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I need some help in understanding the integration performed in below equation. My question is how step 2 is obtained from the first step (i.e., how integration of exponential and dirac delta functions is performed). Thanks in advance.

$$\begin{align*} \mathsf{P}(0<Y\leq 7) &=\int_{0^+}^7\left[\frac{1}{4}e^{-|y|}+\frac{1}{3}\delta(y)+\frac{1}{6}\delta(y-7)\right]dy\\\\\\ &=\frac{1}{4}\int_0^7e^{-y}\,dy+\frac{1}{6}\\\\\\ &=\frac{1-e^{-7}}{4}+\frac{1}{6}=\frac{5}{12}-\frac{e^{-7}}{4} \end{align*}$$

(original image)

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I understood the exponential integration... now i just want to know how we get 1/6 please. –  Osman Khalid Feb 10 '13 at 18:53
By definition $\int_{-\infty}^{\infty} \delta (y) dy = 1$. the integral borders are 0+ and 7 therefore the integration do not contain value of the function at 0 and contain value at 7. Therefore the peak of the integrand $\frac{1}{3} \delta(y)$ is not in the interval of integration –  OukiDouki Feb 10 '13 at 18:58
@OukiDouki: Then what about the fact that the result of an integral should not depend on altering the interval of integration for a zero measure set of points? I mean, I could well remove the single point $y=7$ and use $y=7^-$ as upper limit. But then - following your argument - also that integral would be zero. –  Andrea Orta Feb 10 '13 at 19:05
@Andrea Orta: Good question. I didn't mastered the measure theory and Lebesque integrals so well to answer it. I followed the simple logic that when is integration interval $(o,7>$ and a Dirac function is non-zero only at point $0$ then it simply you cannot integrate it. Somebody else should anwer this question –  OukiDouki Feb 10 '13 at 19:22
@OukiDouki: Yes, I was writing something along the lines of your reasoning, but then I stopped, thinking about that problem. Osman: Wait a moment, better answers will appear! –  Andrea Orta Feb 10 '13 at 19:28

2 Answers 2

up vote 1 down vote accepted

A delta function satisfies $\int_{-\infty}^\infty \delta(x) dx=1$ but also $\int_{-\epsilon}^\epsilon \delta(x) dx=1$ for every $\epsilon>0$. Intuitively, it can be thought of as a function that is zero everywhere, but jumps quickly to infinity at zero. This is the physical/engineering interpretation.

Another way to think about a delta function, is as the derivative of a step function, usually denoted as $u(x)$ satisfies $u(x)=1$ for $x \ge 0$ and $u(x)=0$ for $x<0$

In your problem, the integral starts from $0^+$, so the integral over the first delta is $0$ (it's like integrating a zero). The second delta is a shifted delta to $x=7$, which its integral is $u(x-7)$. Having said that, we get: $\int_{0^+}^7 \frac{1}{6}\delta(x-7)dx=\frac{1}{6}[u(7-7)-u(0^+-7)]=\frac{1}{6}[u(0)-u(-7)]=\frac{1}{6}[1-0]=\frac{1}{6}$

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The defining properties of $\delta(y)$ are $$\delta(0)\to\infty$$ $$\delta(y)=0\quad y\ne0$$ $$\int_{-\infty}^{\infty}\delta(y)dy=1$$ The second one means that $\delta(y)$ vanishes in the neighbourhood of $0$ however small. It also follows that $$\int_{-\infty}^{\infty}\delta(y)dy=\int_{-\epsilon}^{\epsilon}\delta(y)dy=1\tag{1}$$ for arbitrarily small $\epsilon>0$ The integral could also be written as $$I=\int_{\epsilon}^{7}\left[\frac{1}{4}e^{-|y|}+\frac{1}{3}\delta(y)+\frac{1}{6}\delta(y-7)\right]dy$$ From which it is clear that the $\delta(y)$ term vanishes as the interval of integration does not include the singularity. Now add a small $\epsilon_1$ neighbourhood of 7 to the interval of integration and consider $$I_1=\int_{7-\epsilon_1}^{7+\epsilon_1}\left[\frac{1}{4}e^{-|y|}+\frac{1}{6}\delta(y-7)\right]dy$$ By the man value theorem we can write: $$I_1=\frac{\epsilon_1}{2}e^{-c}+\frac{1}{6}\int_{7-\epsilon_1}^{7+\epsilon_1}\delta(y-7)dy$$ where $c\in [7-\epsilon_1,7+\epsilon_1]$. Now using $(1)$ we can write $$I_1=\frac{1}{6}\int_{-\infty}^{\infty}\delta(y-7)dy+O\left(\epsilon_1\right)=\frac{1}{6}+O\left(\epsilon_1\right)$$ Since $\epsilon_1$ is arbitrary we obtain the final result.

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Could you please explain why the comment I made to the question (about removing the single point $y=7$ from the interval of integration) doesn't raise a problem with your proof? Thanks! –  Andrea Orta Feb 10 '13 at 20:53
In this case we are dealing with a singular function and the singularity is placed exactly at the end of the interval, therefore we cannot simply throw it away as it makes finite difference to the value of the integral. Therefore, I took a complementary approach showing that adding a small neighbourhood does not alter the value much, because the contribution from the regular part of the integrand is negligible –  Valentin Feb 10 '13 at 21:18
All right, thank you. +1 –  Andrea Orta Feb 10 '13 at 22:54

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