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I working through the question in the link and i am stuck on something:

How to solve integration with Dirac Delta function?

How does one solve the equation: $\int^{-5/6}_{11/6}(-\frac{1}{2}-3u)\delta (u)du$

My solution:

$\int^{-5/6}_{11/6}(-\frac{1}{2}-3u)\delta (u)du =$

$\int^{-5/6}_{11/6}(-\frac{1}{2})\delta (u)du-\int^{-5/6}_{11/6}(-3u)\delta (u)du=$

$-\frac{1}{2}\int^{-5/6}_{11/6}\delta (u)du-3\int^{-5/6}_{11/6}u\delta (u)du=$

$-1/2-3\int^{-5/6}_{11/6}u\delta (u)du $

I dont't know how to solve this equation: $-3\int^{-5/6}_{11/6}u\delta (u)du$

Edit:

Would anyone mind explaining why: $\int^{-5/6}_{11/6}u\delta (u)du=0$. Or in wider sence why $\int^{\infty}_{-\infty}x\delta (x)dx=0$

PhysicsQuestion
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2 Answers2

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The delta function has the property that $$\int_{-\infty}^{\infty}f(u)\delta(u-a)du=f(a)$$ or $$\int_{a-\epsilon}^{a+\epsilon}f(u)\delta(u-a)du=f(a)$$

for $\epsilon>0$ and $\delta(u-a)=0$ for $u\neq a$. In your case $f(u)=-\frac{1}{2}-3u$ and $a=0$ (which lies in the interval $[\frac{-5}{6},\frac{11}{5}]$), thus we have $$\int_{\frac{11}{6}}^{-\frac{5}{6}}f(u)\delta(u)du=-\int_{-\frac{5}{6}}^{\frac{11}{5}}f(u)\delta(u)du$$ $$=-f(0)=-(-\frac{1}{2})=\frac{1}{2}.$$

Alessio K
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By definition (of the Dirac delta function), $$\delta(u) = 0 \:\: \text{for}\:\: u\ne 0\implies -\int_{11/6}^{5/6}\left(3u+\frac{1}{2}\right)\delta(u) \:du=0$$ since the range of integration does not contain $0$.

EDIT $1$
I just took a look at the link in the post and realized that there could be a bunch of typos in the question. Henceforth, I shall assume that this is what the OP actually wants:

$$\int_{-5/6}^{11/6}\left(\frac{1}{2}-3u\right)\delta(u)\:du$$ We have that $$\boxed{\:\int_c^d f(t)\delta(t-a)\:dt=f(a),\:\:\:\:\:\text{ if $a\in(c,d)$}\:\:}\tag{1}$$ Observe that $0$ is contained in the range of integration. Therefore, $$\int_{-5/6}^{11/6}\left(\frac{1}{2}-3u\right)\delta(u)\:du=\frac{1}{2}-3(0)=\frac{1}{2}$$

EDIT $2$
Observe that $$\int_{11/6}^{-5/6}u\delta(u)\:du=\int_{-5/6}^{11/6}u\delta(u)\:du $$ After an application of $(1)$, we arrive at the desired answer.

Alan
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