# Why does $\int_0^T ne^{n(x-T)}f(x)$ converge to $f(T)$?

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.

• 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. – Kaynex Feb 14 '17 at 2:53
• Can you show us your work in using integration by parts, because that is exactly what I would do. – Doug M Feb 14 '17 at 3:03
• 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. – Helmut Feb 14 '17 at 3:12
• What's a nice function.? – Nosrati Feb 14 '17 at 3:35
• @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. – RRL Feb 14 '17 at 5:51

## 2 Answers

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.

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.

• Nice answer. Much more indirect than I thought of. – Helmut Feb 14 '17 at 4:19