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I have been trying to evaluate this integral:$$\int_0^\infty xe^{-x^2}\sin(\xi x)\ dx$$ But I seem to be a little bit stuck on how to do this. I have tried partial integration by taking derivatives of $x\sin(\xi x)$ or $xe^{-x^2}$, but both times I arrived at an integral that seemed even harder to solve. An idea that kind of worked was to use the series representation of $\sin(\xi x)$ and then interchanging the summation with the integral. In that case, I would not know if the interchange is viable though ( I could not really find an integrable upper bound to use dominated convergence).

Any hints on how the interchange with summation would work or any other approach would be greatly appreciated!

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  • $\begingroup$ hint: your integral is the derivative of the real part of the fourier transform of a gaussian with respect to frequency $\endgroup$ – tired Nov 25 '17 at 10:56
  • $\begingroup$ @tired Actually, thats how I arrived here since I wanted to evaluate the fourier transform. Was that a step backwards? $\endgroup$ – Jack4t3 Nov 25 '17 at 10:57
  • $\begingroup$ the searched resuld should be this here $$\frac{1}{4} \sqrt{\pi } e^{-\frac{\xi ^2}{4}} \xi$$ $\endgroup$ – Dr. Sonnhard Graubner Nov 25 '17 at 10:57
  • $\begingroup$ @Jack4t3 the FT of a Gaussian is usually shown using the completition of the square in the exponent. and yes i think your approach might go a little bit in a not so useful direction ;) $\endgroup$ – tired Nov 25 '17 at 11:01
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With integrating by parts \begin{align} \int_0^\infty xe^{-x^2}\sin(\xi x)\ dx &= \dfrac{\xi}{2}\int_0^\infty e^{-x^2}\cos(\xi x)\ dx \\ &= \dfrac{\xi}{2}{\bf Re}\int_0^\infty e^{-x^2}e^{i\xi x}\ dx \\ &= \dfrac{\xi}{2}{\bf Re}\int_0^\infty e^{-(x-\frac12\xi i)^2-\frac14\xi^2}\ dx \\ &= \dfrac{\xi}{2}e^{-\frac14\xi^2}{\bf Re}\int_0^\infty e^{-(x-\frac12\xi i)^2}\ dx \\ &= \dfrac{\xi}{2}e^{-\frac14\xi^2}\dfrac{\sqrt{\pi}}{2} \end{align}

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Here is a real way of arguing, that is a bit longer, but still. If you define $$ f(\xi)=-\int_0^{+\infty}e^{-x^2}\cos(\xi x)\,dx $$ then your integral is $f'(\xi)$.

But using parity arguments, we find that $$ \begin{align} f'(\xi)&=\int_0^{+\infty} xe^{-x^2}\sin(\xi x)\,d x\\ &=\frac{1}{2}\int_{-\infty}^{+\infty} xe^{-x^2}\bigl(\sin(\xi x)+\cos(\xi x)\bigr)\,dx\\ &=\frac{1}{2}\int_{-\infty}^{+\infty}\biggl[\frac{d}{dx}\Bigl(-\frac{1}{2}e^{-x^2}\bigl(\sin(\xi x)+\cos(\xi x)\bigr)\Bigr)+\frac{\xi}{2}e^{-x^2}\bigl(\cos(\xi x)-\sin(\xi x)\bigr)\biggr]\,dx\\ &=\frac{\xi}{4}\int_{-\infty}^{+\infty}e^{-x^2}\bigl(\cos(\xi x)-\sin(\xi x)\bigr)\,dx\\ &=\frac{\xi}{4}\int_{-\infty}^{+\infty}e^{-x^2}\cos(\xi x)\,dx\\ &=-\frac{\xi}{2}f(\xi). \end{align} $$ Hence $$ f(\xi)=Ce^{-\xi^2/4} $$ for some constant $C$. But $$ C=f(0)=-\int_0^{+\infty}e^{-x^2}\,dx=-\frac{\sqrt{\pi}}{2}, $$ so $$ f(\xi)=-\frac{\sqrt{\pi}}{2}e^{-\xi^2/4}. $$ Differentiating, we find that $$ f'(\xi)=\frac{\sqrt{\pi}}{4}\xi e^{-\xi^2/4}. $$

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