Delta Function Integration: Justified or Fudging?

[BOAS]

Hello,

I feel like I am fudging these integrals a bit and would like some concrete guidance about what's going on.

1. Homework Statement
Evaluate $I = \int_{-1}^{1} dx \delta'(x)e^3{x}$

Homework Equations

The Attempt at a Solution

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I use integration by parts as follows,
$u = e^{3x}$
$du = 3e^{3x} dx$

$dv = \delta'(x)$
$v = \delta(x)$

$I = uv|^{1}_{-1} - \int^{1}_{-1} v du$

Evaluating $uv$ gives zero because the delta function is zero at both limits.

$I = - \int^{1}_{-1} dx \delta(x) 3e^{3x}$

Clearly, the only non zero value of this function is at $x = 0$, making me think the answer to my problem is $I = -3$, but this is the step that I think I'm 'fudging'. I feel like I'm disregarding the integration sign...

Is this justified?

Thanks for any guidance you can give.

 

[blue_leaf77]

That's justified because $\delta(x)$ is centered in $x=0$, which is contained by the integration limits.

 

[PeroK]

That looks correct. It sometimes helps to think of a finite spiked function instead of delta. Let's say it's 0 outside $[-a, a]$ for some small a, and has a value of $1/2a$ in that interval. The integral is then approx: $2a \frac{1}{2a} e^0 = e^0$

 

[BOAS]

That looks correct. It sometimes helps to think of a finite spiked function instead of delta. Let's say it's 0 outside $[-a, a]$ for some small a, and has a value of $1/2a$ in that interval. The integral is then approx: $2a \frac{1}{2a} e^0 = e^0$

I don't quite follow.

Where do you evaluate the limits?

I see that the spiked function is $\frac{1}{2a}$ inside this interval, and since (I assume) it's centered on $0$, I see where $e^0$ comes from, but not the $2a$.

 

[PeroK]

I don't quite follow.

Where do you evaluate the limits?

I see that the spiked function is $\frac{1}{2a}$ inside this interval, and since (I assume) it's centered on $0$, I see where $e^0$ comes from, but not the $2a$.

2a is the width of the interval. An integral is the area under a curve!

 

[BOAS]

2a is the width of the interval. An integral is the area under a curve!

Ah!

I'm with you now. So an integral of a function multiplied by the dirac delta function is like evaluating the function itself at the point where the delta function is non-zero?

 

[PeroK]

Ah!

I'm with you now. So an integral of a function multiplied by the dirac delta function is like evaluating the function itself at the point where the delta function is non-zero?

That's exactly what it is. You can think of the delta function as the limit of a sequence of rectangular functions with area 1. That gives some justification for things.

 

[BOAS]

That's exactly what it is. You can think of the delta function as the limit of a sequence of rectangular functions with area 1. That gives some justification for things.

Cool - Thanks for your help!