# Rigorous proof of $\lim\limits_{n \to \infty } \int_0^n f(n,t) dt = \int_0^\infty \lim\limits_{n \to \infty } f(n,t) dt$

I faced with this equality $$\lim _{n \to \infty } \int_0^n \left( {1 - \frac{m}{n}}\right) ^n \log(m)dm= \int_0^\infty {{e^{ - m}}} \log mdm.$$ I know a rigorous proof for $\displaystyle \lim_{n\to\infty} \left(1+\frac {x}{n}\right)^n=\exp x$ but the problem is a general effect of limit on both integrand and upper bound and my question is proving the equality in the title and not just the example above; that is I can't rigorously prove why $$\lim _{n \to \infty } \int_0^n f(n,t) dt = \int_0^\infty \lim_{n \to \infty } f(n,t) dt.$$

Unfortunately, I don't know more than undergraduate real analysis. A simpler and clear proof would be much appreciated.

• Something like this is probably easiest to handle using Lebesgue theory, e.g. dominated convergence theorem. Something simple like uniform convergence is going to struggle with the singularity, I think. – Ian Aug 30 '18 at 12:59
• I thought that the Dominated convergence theorem only applied to integration over fixed sets? – MSobak Aug 30 '18 at 13:03
• @Sobi $\int_A f(x) dx = \int_D f(x) \chi_A(x) dx$ if $A \subset D$. Here, $D=[0,\infty),A=[0,n]$. – Ian Aug 30 '18 at 13:05
• @Ian Nice! Never thought of that! – MSobak Aug 30 '18 at 13:07
• @Ian unfortunately I don't know more than undergraduate real analysis which doesn't cover Lebesgue theory.. – user231343 Aug 30 '18 at 13:10

We can fashion a dominated convergence theorem of the kind that the old masters, pre measure-theory, would have known. I'll present an argument for your particular case, but it will be clear that there is a general theorem here.

Sketch of main ideas: First make all the domains of integration $(0,\infty).$ Just set

$$f_n(x) = (1-x/n)^n\log x \,\chi_{(0,n)}(x),\,\,f(x)=e^{-x}\log x.$$

Your problem is then to show $\int_0^\infty f_n \to \int_0^\infty f.$

Note that $|f_n(x)| \le |f(x)|$ for all $x\in (0,\infty).$ Furthermore, $\int_0^\infty |f| <\infty.$ And very importantly, $f_n \to f$ uniformly on any $(a,b)$ with $0<a<b<\infty.$ (If you haven't seem the last result don't despair; it's provable with the tools of undergraduate real analysis, give it a try.)

We then proceed:

$$|\int_0^\infty f-\int_0^\infty f_n| =|\int_0^\infty (f-f_n) |\le \int_0^\infty|f-f_n|$$ $$= \int_0^a |f-f_n| + \int_a^b |f-f_n| + \int_b^\infty |f-f_n|$$ $$\le \int_0^a 2|f| + \int_a^b |f-f_n| + \int_b^\infty 2|f|.$$

We can choose $a,b$ so that the first and third integrals are as small as we like. Uniform convergence shows the second integral $\to 0.$ We're in a good spot now. It will lead to the result you're after.

• From what I understood, am I right to say that the necessary and sufficient condition for interchange of lim and integral (for fixed upper and lower bound of integration) is that $f_n$ converges uniformly to $f$ for all $x \in (a,b)$? – user231343 Aug 30 '18 at 16:06
• It's a sufficient condition; it's not necessary. – zhw. Aug 30 '18 at 16:13

Assuming everything has sufficient decay (i.e. the integrals converges), your question clearly depends on the growth of the area $$\int_{0}^{n}(f(n,t)-\lim_{m\to\infty}f(m,t))dt.$$ Your equality will hold if an only if the above goes to $0$ as $n\to \infty$. For instance, this fails for $f(n,t)=e^t+\frac{1}{n}$.

To prove your particular example (with the domain $[1,\infty)$), just divide the range of the above integration as $[1,n^{1/4}]$ and $[n^{1/4},n]$ and show that each goes to zero. (In the first range the difference $(f(n,t)-\lim_{m\to\infty}f(m,t))$ decays rapidly and in the second range $f(n,t)$ and $\lim_{m\to\infty}f(m,t)$ decay individually.)

• Why does $(1-n^{-3/4})^n$ decay faster than $1/n$? – Ian Aug 30 '18 at 15:27
• In fact $m(n)=n^{1/4}$ seems rather arbitrary...all you really need is $(1-m(n)/n)^n=o(1/n)$ and $m(n)^2/n=o(1)$, so it is enough to have $m=o(n^{1/2})$ and $m(n)=\Omega(1)$. – Ian Aug 30 '18 at 15:40

Apply Monotone Convergence Theorem https://en.wikipedia.org/wiki/Monotone_convergence_theorem: Since the successive terms $(1-\frac{m}{n})^{n}$ increases as proved in the following: (Using $\log(x) \leq x-1$) $$f(x) = (1-\frac{m}{x})^x \\ \log(f(x)) = x \log(1-\frac{m}{x})\\ f'(x) = f(x) (\log(1-\frac{m}{x}) + \frac{x}{1-\frac{m}{x}} \frac{m}{x^2}) \\ f'(x) = f(x) (\log(1-\frac{m}{x}) + \frac{m}{x-m}) \\ f'(x) = f(x) (-\log(\frac{x}{x-m}) + \frac{m}{x-m}) \\ f'(x) \geq f(x) (-(\frac{x}{x-m}-1) + \frac{m}{x-m}) \\ \geq 0 \\$$