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This is a really short and kind of an interesting thing that popped out in my mind.

Question. Consider $$\int^\infty_0 \frac{\sin^2(x)}{x^2(x^2+1)}dx$$ I have tried with integration by parts, substitutions and a lot of ways to try to tackle this, in a way that no Laplaces are used or Residue theorem or Feynman's integration or any advanced deep analysis theory. Because there are already videos about it, but since I'm not yet that far into math, I went and gave it some more last thoughts with everything that I've learned. And I thought about the Series Expansion of $\sin^2(x)$, is there a way to use the series expansion of $\sin^2(x)=x^2+\frac{x^4}{3}-\frac{2x^6}{45}-\frac{x^8}{315}+O(x^9)$ in order to integrate this? This is just an idea and I don't even know if it's possible, it was just a curious thought.

Acyex
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  • https://math.stackexchange.com/questions/803954/surely-youre-joking-mr-feynman-int-0-infty-frac-sin2xx21x2-dx?noredirect=1&lq=1 See here for a way of doing it with Feynman's technique. – A-Level Student Jul 07 '21 at 22:14
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    You cannot use the series expansion for $\sin^2(x)$ as the upper limit of integration is $\infty$. You'll need to relax your restriction here if you'd like to see ways to evaluate this integral. – Mark Viola Jul 07 '21 at 22:41
  • yeah, just no complex analysis, I haven't gone into that yet so I can't really tell which method is from complex analysis and which isn't, but as long as I get an answer if it's really possible to evaluate this with the idea I've came up it would be great – Acyex Jul 07 '21 at 22:42
  • oh okay Mark Viola, if it would be without limits, just the indefinite integral, would it be possible? I'd still be satisfied with so – Acyex Jul 07 '21 at 22:45
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    Note that there's no closed form for the antiderivative of $\sin^2x/x^2(x^2+1)$, so there's no combination of substitution / integration by parts that will let you compute the integral (any such combo would eventually lead you to a closed form). So any computational trick must use some of the other tools that you mentioned. – csch2 Jul 08 '21 at 01:00
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    This is evidently a duplicate question (See Here and HERE). – Mark Viola Jul 08 '21 at 04:02

2 Answers2

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This integral can be rewritten as $$\int_0^\infty \dfrac{\sin^2(x)dx}{x^2}-\int_0^\infty \dfrac{\sin^2(x)dx}{x^2+1}$$ For the first integral, let $$I(a)=\int_{0}^{\infty}\frac{\sin^2(ax)dx}{x^2}$$
Differentiating both sides yields $$\frac{dI}{da}=\int_{0}^{\infty}\frac{\sin(2ax)dx}{x}=\dfrac{\pi}{2}$$ Thus, $I(a)=\dfrac{\pi a}{2}+C$ and $I(a)=\dfrac{\pi a}{2}$ because $I(0)=0$. Setting $a=1$ gives us $I(1)=\dfrac{\pi}{2}$.

We now have $$\displaystyle\dfrac{\pi}{2}-\dfrac{1}{2}\int_{0}^\infty\dfrac{dx}{x^2+1}+\dfrac{1}{2}\int_0^\infty \dfrac{\cos(2x) dx}{x^2+1}=\dfrac{\pi}{4}+\dfrac{1}{2}\int_0^\infty \dfrac{\cos(2x)dx}{x^2+1}$$

This integral evaluates to $\dfrac{\pi e^{-2}}{2}$ (see https://math.stackexchange.com/a/9409/877722) and so your integral is equal to $\dfrac{\pi}{4}(1+e^{-2})$.

XIC22
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    OP asked not to use Feynman's technique, for the record. – Moni145 Jul 07 '21 at 21:04
  • yeah but after being told I couldn't do anything else this is fine, and they made a new unfamiliar method look intuitive – Acyex Jul 09 '21 at 16:23
  • this may be an extremely basic question but what justifies decomposing the original integrand the way you did? does it mean a/bc = a/b - a/c ? – Hisham Mar 07 '23 at 02:35
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It is possible to compute the antiderivative.

Use this tool (which uses Maxima) and look at all the intermediate steps (they are very instructive to read). Look here for Wolfram Alpha solution.

So, omitting the integration constant, the final result is

$$4x \int\frac{\sin^2(x)}{x^2(x^2+1)}dx=2 \left(\cos (2 x)-x \tan ^{-1}(x)-1\right)+$$ $$x (-i \cosh (2) (\text{Ci}(2 i-2 x)-\text{Ci}(2 x+2 i))+$$ $$2 i (\text{Ei}(-2 i x)-\text{Ei}(2 i x))+\sinh (2) (\text{Si}(2 i-2 x)-\text{Si}(2 x+2 i)))$$ where appear the exponential, sine and cosine integral functions.

Claude Leibovici
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