Does the delta function integral still hold true for non-continuous functions?

[nicksauce]

Evaluate:

\int^{\infty}_{-\infty} f(x)\delta(x-x_0)dx

Where

f(x)=ln(x+3), x_0=-2

Ordinarily, you would just evaluate f(x_0), so it would be 0, but in this case, since f(x) is -\infty at x=-3, does that make a difference?

 

[d_leet]

But you have x₀=-2 and f(x) is certainly not negative infinity there. What is f(-2)?

 

[nicksauce]

Right, f(-2) = ln(1) = 0

I'm just wondering if for the formula:

\int_{-\infty}^{\infty}f(x)\delta (x-x_0) = f(x_0)

there need to be certain conditions on f(x), that would make a logarithmic function not suitable? My textbook was very vague in this area.

 

[Hurkyl]

You're missing the + and the - on your limits...


There are ways to define integration of generalized functions so that this integral makes sense for any (partial) function f that is well-behaved near x₀. It might even be possible to make this integral well-defined when f is a generalized function that is well-behaved near x₀!

 

[HallsofIvy]

Right, f(-2) = ln(1) = 0

I'm just wondering if for the formula:

\int_{-\infty}^{\infty}f(x)\delta (x-x_0) = f(x_0)

there need to be certain conditions on f(x), that would make a logarithmic function not suitable? My textbook was very vague in this area.

No. For any function defined at x₀,
\int_{-\infty}^\infty f(x)\delta (x)dx= f(x_0)

 

[Hurkyl]

No. For any function defined at x₀,
\int_{-\infty}^\infty f(x)\delta (x)dx= f(x_0)

I think I won't feel comfortable unless you insist on continuity in a neighborhood of x₀.