# Evaluating the integral $\int_0^\infty xe^{-x^2}\sin(\xi x)\ dx$

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!

• hint: your integral is the derivative of the real part of the fourier transform of a gaussian with respect to frequency – tired Nov 25 '17 at 10:56
• @tired Actually, thats how I arrived here since I wanted to evaluate the fourier transform. Was that a step backwards? – Jack4t3 Nov 25 '17 at 10:57
• the searched resuld should be this here $$\frac{1}{4} \sqrt{\pi } e^{-\frac{\xi ^2}{4}} \xi$$ – Dr. Sonnhard Graubner Nov 25 '17 at 10:57
• @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 ;) – tired Nov 25 '17 at 11:01

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 $$f'(\xi)=\frac{\sqrt{\pi}}{4}\xi e^{-\xi^2/4}.$$