\frac{1-\cos 2x}{2x^2}dx}_{\textrm{let }x=t/2} - 6 \underbrace{\int_0^\infty\frac{1-\cos 3x}{3x^2} dx}_{\textrm{let }x=t/3} \\ &= 6\int_0^\infty\frac{1-\cos t}{t^2} - 6\int_0^\infty\frac{1-\cos t}{t^2} \\ &= 0. \end{align*} It remains to show that $(1-\cos 2x)/(2x^2)$ and $(1-\cos 3x)/(3x^2)$ are integrable on $[0,\infty)$, which is a straightforward exercise.

user26872
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