Questions concerning an improper Riemann integrable funciton

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

• Maybe integration by parts. – xbh Oct 3 '18 at 10:54
• Also there seems a typo: $|-t\mathrm e^{-t^2} \sin(xt)|\leqslant t\mathrm e^{\color{red}{-t^2}}$. However this is not correct either. It should be $|t|\mathrm e^{-t^2}$ instead. – xbh Oct 3 '18 at 10:59

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 $$-\infty$$).

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,

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

$$f(x)=2\int_0^\infty\cos(xt)e^{-t^2}dt=2\sum_{n=0}^\infty\frac{(-1)^n x^{2n}}{(2n)!}\int_0^\infty t^{2n}e^{-t^2}dt$$ 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: $$\Gamma(n)=(n-1)!=\int_0^\infty x^{n-1}e^{-x}dx$$ $$\Gamma(n+\frac{1}{2})=\frac{\Gamma(2n)\sqrt{\pi}}{2^{2n-1}\Gamma(n)}$$