Let $f(x)=\delta(x) $ if $ x\in (-\infty,0] $ and $f(x)=0$ if $x\in (0,\infty)$. Then $I=\int_{-\infty}^{\infty}f(x)dx=?$

Question

Let $\delta$ be the Dirac delta function.

Then, let $f(x)=\delta(x) $ if $ x\in (-\infty,0] $ and $f(x)=0$ if $x\in (0,\infty)$.Then $I=\int_{-\infty}^{\infty}f(x)dx=?$

Is $I=1/2$? We know $\int_{-\infty}^{\infty}\delta(x)dx=1$. But we can not say that $\delta$ is symmetric about $y-axis$. So we can not conclude that $I=1/2$. Then what will be $I$?

Answer 1

Since Delta is defined with $\int_{-\infty}^{\infty}$ you cannot change those limits. What you can do is construct test function as $\phi(x)=1$ if $x\in(-\infty,0]$ and $\phi(x)=0$ if $x\in(0,\infty)$ and ask what is $\int_{-\infty}^{\infty}\phi(x)\delta(x)dx$. The answer is obviously $\phi(0)=1$.

Answer 2

1. OP is essentially describing an ill-defined product of two distributions$^1$ $$ u~=~\theta \delta, $$ where $\theta$ denotes the Heaviside step function and $\delta$ is the Dirac delta distribution. Products of distributions is a well-known delicate topic, see e.g. this mathoverflow post and Colombeau algebra.

   To see what slippery slope lies ahead, consider the following ill-defined chains of formal manipulations: $$u~=~\theta \delta~=~\theta \theta^{\prime}~=~\frac{1}{2}(\theta^2)^{\prime} ~=~\frac{1}{2}\theta^{\prime}~=~\color{red}{\frac{1}{2}} \delta $$ versus $$u~=~\theta \delta~=~\theta^2 \delta~=~\theta^2 \theta^{\prime}~=~\frac{1}{3}(\theta^3)^{\prime} ~=~\frac{1}{3}\theta^{\prime}~=~\color{red}{\frac{1}{3}} \delta~! $$

2. Later OP asks to evaluate the distribution $u[\varphi]$ for the constant function $\varphi=1$. Technically speaking, a constant function $\varphi=1$ is not a valid test function.

--

$^1$ We consider in this answer the "mirror" distribution $\theta(x) \delta(x)$ rather than OP's distribution $\theta(-x) \delta(x)$ to simplify notation. The conceptional challenges will be the same.