After reading articles on differentiation under the integral sign, I hit this post from mit, where after introducing the power tool, it challenges reader to do

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

Obviously I have no clue where to start. Could any one give a hint?

  • 1
    $\begingroup$ I think you can simplify first the integral using partial fractions since $\frac{1}{x^2 \left(x^2+1\right)}=\frac{1}{x^2}-\frac{1}{x^2+1}$. The first integral is simple; the second one is more problematic to me. Good luck. $\endgroup$ – Claude Leibovici Aug 4 '14 at 7:18
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    $\begingroup$ The definite integral of $\frac{1}{x^2+1}$ is simple: it is $\arctan(x)$. Remember that $\arctan(0)=0$ and $\arctan(\infty)=\pi/2$. $\endgroup$ – Steven Van Geluwe Aug 4 '14 at 7:27
  • $\begingroup$ This question is the same as the problem in this link math.stackexchange.com/questions/691798/… $\endgroup$ – xpaul Aug 4 '14 at 23:46

This is a possible way to evaluate the integral. Partial fraction decomposition and the double angle formula yield $$\int^\infty_0\frac{\sin^2{x}}{x^2(1+x^2)}dx=\frac{1}{2}\int^\infty_0\frac{1-\cos{2x}}{x^2}dx-\frac{1}{2}\int^\infty_0\frac{1-\cos{2x}}{1+x^2}dx$$ The first integral can be evaluated in many ways, differentiation under the integral sign is one of them. I prefer to proceed with a simple fact that follows from the definition of the gamma function. $$\int^{\infty}_0t^{n-1}e^{-xt} \ dt=\frac{\Gamma(n)}{x^n}$$ Hence the first integral is \begin{align} \frac{1}{2}\int^\infty_0\frac{1-\cos{2x}}{x^2}dx &=\frac{1}{2}\int^\infty_0(1-\cos{2x})\int^\infty_0te^{-xt} \ dt \ dx\\ &=\frac{1}{2}\int^\infty_0t\int^\infty_0e^{-xt}(1-\cos{2x}) \ dx \ dt\\ &=\int^\infty_0\left(\int^\infty_0e^{-xt}\sin{2x} \ dx\right)dt\\ &=\int^\infty_0\frac{2}{t^2+4}dt\\ &=\frac{\pi}{2}\\ \end{align} The second integral can be broken up further and evaluated using the residue theorem. \begin{align} \frac{1}{2}\int^\infty_0\frac{1-\cos{2x}}{1+x^2}dx &=\frac{\pi}{4}-\frac{1}{4}\Re\oint_{\Gamma}\frac{e^{2iz}}{1+z^2}dz\\ &=\frac{\pi}{4}-\frac{1}{2}\Re\left(\pi i\operatorname{Res}(f,i)\right)\\ &=\frac{\pi}{4}-\frac{1}{2}\Re\left(\pi i\frac{e^{-2}}{2i}\right)\\ &=\frac{\pi}{4}-\frac{\pi}{4e^2} \end{align} Hence $$\int^\infty_0\frac{\sin^2{x}}{x^2(1+x^2)}dx=\frac{\pi}{4}\left(1+e^{-2}\right)$$

  • $\begingroup$ thanks a lot and.. the trick with $\int^{\infty}_0t^{n-1}e^{-xt} \ dt=\frac{\Gamma(n)}{x^n}$ is brilliant! i only saw $n=1$ case before, never realized that could utilize $n>1$! $\endgroup$ – athos Aug 4 '14 at 10:35

Could any one give a hint?

Partial fraction decomposition, together with the fact that

  • $\displaystyle\int_0^\infty\frac{\sin^2x}{x^2}dx=\frac\pi2$

  • $\sin^2x=\dfrac{1-\cos2x}2$

  • $\displaystyle\int_0^\infty\frac{\cos x}{x^2+a^2}dx=\frac\pi{2a~e^a}$


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