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How would I integrate the following?

$\int_a^\infty (w-a)dF(w)$

for any fixed $a$, where $F(0)=0$ and $F(w)$ is strictly increasing and converges to $1$ as $w\to \infty$ .

I've started by using the formula I've seen for integration by parts for Riemann-Stieltjes, modifying it slightly to be defined for unbounded intervals, so I've written,

$\int_a^\infty (w-a)dF(w)$=$\lim_{w \to \infty} (w-a) F(w) - (a-a)F(a) -\int_a^\infty F(w)dw$

$\int_a^\infty (w-a)dF(w)$=$\lim_{w \to \infty} (w-a) F(w) -\int_a^\infty F(w)dw$

Clearly, the limit of the first object on the right goes to $\infty$, so where have I gone wrong?

Thanks.

Tom-Tom
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Riemann-Stieltjes
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1 Answers1

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Your mistake is that you applied integration by parts to an integral with $\infty$ in the upper limit. But this is an improper integral defined as

$$\int_a^\infty (w-a)dF(w)=\lim_{L\rightarrow\infty}\int_a^L (w-a)dF(w)$$

Applying integration by parts to that truncated integral gives

$$\int_a^L (w-a)dF(w)=(L-a)F(L)-\int_a^L F(w)dw$$

Now as you observed correctly, $(L-a)F(L)\rightarrow\infty$ as $L\rightarrow\infty$. But that does not mean the whole expression goes to $\infty$ (the integral being subtracted from it might still save the day!).

J.R.
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