Computation of $\int_0^\infty xe^{-\frac{x^2}{2}}\sin(x \xi) \, dx$

Question

How to compute $$\int_0^\infty xe^{-\frac{x^2}{2}}\sin(x \xi) \, dx$$

I tried to integrate by part but it was bad idea.

Answer 1

Fourier transform, contour shifting, differential equation and other nice techniques work nicely here. Since they are already explained by other users, let me show a brutal-force computation.

Using the substitution $u = x^2/2$,

\begin{align*} \int_{0}^{\infty} x \sin(\xi x) e^{-x^2/2} \, dx &= \int_{0}^{\infty} \sin(\xi\sqrt{2u}) e^{-u} \, du \\ &= \int_{0}^{\infty} \left( \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \xi^{2n+1} (2u)^{n+\frac{1}{2}} \right) e^{-u} \, du \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\xi^{2n+1} 2^{n+\frac{1}{2}} \Gamma\left(n + \frac{3}{2}\right) \\ &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\xi^{2n+1} 2^{n+\frac{1}{2}} \cdot \frac{\sqrt{\pi}}{2}\prod_{k=1}^{n} \left(k + \frac{1}{2}\right) \\ &= \sqrt{\frac{\pi}{2}} \sum_{n=0}^{\infty} \frac{(-1)^n}{n! 2^n}\xi^{2n+1} \\ &= \sqrt{\frac{\pi}{2}} \xi e^{-\xi^2/2}. \end{align*}

Here, we utilized the following identities

$$ \Gamma(s) = \int_{0}^{\infty} u^{s-1}e^{-u} \, du, \qquad \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}. $$

Also, interchanging the summation and integral is justified by Fubini's theorem together with the estimation

$$\int_{0}^{\infty} \left( \sum_{n=0}^{\infty} \frac{|\xi|^{2n+1}}{(2n+1)!} (2u)^{n+\frac{1}{2}} \right) e^{-u} \, du = \int_{0}^{\infty} \sinh(|\xi|\sqrt{2u}) e^{-u} \, du < \infty. $$

Answer 2

HINT:

Let $F(\xi)$ be given by

$$F(\xi)=\int_0^\infty xe^{-x^2/2}\sin(x\xi)\,dx \tag1$$

Integrating by parts the right-hand side of $(1)$ with $u=\sin(x\xi)$ and $v=-e^{-x^2/2}$ reveals

$$\begin{align} F(\xi)&=\xi \int_0^\infty e^{-x^2/2}\cos(x\xi)\,dx\\\\ &=\frac\xi2 \int_{-\infty}^\infty e^{-x^2/2}\cos(x\xi)\,dx\\\\ &=\frac\xi2 \text{Re}\left(\int_{-\infty}^\infty e^{-x^2/2}e^{ix\xi}\,dx\right) \end{align}$$

Now, complete the square, deform the contour back to the real line using Cauchy's Integral Theorem, and evaluate the resulting Gaussian integral.

Answer 3

Method 1

Consider the integral \begin{align} \int_{0}^{\infty} e^{-x^2/2} \, x^{2n} \, dx &= \frac{2^{n} \, \Gamma\left(n + \frac{1}{2}\right)}{\sqrt{2}} \end{align} then $$\int_{0}^{\infty} e^{-x^2/2} \, \cos(a x) \, dx = \sqrt{\frac{\pi}{2}} \, e^{-a^{2}/2}.$$ Differentiation with respect to $a$ leads to the desired result, namely, $$\int_{0}^{\infty} x \, e^{-x^2/2} \, \sin(a x) \, dx = \sqrt{\frac{\pi}{2}} \, a \, e^{-a^{2}/2}$$

Method 2

Integration by parts: $dv = x \, e^{-x^2/2}$, $u = sin(ax)$ leads to \begin{align} \int_{0}^{\infty} x \, e^{-x^2/2} \, \sin(a x) \, dx &= \left[ - \sin(ax) \, e^{-x^2/2} \right]_{0}^{\infty} + a \, \int_{0}^{\infty} e^{-x^2/2} \, \cos(ax) \, dx \\ &= \sqrt{\frac{\pi}{2}} \, a \, e^{-a^{2}/2}. \end{align}