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Use differentiation with respect to parameter obtaining a differential equation to solve

$$ \int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)}dx $$ No complex variables, only this approach. Interesting integral and it should have a nice ODE. I have not found the right way yet. we have singularities at $x=\pm i$.

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  • $\begingroup$ Are we able to use $\int_0^\infty \frac{\cos(kx)}{x^2+1}dx= \frac{\pi}{2e^k}$ as a given? $\endgroup$ – Meow Feb 26 '14 at 21:37
  • $\begingroup$ @Alyosha No. because that is just a fourier transform which is easy to do because of the 2 simple poles, but no complex methods. $\endgroup$ – Jeff Faraci Feb 26 '14 at 21:50
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Consider for $a>0$ $$ I(a)=\int_0^\infty\frac{\sin^2(ax)}{x^2(x^2+1)}dx $$ Differentiate it twice. Since $$ \int_0^\infty\frac{\cos(kx)}{x^2+1}dx=\frac{\pi}{2e^k} $$ for $k>0$ we get $I''(a)=\pi e^{-2a}$. Note that $I'(0)=I(0)=0$, so after solving respective IVP we get $$ I(a)=\frac{\pi}{4}(-1+2a+e^{-2a}) $$ It is remains to substitute $a=1$.

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  • $\begingroup$ This is the approach I was looking for. Very nice, thanks $\endgroup$ – Jeff Faraci Feb 27 '14 at 3:24
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    $\begingroup$ @Jeff. there were some typos, see edis. $\endgroup$ – Norbert Feb 27 '14 at 5:24
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    $\begingroup$ @Norbert $I(a)$ is an even function of $a$. $\endgroup$ – Felix Marin May 30 '14 at 15:52
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Well, as an alternative (and I promise, none of the dreaded complex analysis stuff), we could use Parseval's theorem for Fourier transforms:

For example, the FT of $(\sin{x}/x)^2$ is

$$\int_{-\infty}^{\infty} dx \: \frac{\sin^2{x}}{x^2} e^{i k x} = \begin{cases} \\\pi \left (1 - \frac{|k|}{2} \right ) & |k| \le 2 \\ 0 & |k| > 2 \end{cases}$$

The FT of $1/(1+x^2)$ is

$$\int_{-\infty}^{\infty} dx \: \frac1{1+x^2} e^{i k x} = \pi \, e^{-|k|}$$

By Parseval's theorem,

$$\begin{align}\int_0^{\infty} dx \: \frac{\sin^2{x}}{x^2} \frac1{1+x^2} &= \frac{\pi}{2} \int_0^2 dk \, \left (1 - \frac{k}{2} \right ) e^{-k}\\ &= \frac{\pi}{2} \left (1-e^{-2}- \frac12 (1-3 e^{-2}) \right )\\ &= \frac{\pi}{4} \left (1+\frac1{e^2} \right )\end{align}$$

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  • $\begingroup$ this is nice approach thanks $\endgroup$ – Jeff Faraci Feb 26 '14 at 21:56
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Well, as an alternative (like Mr. Ron Gordon did). \begin{align} \int_0^\infty\frac{\sin^2x}{x^2(1+x^2)}dx&=\int_0^\infty\left[\frac{\sin^2x}{x^2}-\frac{\sin^2x}{1+x^2}\right]dx\\ &=\int_0^\infty\frac{\sin^2x}{x^2}dx-\frac{1}{2}\int_0^\infty\frac{1-\cos2x}{1+x^2}dx\\ &=\frac{\pi}{2}-\frac{1}{2}\int_0^\infty\frac{1}{1+x^2}dx+\frac{1}{2}\int_0^\infty\frac{\cos2x}{1+x^2}dx\\ &=\frac{\pi}{2}-\frac{1}{2}\frac{\pi}{2}+\frac{1}{2}\frac{\pi}{2e^2}\\ &=\frac{\pi}{4}+\frac{\pi}{4e^2} \end{align} where I use these links: $\displaystyle\int_0^\infty\frac{\sin^2x}{x^2}dx$ and $\displaystyle\int_0^\infty\frac{\cos2x}{1+x^2}dx$ to help me out.

Unfortunately, this is not differentiation with respect to parameter method (the Feynman way) but I still love this method. (>‿◠)✌

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  • $\begingroup$ ...is that symbol giving us the proverbial? $\endgroup$ – JP McCarthy May 30 '14 at 10:59
  • $\begingroup$ @JpMcCarthy What did you mean? Please use a basic word for 'proverbial'? Sorry, English is not my native language $\endgroup$ – Anastasiya-Romanova 秀 May 30 '14 at 11:26
  • $\begingroup$ Is the last symbol a V finger salute?? $\endgroup$ – JP McCarthy May 30 '14 at 11:32
  • $\begingroup$ @JpMcCarthy In my country it means 'peace' symbol. It often uses when we want to express our idea/ mind without making the other to be offended (ô‿ô) $\endgroup$ – Anastasiya-Romanova 秀 May 30 '14 at 13:10
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How about this method? Denote the integral by $I$. Let $$ I(a,b)=\int_0^\infty \frac{\sin(ax)\sin(bx)}{x^2(1+x^2)}dx. $$ Then $I(0,b)=I(a,0)=0$, $I(1,1)=I$ and \begin{eqnarray} \frac{\partial^2I}{\partial a\partial b}&=&\int_0^\infty\frac{\cos(ax)\cos(bx)}{1+x^2}dx\\ &=&\frac{1}{2}\int_0^\infty\frac{\cos((a+b)x)}{1+x^2}dx+\frac{1}{2}\int_0^\infty\frac{\cos((a-b)x)}{1+x^2}dx\\ &=&\frac{\pi}{4}(e^{-(a+b)}+e^{-|a-b|}) \end{eqnarray} where we use $$ \int_0^\infty\frac{\cos(\alpha x)}{1+x^2}dx=\frac{\pi}{2}e^{-|\alpha|}. $$ Thus $$ I=\frac{\pi}{4}\int_0^1\int_0^1(e^{-(a+b)}+e^{-|a-b|})dadb=\frac{(1+e^2)\pi}{4e^2}. $$

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