Formally derive $\displaystyle\int_{x=-\infty}^{x=\infty} f(x) \delta(x) \, \mathrm{d}x = f(0)$

I have been searching for a derivation of the defining property for the Dirac-delta function: $\displaystyle\int_{x=-\infty}^{x=\infty} f(x) \delta(x) \, \mathrm{d}x = f(0)$ and found this derivation on the first 2 pages:

Defining the Heaviside (step) function $H(x)$ as

$$H(x) = \begin{cases} 0 & \text{for } x \lt 0 \\1& \text{for } x \gt 0 \end{cases}$$ The derivative of the Heaviside function is zero for $x \ne 0$ and undefined for $x=0$ so the $\delta$ function can represent the derivative of the Heaviside function

$$\delta(x) = \begin{cases} 0 & \text{for } x \ne 0 \\\infty& \text{for } x = 0 \end{cases}$$ and $$\int_{x=-\infty}^{x=\infty} \delta(x) \, \mathrm{d}x=1$$

Let $f(x)$ be any continuous function that vanishes at $x=\pm\infty$ and integrating by parts

\begin{align} & \int_{x=-\infty}^{x=\infty} f(x)\delta(x) \, \mathrm{d}x = \color{green}{\left.\vphantom{\frac 1 1} f(x)H(x) \right|_{x=-\infty}^{x=\infty}} - \int_{\color{red}{x=-\infty}}^{x=\infty} f^\prime(x)H(x) \, \mathrm{d}x \\[10pt] = {} &0-\int_{\color{red}{x=0}}^{x=\infty} f^\prime(x)H(x) \, \mathrm{d}x= \left.\vphantom{\frac 1 1}-f(x) \right|_{x=0}^{x=\infty}=f(0) \end{align}

Could someone please explain how the limits marked $\color{red}{\mathrm{red}}$ were changed?

Thank you.

(and yes I know from last time, it's an abuse of notation placing the Dirac measure/distribution inside an integral)

• Heaviside – Jyrki Lahtonen Oct 9 '15 at 17:39
• @JyrkiLahtonen That was helpful, thank you – BLAZE Oct 9 '15 at 17:49

The way the limits in red were changed is simply that the piecewise definition of $H$, stated earlier in the question, says that $H(x)=0$ when $x<0$. Thus $$\int_{-\infty}^0 (\text{anything}\times H(x))\, dx = 0.$$
The idea that $\delta(0)=\infty$ should not be taken too literally. Notice that $$\int_{-\infty}^\infty 3.4\delta(x) f(x)\,dx = 3.4f(0),$$ so one would then say that this "infinity" is $3.4$ times as big as the earlier "infinity". But no attempt is made to give such concepts any precise definition. If one wants to show that $H'(x)=\delta(x)$, one can observe that $H$ has a vertical jump at $0$ and therefore say there is an infinite slope, but again no attempt is made to make that precise. Rather, one precisely defines $H'$ by integrating by parts, as you did.
• +1 For the comments - that is not really a "proof", rather a motivation. Besides the property $\int_{x=-\infty}^{x=\infty} f(x) \delta(x) \, \mathrm{d}x = f(0)$ does not require that the function vanishes at $\infty$ – leonbloy Oct 4 '15 at 2:44
• The proofs were first given by the French mathematician Laurent Schartz in the 1940s. There are a number of books on the subject, including Theory of Distributions by Richards and Youn. ${}\qquad{}$ – Michael Hardy Oct 4 '15 at 2:50
• @BLAZE : I wouldn't use the word "nonsense" but I would say the definition of the $\delta$ function stated on the first page should not be taken literally. Perhaps one should say that to take it literally would be nonsense, but if you know it's not meant literally, then it might not be nonsense. ${}\qquad{}$ – Michael Hardy Oct 4 '15 at 3:15
Probably not as much rigorous: $$\int_{-\infty}^{+\infty}f(x)\delta(x)\,dx=\int_{-\infty}^{+\infty}(f(x)-f(0))\delta(x)\,dx+\int_{-\infty}^{+\infty}f(0)\delta(x)\,dx\\=\int_{-\infty}^{+\infty}(f(x)-f(0))\delta(x)\,dx+f(0)\int_{-\infty}^{+\infty}\delta(x)\,dx\\=\int_{-\infty}^{+\infty}(f(x)-f(0))\delta(x)\,dx+f(0)$$ by definition of $\delta(x)$ it follows that $(f(x)-f(0))\delta(x)=0$ for all $x\in(-\infty,+\infty)$. So the last integral vanishes and we are left with $$\int_{-\infty}^{+\infty}f(x)\delta(x)\,dx=f(0)$$