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What is the problem with this reasoning?

$$ \intop_0^K f(x)dx = \intop_0^{\infty}I(x\leq K)f(x)dx.$$

Using integration by parts with $u=I(x\leq K)$ and $\frac{dv}{dx}=f(x)$, we have:

$$\intop_0^K f(x)dx = [uv]_0^{\infty} - \intop_0^{\infty}v\frac{du}{dx} dx.$$

Since $u$ is the indicator function, its derivative is the negative dirac delta function $-\delta(x-K)$, and we have.

$$\intop_0^K f(x)dx = [uv]_0^{\infty} + \intop_0^{\infty}v\delta(x-K)dx = [uv]_0^{\infty} + v(K),$$

where $v(K)$ is the function $v$ evaluated at $K$.

David Simmons
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  • Have you tried it for some easy $f$ to see whether it works? – Gerry Myerson Sep 05 '13 at 12:49
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    If you're going to differentiate the indicator function, at the very least you should say that its derivative is a Dirac delta (concentrated at the appropriate point)! – Zhen Lin Sep 05 '13 at 12:53

1 Answers1

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The derivative of $I(x \le K)$ is in fact $-\delta(x-K)$, because it jumps downwards. Thus using your $u(x)=I(x\le K)$ and $dv=f$ we have $$v(x) = \int_0^x f \ \ \textrm{ and }\ \ (uv)_0^\infty = I(\infty < K)v(\infty)-v(0)=0-0=0,$$ and integration by parts gives $$\int_0^K f = \int_0^\infty u dv = (uv)_0^\infty - \int_0^\infty v du = 0 - \int_0^\infty -v\delta(x-K)dx=v(K),$$ which is exactly what we started with.

Anthony Carapetis
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