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I have no idea of variable changement to use or other to calculate this integral :

$$ \int_0^{\infty}\frac{(\sin(x))^2}{x^2+1}\,dx $$

Wolfram alpha gives me the result but really no idea ...

I seek a calculus without residues theorem or other big theorems like it because I didn't learn them.. so just by part or by variable changement if possible sure..

Thanks in advance,

Shadock

jimjim
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ParaH2
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    The integrand function does not have a primitive in terms of elementary functions, but the integral can be computed through the residue theorem or the Laplace transform. – Jack D'Aurizio Jan 13 '15 at 00:56
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    This looks like a good candidate for differentiation under the integral. – Euler....IS_ALIVE Jan 13 '15 at 00:57
  • @Euler....IS_ALIVE how to differentiate under the integral with just one variable? – ParaH2 Jan 13 '15 at 01:01
  • This might be useful if you can reduce the power of sine. The linked problem is also worked out using the Laplace transform, which might be easier to follow. – user170231 Jan 13 '15 at 01:03
  • Ok so I just have to learn the Laplace transform and his inverse ? Is there just one transformation and inverse or a lot of possibilities ? @Euler....IS_ALIVE how to differentiate with just one variable ? – ParaH2 Jan 13 '15 at 17:28
  • @Shadock You would have to cleverly introduce a parameter. – Euler....IS_ALIVE Jan 13 '15 at 20:38
  • @Euler....IS_ALIVE like sin(xt)^2 or sin²(x)/(x^2+t^2) ? :) – ParaH2 Jan 13 '15 at 21:02

2 Answers2

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Here it is a proof with the residue theorem. We have: $$ I = \int_{0}^{+\infty}\frac{1-\cos(2x)}{2(1+x^2)}\,dx = \frac{\pi}{4}-\frac{1}{2}\int_{0}^{+\infty}\frac{\cos(2x)}{1+x^2}\,dx=\frac{\pi}{4}-\frac{1}{4}\int_{\mathbb{R}}\frac{\cos(2x)}{1+x^2},$$ but: $$ \int_{\mathbb{R}}\frac{\cos(2x)}{1+x^2}\,dx = \Re\int_{-\infty}^{+\infty}\frac{e^{2iz}}{1+z^2}\,dz=\Re\left(2\pi i\cdot\operatorname{Res}\left(\frac{e^{2iz}}{1+z^2},z=i\right)\right),$$ so: $$\int_{\mathbb{R}}\frac{\cos(2x)}{1+x^2}\,dx = \frac{\pi}{e^2}\tag{1} $$ and: $$ I = \int_{0}^{+\infty}\frac{\sin^2 x}{1+x^2}\,dx = \color{red}{\frac{\pi}{4}\left(1-\frac{1}{e^2}\right)}=\frac{\pi}{2e}\sinh 1.$$ With a probabilistic argument, $(1)$ follows from the fact that the CF of the standard Cauchy distribution is $e^{-|t|}$. To prove it, it is enough to show that: $$\int_{0}^{+\infty}e^{-t}\cos(nt)\,dx = \frac{1}{1+n^2}\tag{2}$$ holds by integration by parts.

Jack D'Aurizio
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I particularly like Jack's hint concerning the Laplace transform. Write $$I(a)=\int_0^\infty \frac{\sin^2(ax)}{x^2+1}\,dx$$ Then, $$\mathcal{L}\left\{I(a)\right\}=\int_0^\infty\int_0^\infty\frac{\sin^2(ax)}{x^2+1}e^{-as}\,da\,dx=\int_0^\infty\frac{2x^2}{(x^2+1)(s^3+4sx^2)}\,dx$$

user170231
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    (+1) This further simplifies if we take $\cos(ax)$ in place of $\sin^2(ax)$. – Jack D'Aurizio Jan 13 '15 at 01:13
  • Perhaps I'm missing something, but it's not clear to me how working out the last equality is any easier that evaluating the original integral. – Travis Willse Jan 13 '15 at 01:22
  • Partial fraction decomposition yields $$-\frac{2}{s(s^2+4)}\int_0^\infty\frac{dx}{x^2+1}+\frac{2s}{s^2+4}\int_0^\infty\frac{dx}{4x^2+s^2}=\frac{\pi}{2(s^2+4)}-\frac{\pi}{s(s^2+4)}$$ Finding the inverse transform is fairly simple, and that gives an expression equivalent to $I(a)$. (Barring any silly mistakes on my part) – user170231 Jan 13 '15 at 15:41
  • @user170231 what is the inverse Laplace of $\pi / (2s^2+4s)$ ? – ParaH2 Jan 14 '15 at 22:30
  • You can determine that with some more partial fraction decomp – user170231 Jan 14 '15 at 23:34