Dirac delta function integral over given limits I know that $\int$$λδ(x-a)=1.λ$ with the limits $$-inf$$ to $$+inf$$ but what if the limits are $$a-ε$$ to $$a+ε$$??
I am guessing it's the same as the former limits but still not sure
Can someone add a homework tag?
 A: The Heaviside function $H(x)$ is defined as
$$H(x)=
  \begin{cases}
                                   0 & \text{for $x < 0$} \\
                                   1 & \text{for $x$ > 0} 
  \end{cases}$$
The derivative of the Heaviside function is zero for $x \neq 0$. At $x = 0$ the derivative is undefined. We can represent the derivative of the Heaviside function by the Dirac delta function, $\delta(x)$. The delta function is zero for $x \neq 0$ and infinite at the point $x = 0$. Since the derivative of $H(x)$ is undefined, $\delta(x)$ is not a function in the conventional sense of the word. Heuristically, the delta function is defined by the properties
$$\delta(x)=
  \begin{cases}
                                   0 & \text{for $x \neq 0$} \\
                                   \infty& \text{for $x = 0$} 
  \end{cases}$$
and
$$\int_{-\infty}^{\infty}\delta(x)dx =1$$
since $\delta(x)$ represents the derivative of $H(x)$. Physically, this function looks like a bump at the origin where the height of the bump becomes arbitrarily large.

In terms of your homework problem, you have that
$$\int_{-\infty}^{\infty}\delta(x-a)dx =1$$
so the above bump is shifted from $x=0$ to $x=a$ (I'm not sure why you include $\lambda$). From the picture above, it is clear that for some small $\epsilon > 0$
$$\int_{-\epsilon}^{\epsilon}\delta(x)dx =1$$
So, you should calculate
$$\int_{a-\epsilon}^{a+\epsilon}\delta(x-a)dx$$
and reach the same conclusion by drawing a similar picture. This approach is entirely heuristic and can be made rigorous through the use of measure theory and the theory of distributions.
