Derivation of the Heaviside Step Function to Dirac Delta Function Relation with Distribution Definition

Here we derive the definition of the distributional derivative and integral $⟨T,\phi ⟩$. Dirac Delta and Heaviside Step Function are distributions and general calculus breaks down due to their conditional range and discontinuities. We can however multiply discontinuous distribution functions with totally fine continuous ones to leverage properties such as integration by parts. That is what we’ll do.

Distributional Derivative Definition

We choose $\phi (x)$ to be a smooth, continuous, compactly supported test function. The distributional angle brackets are defined as

$$\int \_{-\infty }^{\infty }{f}^{\prime }(x)\phi (x)dx=⟨f^{\prime },\phi ⟩$$

Now we find the identity by using integration by parts

$$\int \_{-\infty }^{\infty }{f}^{\prime }(x)\phi (x)dx=-\int \_{-\infty }^{\infty }f(x){\phi }^{\prime }(x)dx+\underset{0}{\underbrace{\text{[}f(x)\phi (x){]}_{-\infty }^{\infty }}}$$

Notice, that the last term is zero because per compact support definition $\phi (x)=0$ for every $|x|\ge R$. Since $R$ must be finite, $|\pm \infty |\ge R$ infinity is always larger and therefore $\phi (\pm \infty )=0$. The term vanishes. We remain with

$$\int \_{-\infty }^{\infty }{f}^{\prime }(x)\phi (x)dx=-\int \_{-\infty }^{\infty }f(x){\phi }^{\prime }(x)dx$$

Which is expressed as the general governing definition

$$⟨f^{\prime },\phi ⟩=⟨f,{\phi }^{\prime }⟩$$

Derivation of the Relationship between Heaviside Step Function and Dirac Delta

$$H^{\prime }(x)=\delta (x)$$

We again choose a compactly supported test function $\phi$ and mix it with $H^{\prime }$ into an integrand.

$$⟨H^{\prime },\phi ⟩=\int \_{-\infty }^{\infty }{H}^{\prime }(x)\phi (x)dx$$

We then use $⟨f^{\prime },\phi ⟩=⟨f,{\phi }^{\prime }⟩$, the identity from before

$$⟨H^{\prime },\phi ⟩=⟨H,{\phi }^{\prime }⟩=\int \_{-\infty }^{\infty }H(x){\phi }^{\prime }(x)dx$$

For $x\le 0$ we notice that $H(x)=0$. The portion of the integral from $-\infty$ to 0 is all 0! Also, for all $x>0$ the step function is exactly $H(x)=1$. Let’s rewrite the summation then!

$$\int \_0^{\infty }{\phi }^{\prime }(x)dx$$

Integrate

$$\int \_0^{\infty }{\phi }^{\prime }(x)dx=-\text{[}\phi (\infty )-\phi (0)]=\phi (0)$$

Notice again that due to compact support our $\phi (\infty )=0$. Okay, recall now the few properties of the Dirac Delta function. We will use the identity

$$\int \_{-\infty }^{\infty }f(x)\delta (x)=f(0)$$

such that our

$$\phi (0)=\int \_{-\infty }^{\infty }\phi (x)\delta (x)=⟨\phi ,\delta ⟩$$

Let’s compare to the Heaviside step function brackets from before and notice that $H^{\prime }=\delta$!

$$⟨H^{\prime },\phi ⟩=⟨\delta ,\phi ⟩$$

Therefore

$$\boxed{H^{\prime }=\delta }$$