Past exam of the short options course of the University of Oxford / 2023 / 5 (Integral transforms)

The distributional derivative is defined by $⟨F^{\prime },\varphi ⟩=-⟨F,{\varphi }^{\prime }⟩$ for every test function. For the Heaviside function, compact support givesThus ${\mathcal{H}}^{\prime }=\delta$. Splitting the integral for $|x|$ at zero and using integration by parts gives its first derivative as the sign function. The jump of that function at zero is two, soas distributions. The value assigned to the sign function at the single point zero does not affect its associated distribution.

For real $p>0$, substitute $t=px$ in the Laplace transform:The Gamma function integral is finite because $a>0$. For complex $p$, absolute convergence holds exactly when $\mathrm{Re}p>0$: the modulus of the integrand is $x^{a-1}{e}^{-(\mathrm{Re}p)x}$. On this half-plane the same answer holds with the branch $p^{-a}=\exp (-a\mathrm{Log}p)$ whose logarithm is real on the positive axis, by holomorphic continuation of the two sides.

If “exists” permits conditionally convergent ordinary improper integrals, there are additional boundary points when $0<a<1$: for $p=i\omega \ne 0$, the Dirichlet test gives convergence, with value $\Gamma (a)|\omega {|}^{-a}{e}^{-i\mathrm{sgn}(\omega )\pi a/2}$. For $a\ge 1$ these boundary integrals do not converge in that sense. At $p=0$, and for $\mathrm{Re}p<0$, the integral diverges. Thus $\mathrm{Re}p>0$ is the region of absolute convergence of a Laplace transform, with the stated conditional boundary qualification.

The integral $I$ is the convolution of the positive-half-line kernels $x^{a-1}$ and $x^{b-1}$. The convolution theorem for Laplace transforms therefore gives, for $p>0$,On the other hand, the substitution $t=xu$ gives $I(x)=x^{a+b-1}B(a,b)$, where $B(a,b)={\int }_0^1{u}^{a-1}(1-u{)}^{b-1}du$ is the beta function. Applying part (i) again yields $\mathcal{L}I(p)=B(a,b)\Gamma (a+b)p^{-(a+b)}$. Comparing the two expressions provesAll integrands used in the real-$p$ convolution argument are nonnegative, so Tonelli's theorem justifies the interchange; the endpoint singularities are integrable because $a,b>0$.

Use the Fourier transform convention $\hat{f}(s)={\int }_{\mathbb{R}}{e}^{-isx}f(x)dx$. First observe that $f(x)=(2\pi {)}^{-1/2}{e}^{-x^2/2}$ satisfies the differential equation, decay, and normalization, using the Gaussian integral. The uniqueness allowed in the question identifies it with $f$. In particular it is a Schwartz function, so the derivative and multiplication rules below are justified.

Taking the Fourier transform of the differential equation givesThe Riemann-Lebesgue lemma gives decay of $\hat{f}$. Every decaying solution $h$ of this equation is proportional to $e^{-s^2/2}$: setting $w=h^{\prime }+sh$ gives $w^{\prime }=sw$, and integration yieldsThe term with $C\ne 0$ grows like $C{e}^{s^2/2}/(2s)$ as $s\to +\infty$, so decay forces $C=0$. Thus $\hat{f}=\lambda f$ for a scalar $\lambda$. This establishes proportionality before using Fourier inversion to calculate the scalar.

At zero, the Fourier transform is $\hat{f}(0)=\int f=1$, so $\lambda f(0)=1$. In particular $\lambda \ne 0$ and $f(0)=1/\lambda$. The Fourier inversion theorem at zero also givesEquating the two values yields $\boxed{{\lambda }^2=2\pi }$. Since the explicit normalized Gaussian function in part (i) has $f(0)>0$, here $\lambda =\sqrt{2\pi }$.

In the space of tempered distributions, this Fourier transform convention gives $\hat{1}=2\pi \delta$ and $\hat{\delta }=1$. ThereforeThe constant distribution and the Dirac delta distribution are linearly independent: testing away from zero first determines the constant coefficient, and then testing at zero determines the delta coefficient. Hence $c=\lambda$ and $\lambda c=2\pi$, giving