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

Question

After reading articles on differentiation under the integral sign, I hit this post from MIT OCW, 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?

Answer 1

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 $$\int^\infty_0\frac{\sin^2{x}}{x^2(1+x^2)}dx=\frac{\pi}{4}\left(1+e^{-2}\right)$$

Answer 2

Could any one give a hint?

Partial fraction decomposition, together with the fact that

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

• $\sin^2x=\dfrac{1-\cos2x}2$

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

Answer 3

We first split the integral into two as

$$ \int_0^{\infty} \frac{\sin ^2 x}{x^2\left(1+x^2\right)} d x=\int_0^{\infty} \frac{\sin ^2 x}{x^2} d x-\int_0^{\infty} \frac{\sin ^2 x}{1+x^2} d x $$

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For the first one, we can use integration by parts and double integral.

$$ \begin{aligned} \int_0^{\infty} \frac{\sin ^2 x}{x^2} d x & =\int_0^{\infty} \sin ^2 x d\left(-\frac{1}{x^2}\right) \\ & =\int_0^{\infty} \frac{\sin (2 x)}{x} d x \\ & =\int_0^{\infty} \frac{\sin x}{x} d x \\ & =\int_0^{\infty} \int_0^{\infty} e^{-x t} \sin x d t d x \\ & =\int_0^{\infty} \int_0^{\infty} e^{-x t} \sin x d x d t \\ & =\int_0^{\infty} \frac{1}{t^2+1} d t \\ & =\left[\tan ^{-1} t\right]_0^{\infty} \\ & =\frac{\pi}{2} \end{aligned}$$

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For the second integral, we use contour integration by the path $\gamma$ in anti-clockwise direction.

$$ \gamma=\gamma_{1} \cup \gamma_{2} \textrm{ where } \gamma_{1}(t)=t+i 0(-R \leq t \leq R) \textrm{ and } \gamma_{2}(t)=R e^{i t} (0<t<\pi). $$

$$ \begin{aligned} \int_0^{\infty} \frac{\sin ^2 x}{1+x^2} d x&=\frac{1}{2} \int_0^{\infty} \frac{1-\cos 2 x}{1+x^2} d x \\& =\frac{\pi}{4}-\frac{1}{2} \int_0^{\infty} \frac{\cos 2 x}{1+x^2} d x \end{aligned} $$ $$ \begin{aligned} \int_0^{\infty} \frac{\cos 2 x}{1+x^2} d x&=\frac{1}{2} \Re \int_{-\infty}^{\infty} \frac{e^{2 x i}}{1+x^2} d x \\& =\frac{1}{2} \Re\left(\lim _{R \rightarrow \infty} \oint_\gamma \frac{e^{2 z i}}{1+z^2} d x\right) \\ & =\frac{1}{2} \Re\left(2 \pi i \lim _{z \rightarrow i} \frac{e^{-2}}{2 i}\right)\\&=\frac{\pi}{2 e^2} \end{aligned} $$

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Now we can conclude $$ \boxed{\int_0^{\infty} \frac{\sin ^2 x}{x^2\left(1+x^2\right)} d x =\frac{\pi}{4}\left(1+\frac{1}{e^2}\right)} $$

Answer 4

Solution using Fourier transform

The integral can be seen as a Fourier transform: $$ I := \int_0^\infty \frac{\sin^2(x)}{x^2(x^2+1)} \, dx = \frac12 \int_{-\infty}^{\infty} \frac{\sin(x)}{x} \frac{\sin(x)}{x} \frac{1}{x^2+1} e^{-i0x} \, dx \\ = \frac12 \mathcal{F}\{ \frac{\sin(x)}{x} \frac{\sin(x)}{x} \frac{1}{x^2+1} \}(0) . $$

By the duality of product and convolution under Fourier transform we can rewrite this as $$ I = = \frac12 \frac{1}{(2\pi)^2} \left(\mathcal{F}\{ \frac{\sin(x)}{x} \} * \mathcal{F}\{ \frac{\sin(x)}{x} \} * \mathcal{F}\{ \frac{1}{x^2+1} \} \right)(0) . $$

Now, $$ \mathcal{F}\{ \frac{\sin(x)}{x} \} = \pi\mathbf{1}_{[-1,1]}(\xi), $$ where $\mathbf{1}_A$ is the indicator function on the set $A,$ and $$ \mathcal{F}\{ \frac{1}{x^2+1} \} = \pi e^{-|\xi|} . $$

Thus, $$ I = \frac{1}{8\pi^2} \left( \pi\mathbf{1}_{[-1,1]} * \pi\mathbf{1}_{[-1,1]} * \pi e^{-|\bullet|} \right)(0) \\ = \frac{\pi}{8} \left( \mathbf{1}_{[-1,1]} * \mathbf{1}_{[-1,1]} * e^{-|\bullet|} \right)(0) . $$

Here, $$ (\mathbf{1}_{[-1,1]} * \mathbf{1}_{[-1,1]})(x) = (2-|x|) \, \mathbf{1}_{[-2,2]}(x) $$ so $$ \left( \mathbf{1}_{[-1,1]} * \mathbf{1}_{[-1,1]} * e^{-|\bullet|} \right)(\xi) = \int_{-2}^{2} (2-|\eta|) e^{-|\xi-\eta|} \, d\eta $$ However, we only need to calculate this at $\xi=0$: $$ I = \left( \mathbf{1}_{[-1,1]} * \mathbf{1}_{[-1,1]} * e^{-|\bullet|} \right)(0) = \int_{-2}^{2} (2-|\eta|) e^{-|\eta|} \, d\eta = 2 \int_{0}^{2} (2-\eta) e^{-\eta} \, d\eta = 2 (1 + e^{-2}). $$

Thus, finally, $$ I = \frac{\pi}{8}\cdot 2 (1 + e^{-2}) = \frac{\pi}{4}(1 + e^{-2}). $$