Questions concerning an improper Riemann integrable funciton

Question

I have to show that \begin{equation}f(x)=\int_{-\infty}^{\infty}e^{-t^{2}}\cos(xt) \ dt\end{equation} is a continuous differentiable function and that $xf(x)=-2f'(x)$ for all $x\in\mathbb{R}$.

We see that $e^{-t^2}\cos(xt)$ is improper Riemann integrable for all $x$ and $e^{-t^2}\cos(xt)$ is partially differentiable to the first variable. We note that $\frac{\partial}{\partial x}e^{-t^2}\cos(xt)=-t e^{-t^2}\sin(xt)$. Because $|-t e^{-t^2}\sin(xt)|\leqslant t e^{t^2}$ for all $(x,t)$, with $t e^{t^2}$ a improper Riemann integrable function, we have that $f(x)$ is continuous and differentiable.

I know that \begin{equation}-2f'(x)=-2\int_{-\infty}^{\infty}\frac{\partial}{\partial x}e^{-t^{2}}\cos(xt)\ dt=\int_{-\infty}^{\infty}2t e^{-t^2}\sin(xt)\ dt.\end{equation} But \begin{equation}xf(x)=x\int_{-\infty}^{\infty}e^{-t^{2}}\cos(xt)\ dt.\end{equation} I don't know why this equality holds. Also, I don't know if what I've done to show that $f$ is continuous and differentiable is correct.

Answer 1

Consider $$ \int 2t e^{-t^2}\sin(xt)\,dt=-2e^{-t^2}\sin(xt)+x\int e^{-t^2}\cos(xt)\,dt $$ and the fact that $$ \lim_{t\to\infty}e^{-t^2}\sin(xt)=0 $$ (the same at $-\infty$).

Answer 2

$$f(x)=2\int_0^\infty\cos(xt)e^{-t^2}dt=2\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{(2n)!}\int_0^\infty t^{2n}e^{-t^2}dt$$ Let $u=t^2$, $t=\sqrt{u}$, $dt=\frac{du}{2\sqrt{u}}$

$$f(x)=\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{(2n)!}\int_0^\infty u^{n-\frac{1}{2}}e^{-u}du=\sum_{n=0}^\infty\frac{(-1)^n x^{2n}\Gamma(n+\frac{1}{2})}{(2n)!}=\sqrt{\pi}\sum_{n=0}^\infty\frac{(-1)^n x^{2n}(2n-1)!}{2^{2n-1}\Gamma(n)(2n)!}=\sqrt{\pi}\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{2^{2n}n\Gamma(n)}=\sqrt{\pi}\sum_{n=0}^\infty\frac{(-1)^n (\frac{x}{2})^{2n}}{n!}=\sqrt{\pi}e^{-\frac{x^2}{4}}$$ The Guassian Function is continuous and differentiable, thus $f(x)$ is as well.

Note: $$\Gamma(n)=(n-1)!=\int_0^\infty x^{n-1}e^{-x}dx$$ $$\Gamma(n+\frac{1}{2})=\frac{\Gamma(2n)\sqrt{\pi}}{2^{2n-1}\Gamma(n)}$$

Answer 3

As you showed, the Weierstrass test implies uniform convergence, for all $x$, of the improper integral

$$\int_{-\infty}^{\infty} \frac{\partial}{\partial x} e^{-t^2} \cos(xt) \, dt$$

In turn this implies that $f$ is differentiable and the interchange of derivative and integral is permissible -- by a widely known theorem (see, for example, The Elements of Real Analysis by Bartle).

Regarding the identity $xf(x) = -2f'(x)$, note that by taking the real part of a complex exponential and completing a square we have

$$f(x) = \int_{-\infty}^{\infty}e^{-t^2} \cos(xt) \, dt = \Re\int_{-\infty}^{\infty}e^{-t^2} e^{ixt}\, dt = e^{-x^2/4} \,\Re \int_{-\infty}^{\infty}e^{-(t-ix/2)^2}\, dt \\ = e^{-x^2/4} \,\Re \int_{-\infty-ix/2}^{\infty+ix/2}e^{-z^2}\, dz \\ = e^{-x^2/4} \int_{-\infty}^{\infty}e^{-t^2}\, dt $$

The reduction of the complex contour integral to the real integral $\int_{-\infty}^{\infty} e^{-t^2} \,dt$ follows because $z \to e^{-z^2}$ is analytic. By integrating around a rectangular contour, where one side is the interval $[-R,R]$ on the real axis, we get, with $z = t + iu$,

$$0 = \int_{-R}^Re^{-t^2} \, dt - \int_{-R-ix/2}^{R - ix/2}e^{-z^2} \, dz + i\int_{0}^{ -x/2}e^{-(R+iu)^2} \, du + i\int_{-x/2}^{0}e^{-(R+iu)^2} \, du,$$

and the contributions from the third and fourth integrals vanish as $R \to \infty$.

Thus,

$$-2f'(x) = -2\frac{d}{dx}e^{-x^2/4} \int_{-\infty}^{\infty}e^{-t^2}\, dt = xe^{-x^2/4} \int_{-\infty}^{\infty}e^{-t^2}\, dt = xf(x) $$