# Understanding an integration

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*}

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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
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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.
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