I have to evaluate a few integrals with the dirac delta in them. So far, I got all of them (hopefully right), but I am stuck at the last one, mixing the two methods of the (kx) and (x-a)... help?

 

here it is:

 

[math]

\int_{-\infty}^{\infty} \ln x \delta (1-2x), dx

[/math]

(couldn't find dirac-delta sign in LaTeX, settling for a simple delta)

 

Anyhoo, I figured that the trick is to mentally switch this 'around' so it resembles the basic formula with a kx instead of x:

 

[math]

\int_{-\infty}^{\infty} \ln x \delta (-2x+1), dx

[/math]

So, if I only had -2, I know that the expression of the Dirac Delta equals

 

[math]

\delta (-2x)=\frac{1}{|-2|} \delta (x)

[/math]

 

But I don't have it in that format, I have an "extra" +1 there, and I'm not sure I can remove the -2 if I have that (and can't find anything about it)..

 

Anyone? Help?

 

Thanks!

 

~moo

Edited September 10, 200817 yr by mooeypoo

The general expression for delta functions is:

 

[math]\delta[g(x)] = \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}[/math]

 

where the [math]x_i[/math] s are the roots of g(x).

 

I think you can take it from here.

I hate sums.

 

Wait, I think I can be a bit more accurate.. oh, yes: I hate sums.

 

meh. Now that I got that off my chest, I must be a real nag and ask how do I transfer a sum of i with no end (infinity?) to an integral between infinity and infinity.

 

Do excuse me, these things are awefully new.. and I hate sums. And groups. Ew.

 

~moo

Thinking out loud (it's been a while since I've done this)

 

Can you do a substitution for 1-2x?

 

u=1-2x (so x = (1-u)/2)

du = -2 dx

 

That moves the ugliness out of the delta, and the integral limits are flipped, so I think that adds another - sign to keep them the same.

Thinking out loud (it's been a while since I've done this)

 

Can you do a substitution for 1-2x?

 

u=1-2x (so x = (1-u)/2)

du = -2 dx

 

That moves the ugliness out of the delta, and the integral limits are flipped, so I think that adds another - sign to keep them the same.

 

isn't there some trick with integrating delta functions?

 

It's been a while since I did them too!

 

*goes to read text books*

 

-------------------

 

I agree with swansont's method.

isn't there some trick with integrating delta functions?

 

The "trick" is that you basically get to ignore the integral (pragmatically speaking)

 

[math]

\int_{-\infty}^{\infty} f(x) \delta (x-a) dx = f(a)

[/math]

The "trick" is that you basically get to ignore the integral (pragmatically speaking)

 

[math]

\int_{-\infty}^{\infty} f(x) \delta (x-a) dx = f(a)

[/math]

 

Yes but what happens when you have (kx-a) ?

 

My professor actually solved that in class, but it's not next to me. I'll post the solution tomorrow, it's answering - apparently - another form of this equation.

mooey, the sum (which you hate) is just so that you include all the roots of g(x)

 

So that if g(x) has 5 roots, you turn the [math]\delta(g(x))[/math] in to [math]\frac{\delta(x-x_1)}{c_1} + \frac{\delta(x-x_2)}{c_2} + \frac{\delta(x-x_3)}{c_3} + \frac{\delta(x-x_4)}{c_4} + \frac{\delta(x-x_5)}{c_5}[/math]

 

If g(x) had 5 roots, the sum would be over i = 1 to 5.

 

It was kind of sloppy notation, but I'd suggest you get used to it. It is fairly common to leave the limits off of sums when (in theory) the limits should be obvious. I should have been more explicit.

 

In this case g(x) = 1-2x

 

The root is x=-1/2

 

The derivative is -2

 

So, you can fill in [math]\delta(g(x)) = \delta(1-2x) = \frac{\delta(x+\frac{1}{2})}{-2}[/math]

 

Coincidently, this is the same thing that should happen when you do the change of variables, too. But, change of variables get trickier when g(x) gets trickier. The sum presented above is for any function g(x), polynomial, trigonometric, exponential, etc.

hmmmmm... interesting. Okay, I need to practice that to get used to it.

 

Thanks

The "trick" is that you basically get to ignore the integral (pragmatically speaking)

 

[math]

\int_{-\infty}^{\infty} f(x) \delta (x-a) dx = f(a)

[/math]

 

Aye, I realised that after returning to my text books

Yes but what happens when you have (kx-a) ?

 

My professor actually solved that in class, but it's not next to me. I'll post the solution tomorrow, it's answering - apparently - another form of this equation.

 

Like I suggested — try substituting so that it looks like a form you can solve.

 

u = kx-a

du = kdx