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Let $f(x)$ be a nice function. Consider the integral $$\int_0^T ne^{n(x-T)}f(x)\ dx.$$

Apparently, this converges to $f(T)$ as $n\to\infty$. I can't see why. I've tried doing integration by parts, but it doesn't lead anywhere. I've put this in Wolfram Alpha for various choices of $f(x)$ and it seems to be true.

I can see that the factor $ne^{n(x-T)}$ seems to concentrate its mass around $x=T$ as $n\to\infty$, but then this would seem to suggest that the integral is almost $n\int_0^T \chi([2-\epsilon,2]) f(x)$. Not really sure what's going on!

Why does this integral converge to $f(T)$?

EDIT: I tried integration by parts as follows. Let $v=f(x)$ and $du=ne^{-n(x-T)}$. Then $$\int_0^T ne^{n(x-T)}f(x)\ dx = e^{n(x-T)}f(x)\Bigg|_0^T - \int_0^T e^{n(x-T)}f'(x)$$

$$=f(T) - e^{-nT}f(0)- \int_0^T e^{n(x-T)}f'(x),$$ which doesn't seem to lead anywhere.

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    Have you tried integration by parts? Let $f(x)$ be the derivative, everything else the integral. I can't try it out myself right now, but I can see a lot of reasons it should work. – 2017-02-14
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    Can you show us your work in using integration by parts, because that is exactly what I would do. – 2017-02-14
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    I showed my work. Am I doing something wrong? I also don't know what you mean by "Let $f(x)$ be the derivative". It's confusing because in Integration by Parts a term is both a derivative and an integral in different parts of the formula. – 2017-02-14
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    What's a nice function.? – 2017-02-14
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    If you were able to give an answer based on Laplace's method I would be interest. It's not obvious to me how to apply it. Regarding those who told me that it could be done with IbP, is this possible or are the indirect answers below necessary? – 2017-02-14
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    @Helmut: This is ultimately a consequence of dominated convergence. I purposely avoided the Lebesgue approach to add substance to your correct observation that mass concentrates around $x = T$. All IBP does is shift consideration to the limit of $\int_0^Te^{n(x-T)} f'(x) \, dx$. Under suitable assumptions on $f'$ if it exists and is integrable you would have show that integral converges to $0$. This would require similar consideration of interchanging a limit and an integral as discussed below and brings up the unnecessary details about the derivative. – 2017-02-14

2 Answers 2

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Note that

$$\int_0^T ne^{n(x-T)}f(x)\ dx = \int_0^T ne^{-nx}f(T-x)\ dx. $$

Assuming only that $f$ is continuous, we can show this converges to $f(T)$ using calculus. No assumption about differentiability is required.

We have

$$I_n = \int_0^T ne^{-nx}f(T- x)\, dx= \int_0^T ne^{-nx}[f(T- x) - f(T)] \, dx + f(T)\int_0^T ne^{-nx} \, dx \\ = \int_0^c ne^{-nx}[f(T-x) - f(T)] \, dx + \int_c^T ne^{-nx}[f(T-x) - f(T)] \, dx + f(T)(1- e^{-nT}),$$

and

$$\left| I_n - f(T)\right| \leqslant \int_0^c ne^{-nx}|f(T-x) - f(T)| \, dx + \int_c^T ne^{-nx}|f(T-x) - f(T)| \, dx + |f(T)|e^{-nT}$$

For any $\epsilon > 0,$ choose $c$ sufficiently small such that $|f(T-x) - f(T)| < \epsilon$ for $ 0 < x < c$.

Then

$$|I_n - f(0)| \leqslant \epsilon(1 - e^{-nc}) + (2\sup_{x \in [0,T]} |f(x)|)(e^{-nc}- e^{-nT}) + |f(T)|e^{-nT} .$$

Taking the limit as $n \to \infty$ we get

$$\lim_{n \to \infty}|I_n - f(0)| \leqslant \epsilon.$$

Since $\epsilon$ can be arbitrarily small, it follows that $\lim I_n = f(T).$

Another approach would be to use the dominated convergence theorem.

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this is an approach rather than an answer:

First do the case when $f$ is constant. This should be easy.

Now subtract $f(T)$ to reduce to the case where $f(T)=0.$

Pick any $y < T,$ show that the integral up to $y$ goes to zero. This should follow from the dominated convergence theorem since the integrands are uniformly bounded on such an interval and go to zero point wise.

At $T$ you can write $f(x) = (x-T) g,$ with $g$ bounded by Taylor's theorem. Use this to bound the integrands near 1 and again get it to go zero.

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    Nice answer. Much more indirect than I thought of. – 2017-02-14