# Rational Gaussian-type integral with sin

Here is a tough integral. Does anyone have any clever ideas?. I have tried all sorts of things, but make no real headway.

$\displaystyle \int_{0}^{\infty}\frac{e^{-x^{2}}\sin^{2}(x)}{x^{2}}dx$

Would residues be a consideration with this one?.

Thanks all.

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It's not pretty, but maybe Taylor expand $\sin^2 x / x^2$ about $x=0$, which will contain even powers only of $x$, and then use 27 here?: mathworld.wolfram.com/GaussianIntegral.html – Eric Angle Dec 18 '12 at 23:37

The answer is $$\int_0^\infty dx\,\frac{e^{-x^2}\sin^2 x}{x^2}=\frac{\pi}{2}\,{\rm erf}(1)-\frac{\sqrt{\pi}}{2e}(e-1).$$ Consider $$I(a)=\int_0^\infty dx\,\frac{e^{-ax^2}\sin^2 x}{x^2}$$ for which $$-I'(a)=\int_0^\infty dx\,e^{-ax^2}\sin^2 x.$$ This last integral can be done by expanding sine in terms of exponentials and completing the square in each term. The result is $$-I'(a)=\frac{\sqrt{\pi}}{16a}e^{-1/a}\left(e^{1/a}-1\right)$$ which can be anti-differentiated (I used Mathematica) in terms of the error function. As $a\to\infty$, $I(a)\to 0$, setting the constant of integration. Plugging in $a=1$ gives your answer.
Note that $$\dfrac{\sin^2(x)}{x^2} = \dfrac{1-\cos(2x)}{2x^2} = \dfrac{1 - \displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^k (2x)^{2k}}{(2k)!}}{2x^2} = \dfrac{\displaystyle \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1} (2x)^{2k}}{(2k)!}}{2x^2} = \displaystyle \sum_{k=1}^{\infty} \dfrac{(-1)^{k-1} 2^{2k-1}x^{2k-2}}{(2k)!} = \displaystyle \sum_{k=0}^{\infty} \dfrac{(-1)^{k} 2^{2k+1}x^{2k}}{(2k+2)!}$$ \begin{align} \int_0^{\infty} \dfrac{\exp(-x^2)\sin^2(x)}{x^2} dx & = \displaystyle \sum_{k=0}^{\infty} \left(\dfrac{(-1)^{k} 2^{2k+1}}{(2k+2)!} \int_0^{\infty} x^{2k}\exp(-x^2) dx \right)\\ & = \displaystyle \sum_{k=0}^{\infty} \left(\dfrac{(-4)^{k}}{(2k+2)!} \Gamma \left(k + \dfrac12 \right) \right) \end{align} which is same as what Jonathan has obtained.