Dirac delta function integral

Question

How should i integrate this?

$$\int_{0}^{t} \int_{0}^{t} \delta(x-y)dxdy$$ where $\delta$ represents Dirac delta function

My try: $\int_{0}^{t} \int_{0}^{t} \delta(x-y)dxdy = t$ is it right?

Answer 1

As you mention in a comment, $$ \int_\mathbb{R} f(t) \delta(t-a) dt = f(a).$$ So here, $$ \int_0^t 1 \cdot \delta(x-y) dx = 1$$ as long as the point when $x = y$ occurs within the domain of integration [which it does]. Your integral then becomes $$ \int_0^t 1 \;dy = t.$$

Answer 2

1. OP's integral is: $$ I~:=~\int_0^t \! dy ~\int_0^t \! dx ~ \delta(x-y). $$

2. Hint: Divide into cases where $t$ is positive and negative. Let $1_{[a,b]}(x)$ denote the indicator/characteristic function for the interval [a,b].

3. Case $t\geq 0$: $$ I~=~\int_{\mathbb{R}} \! dy~ 1_{[0,|t|]}(y) ~\int_{\mathbb{R}} \! dx ~1_{[0,|t|]}(x)~\delta(x-y)$$ $$~=~\int_{\mathbb{R}} \! dy ~ 1_{[0,|t|]}(y)~1_{[0,|t|]}(y) ~=~\int_{\mathbb{R}} \! dy ~ 1_{[0,|t|]}(y)~=~|t|.$$

4. Case $t\leq 0$: $$ I~=~\int_{\mathbb{R}} \! dy~ 1_{[-|t|,0]}(y) ~\int_{\mathbb{R}} \! dx ~1_{[-|t|,0]}(x)~\delta(x-y)$$ $$~=~\int_{\mathbb{R}} \! dy ~ 1_{[-|t|,0]}(y)~1_{[-|t|,0]}(y) ~=~\int_{\mathbb{R}} \! dy ~ 1_{[-|t|,0]}(y)~=~|t|.$$

Answer 3

The posts above were almost correct, but the answers fell short when the delta function was defined incorrectly.

Definition of delta function

$$f(y) = \int_{-\infty }^{\infty } f(x)\delta (x-y)\; $$

Which can be simplified as

$$1 = \int_{-\infty }^{\infty }\delta (x-y)\; dx$$

Because the delta function is even $\delta(x) = \delta(-x)$, the definition can be extended to

$$1 = 2\int_{0}^{\infty }\delta (x-y)\; dx\; \; \; \; \Rightarrow \; \; \; \; \frac{1}{2} = \int_{0}^{\infty }\delta (x-y)\; dx$$

Assuming that $0<t<\infty $

$$\frac{1}{2} = \int_{0}^{t}\delta (x-y)\; dx + \int_{t}^{\infty }\delta (x-y)\; dx\; \; \; \; \Rightarrow \; \; \; \; \frac{1}{2} = \int_{0}^{t}\delta (x-y)\; dx $$

$$\because \int_{t}^{\infty }\delta (x-y)\; dx = 0$$

Therefore the integral in question simplifies to

$$\int_0^t \int_0^t \delta (x-y) \; dxdy = \int_0^t \frac{1}{2} \; dy = \frac{1}{2}y \; |_0^t = \frac{1}{2}t$$