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Evaluate $\int_{-\infty}^\infty x\exp(-x^2/2)\sin(\xi x)\ \mathrm dx$

The answer given by Wolfram Alpha is $\sqrt{2\pi}\xi\exp(-\xi^2/2)$. Observe how this is related to the Fourier transform of $x\exp(-x^2/2)$:

the part $\int_{-\infty}^{\infty}x\exp(-x^2/2)\cos\xi x \ \mathrm dx=0$ since the integrand is odd.

In addition, what are the Fourier transforms of $x^k\exp(-x^2/2)$ for $k=2,3$?

Related: How do I compute $\int_{-\infty}^\infty e^{-\frac{x^2}{2t}} e^{-ikx} \, \mathrm dx$ for $t \in \mathbb{R}_{>0}$ and $k \in \mathbb{R}$?

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up vote 1 down vote accepted

Using contour integration, then differentiating, we get $$ \begin{align} \int_{-\infty}^\infty e^{-x^2/2}e^{-i\xi x}\,\mathrm{d}x &=e^{-\xi^2/2}\int_{-\infty}^\infty e^{-(x+i\xi)^2/2}\,\mathrm{d}x\\ &=e^{-\xi^2/2}\int_{-\infty}^\infty e^{-x^2/2}\,\mathrm{d}x\\ &=\sqrt{2\pi}e^{-\xi^2/2}\tag{1}\\ \int_{-\infty}^\infty (-ix)^ke^{-x^2/2}e^{-i\xi x}\,\mathrm{d}x &=\left(\frac{\mathrm{d}}{\mathrm{d}\xi}\right)^k\sqrt{2\pi}e^{-\xi^2/2}\tag{2} \end{align} $$ By taking real and imaginary parts, for $k=1$, $(2)$ gives $$ \int_{-\infty}^\infty x\,e^{-x^2/2}\sin(\xi x)\,\mathrm{d}x =\sqrt{2\pi}\,\xi\,e^{-\xi^2/2}\tag{3} $$ and $$ \int_{-\infty}^\infty x\,e^{-x^2/2}\cos(\xi x)\,\mathrm{d}x\tag{4} =0 $$ For larger $k$, $(2)$ says that the Fourier Transforms of $x^ke^{-x^2/2}$ are polynomials in $\xi$, with integer coefficients, times $\sqrt{2\pi}e^{-x^2/2}$.

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robjohn: Why is it true that $\int_{-\infty}^\infty\exp(-(x+i\xi)^2/2)dx=\int_{-\infty}^\infty\exp(-x^2/2)dx$‌​? Note that $i\xi$ isn't real. – Christmas Bunny Nov 12 '13 at 23:04
@YifengXu: Contour integration says that. There are no poles of $e^{-x^2/2}$ anywhere an integral over a closed path is $0$. The difference of the paths is a very long rectangle whose ends have length $|\xi|$ in a region where $e^{-x^2/2}$ vanishes. Thus, the difference of the integrals is $0$. – robjohn Nov 13 '13 at 0:04
Are you saying that $\int_{0}^{\xi}\exp(-(a+ti)^2/2)idt=\int_{0}^{\xi}\exp(-(-a+ti)^2/2)idt$ as $a\to\infty$? – Christmas Bunny Nov 13 '13 at 2:09
@YifengXu: they both tend to $0$ as $a\to\infty$. – robjohn Nov 13 '13 at 4:13
I see. But how do you formally show this? Thank you! – Christmas Bunny Nov 13 '13 at 7:35

First, let us notice that, due of the the oddity of the sine function, we have $(-x)\sin(-x)=(-x)\cdot$ $(-\sin x)=x\sin x\iff I=2\int_0^\infty xe^{-\frac{x^2}2}\sin kx\,dx$. Secondly, since $\sin kx=\Im(e^{ikx})$, we have $I=2\cdot\Im\left[\int_0^\infty xe^{-\left(\frac{x^2}2-ikx\right)}dx\right]$, where $\Im(z)=\Im(a+bi)=b$. Now, let us pay a closer look at the exponent : $\frac{x^2}2-ikx=\frac12(x^2-2ikx)=\frac12\left[(x-ik)^2+k^2\right]=\frac{t^2+k^2}2$, where $t=x-ik$, and $dt$ $=dx$. Then:


$$=2\,\Im\left[\int_0^\infty e^{-u}du+ike^{-\frac{k^2}2}\int_{0-ik}^{\infty-ik}e^{-\frac{t^2}2}dt\right]=2\,\Im\left[1+ike^{-\frac{k^2}2}\sqrt\frac\pi2\left(1+\text{Erf}\left(\tfrac{ki}{\sqrt2}\right)\right)\right]=$$

$=ke^{-\frac{k^2}2}\sqrt{2\pi}$ , since the error function of purely imaginary argument is purely imaginary as well, meaning that i times itself possesses no imaginary part, and hence does not ultimately influence the final result.

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Lucian: After the substitution $t=x-ik$, why are the upper and lower limits of the integral still $\infty$ and $0$, respectively? $t$ is complex in this substitution. – Christmas Bunny Nov 12 '13 at 23:01
Thanks for the tip ! :-) I've fixed it now. – Lucian Nov 13 '13 at 7:37
French impair = English odd. "The oddity of the sine function..." – GEdgar Nov 13 '13 at 13:36
Thanks for the tip, I didn't know that it was used for functions as well. – Lucian Nov 13 '13 at 13:38

For the second integral, complete the square and use cauchy's theorem. For the first integral, notice that $\int _{-\infty} ^{\infty} xe^{-\frac{x^{2}}{2}}Sin(x \xi)=-\partial _{\xi}\int _{-\infty} ^{\infty} e^{-\frac{x^{2}}{2}}Cos(x \xi)=-\partial _{\xi}\int _{-\infty} ^{\infty} e^{-\frac{x^{2}}{2}-i\xi x}dx$

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